PROCESS CALCULATIONS
PROCESS
CALCULATIONS
SECOND EDITION
V. VENKATARAMANI
Formerly Professor
Department of Chemical Engineering
National Institute of Technology
Tiruchirappalli
N. ANANTHARAMAN
Professor
Department of Chemical Engineering
National Institute of Technology
Tiruchirappalli
K.M. MEERA SHERIFFA BEGUM
Associate Professor
Department of Chemical Engineering
National Institute of Technology
Tiruchirappalli
New Delhi-110001
2011
` 195.00
PROCESS CALCULATIONS, Second Edition
V. Venkataramani, N. Anantharaman and K.M. Meera Sheriffa Begum
© 2011 by PHI Learning Private Limited, New Delhi. All rights reserved. No part of this
book may be reproduced in any form, by mimeograph or any other means, without
permission in writing from the publisher.
ISBN-978-81-203-4199-9
The export rights of this book are vested solely with the publisher.
Second Printing (Second Edition)
…
…
February, 2011
Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus,
New Delhi-110001 and Printed by Meenakshi Art Printers, Delhi-110006.
To My Parents
— V. Venkataramani
— K.M. Meera Sheriffa Begum
To My Mother
— N. Anantharaman
Contents
Preface .............................................................................................................. xi
Preface to the First Edition ............................................................................ xiii
Acknowledgements ........................................................................................... xv
1
UNITS AND DIMENSIONS
1–6
1.1 Introduction ..................................................................................... 1
1.2 Basic Units and Notations ............................................................... 1
1.3 Derived Units ................................................................................... 2
1.4 Definitions ....................................................................................... 3
Worked Examples ..................................................................................... 3
Exercises ................................................................................................... 6
2
MASS RELATIONS
7–34
2.1 Mass Relations in Chemical Reaction ............................................. 7
2.2 Conservation of Mass ...................................................................... 8
2.3 Avogadro’s Hypothesis .................................................................... 9
2.4 Limiting Reactant and Excess Reactant .......................................... 9
2.5 Conversion and Yield ....................................................................... 9
2.6 Composition of Mixtures and Solutions ........................................ 10
2.6.1 Weight Percent ................................................................... 10
2.6.2 Volume Percent .................................................................. 10
2.6.3 Mole Fraction and Mole Percent ....................................... 11
2.6.4 Atomic Fraction and Atomic Percent ................................ 11
2.6.5 Composition of Liquid Systems ........................................ 11
2.7 Density and Specific Gravity ......................................................... 12
2.7.1 Baume’ (°Be’) Gravity Scale ............................................. 12
2.7.2 API Scale (American Petroleum Institute) ........................ 12
2.7.3 Twaddell Scale ................................................................... 13
2.7.4 Brix Scale .......................................................................... 13
Worked Examples ................................................................................... 13
Exercises ................................................................................................. 32
vii
viii
3
CONTENTS
IDEAL GASES
35–73
3.1
Relation between Mass and Volume for Gaseous Substances ....... 35
3.1.1 Standard Conditions .......................................................... 35
3.1.2 Ideal Gas Law .................................................................... 35
3.2 Gaseous Mixture ............................................................................ 36
3.2.1 Partial Pressure (PP) .......................................................... 36
3.2.2 Pure Component Volume (PCV) ....................................... 36
3.2.3 Dalton’s Law ..................................................................... 37
3.2.4 Amagat’s Law (or) Leduc’s Law ....................................... 37
3.3 Average Molecular weight ............................................................. 38
3.4 Density of Mixture ......................................................................... 38
Worked Examples ................................................................................... 38
Exercises ................................................................................................. 70
4
VAPOUR PRESSURE
74–86
4.1 Effect of Temperature on Vapour Pressure .................................... 74
4.2 Hausbrand Chart ............................................................................ 75
Worked Examples ................................................................................... 75
Exercises ................................................................................................. 85
5
PSYCHROMETRY
87–110
5.1 Humidity ........................................................................................ 87
5.2 Definitions ..................................................................................... 87
Worked Examples ................................................................................... 90
Exercises ............................................................................................... 106
6
CRYSTALLIZATION
111–121
Worked Examples ................................................................................. 112
Exercises ............................................................................................... 120
7
MASS BALANCE
122–179
Worked Examples ................................................................................. 122
Exercises ............................................................................................... 173
8
RECYCLE AND BYPASS
180–195
8.1 Recycle ........................................................................................ 180
8.2 Bypass .......................................................................................... 180
8.3 Purge ............................................................................................ 180
Worked Examples ................................................................................. 181
Exercises ............................................................................................... 195
CONTENTS
9
ENERGY BALANCE
ix
196–220
9.1
Definitions ................................................................................... 196
9.1.1 Standard State .................................................................. 196
9.1.2 Heat of Formation ............................................................ 196
9.1.3 Heat of Combustion ......................................................... 197
9.1.4 The Heat of Reaction ....................................................... 197
9.1.5 Heat of Mixing ................................................................ 197
9.2 Hess’s Law ................................................................................... 197
9.3 Kopp’s Rule ................................................................................. 198
9.4 Adiabatic Reaction Temperature ................................................. 198
9.5 Theoretical Flame Temperature ................................................... 198
Worked Examples ................................................................................. 198
Exercises ............................................................................................... 217
10
PROBLEMS ON UNSTEADY STATE OPERATIONS
221–228
Worked Examples ................................................................................. 221
Exercises ............................................................................................... 227
Tables .................................................................................................... 229–234
I
Important Conversion Factors ............................................................. 229
II
Atomic Weights and Atomic Numbers of Elements ......................... 231
III(a) Empirical Constants for Molal Heat Capacities of Gases at
Constant Pressure ................................................................................. 234
III(b)Molal Heat Capacities of Hydrocarbon Gases .................................. 234
Answers to Exercises ........................................................................... 235–245
Index ...................................................................................................... 247–248
Preface
The objective of this book is to enrich a budding chemical engineer the
techniques involved in analyzing a process plant by introducing the concepts
on units and conversions, mass and energy balances. This will enable him to
achieve a proper design of process equipment. An attempt has been made to
explain the principles involved through numerical examples. The problems are
not only confined to SI system of units but also worked out in other systems
like FPS, CGS and MKS systems. We feel that our attempt will be more
rewarding if students come across data presented in FPS, CGS and MKS
systems while designing equipment, since different reference books give
standard values and data in various units.
The book covers various interesting topics such as units and dimensions,
mass relations, properties of gases, vapour pressure, psychrometry,
crystallization, mass balance including recycle and bypass, energy balance and
unsteady state operations. The second edition is now enriched with additional
worked examples and exercises to give additional exposure and practice to
students.
The text is designed for a one semester programme as a four credit
course and takes care of the syllabus on ‘Process Calculations’ of most of the
universities in India.
V. Venkataramani
N. Anantharaman
K.M. Meera Sheriffa Begum
xi
Preface to the First Edition
Chemical engineers in process industries generally need to focus on design,
operation, control and management of a process plant. It is, therefore,
absolutely essential for them to be conversant with the mass and energy
conservation techniques at every stage of the process to achieve economy in
the design of process equipment in various units of the plant. This book aims
at imparting knowledge of the basic chemical engineering principles and
techniques used in analyzing a chemical process. By applying the relevant
techniques, a chemical engineer is able to evaluate material and energy
balances in different units and present the information in a proper form so that
the data can be used by the management in taking correct decisions. Keeping
this in mind, an attempt has been made to give a brief theory on the principle
involved and more emphasis on numerical examples.
Since data are generally obtained in different units, the worked examples
are not confined to SI units, but to other systems as well, such as FPS, CGS
and MKS systems of units. The examples incorporated in the text are simple
and concrete to make the book useful for self-instruction.
The text is organized into ten chapters and appends three important
tables. The organization is such that the topics are presented in order of easy
comprehension rather than following a logical sequence, e.g. the chapter on
unsteady state operations has been included as the last chapter so that students
can absorb the problems easily. We strongly feel that once the student
understands the topics presented in this book, he will find other advanced
courses in chemical engineering simple and easy to follow.
The topics covered in this book cater to the syllabi on ‘Process
Calculations’ of most universities offering courses in chemical engineering
and its allied branches at the undergraduate level.
V. Venkataramani
N. Anantharaman
xiii
Acknowledgements
At the outset, we wish to thank the almighty for his blessings.
V. Venkataramani wishes to acknowledge his wife, Prof. (Mrs.) Booma
Venkataramani, sons Mr. V. Ravi Chandar, Mr. V. Hari Sundar and his
daughters-in-law Mrs. Vandana Ravi Chandar and Mrs. Ramya Hari Sundar
and granddaughters Miss Vaishnavi Ravi Chandar and Miss Sadhana Hari
Sundar for their support, cooperation and patience shown during the
preparation of this book.
N. Anantharaman wishes to thank his mother, wife, Dr. Usha
Anantharaman, sons Master A. Srinivas and A. Varun for all their patience,
cooperation and support shown during the preparation of this book. The
encouragement received from his brothers and sisters and their family
members is gratefully acknowledged. He also wishes to place on record the
support received from his brothers-in-law and sisters-in-law and their family
members.
K.M. Meera Sheriffa Begum wishes to acknowledge her mother,
husband Mr. S. Malik Raj and baby M. Rakshana Roshan for all their
encouragement, support and cooperation while preparing this book. She also
wishes to place on record the support received from her parents-in-law. The
support received from her brothers, sisters, in-laws and their families is
gratefully acknowledged.
We also thank Director, NIT, Tiruchirappalli for extending all the
facilities and his words of appreciation.
We wish to acknowledge the support and encouragement received from
the Head of Chemical Engineering Department and all our colleagues during
the course of preparation of this book.
We also wish to place on record the suggestions received from
students, especially those at NIT, Tiruchirappalli, and also the faculty from
other institutions.
xv
xvi
ACKNOWLEDGEMENTS
We gratefully acknowledge all the well-wishers.
Finally, we wish to thank the publishers, PHI Learning, New Delhi for
encouraging us to bring out the second edition of the book.
V. Venkataramani
N. Anantharaman
K.M. Meera Sheriffa Begum
Units and Dimensions
1
1.1 INTRODUCTION
Chemical engineers are concerned with the design and development of
processes which involve changes in the bulk properties of matter. To make a
quantitative estimation of these processes, chemical equations showing the
quantities of reactants and products are used. Though internationally we
follow SI system of units, a chemical engineer is expected to be familiar and
conversant with all the systems so far adopted for measuring and expressing
various quantities. A review of literature and data over the years will be
available in various units. These are used to express properties, process
variables and design parameters in FPS, CGS, MKS and SI systems of
Units. Hence, one has to be conversant with their use and applications. This
chapter deals with the basic notations and conversion of a given quantity
from one system of units to another.
The quantities used in our analysis are classified as fundamental
quantities and derived quantities. The fundamental quantities comprise
length, mass, time and temperature. The quantities such as force, density,
pressure, mass flow rate derived from the fundamental quantities are called
derived quantities. While handling these quantities, we come across different
systems of units as mentioned earlier. Now let us see in detail these systems
of units and their conversion from one unit to another.
1.2 BASIC UNITS AND NOTATIONS
English
Engineering,
FPS
Mass (m)
Length (L)
Time (t)
Temperature (T)
Metric Engineering
lb
ft
s
°F
1
CGS
MKS
g
cm
s
°C
kg
m
s
°C
System
International,
SI
kg
m
s
K
2
PROCESS CALCULATIONS
Mass (m)
1 kg = 2.205 lb
Length (L)
1 ft = 30.48 cm
= 0.3048 m
Time (t)
1 h = 3600 s
Temperature (T )
°C
=
°F 32
(Celsius and Centigrade are same)
1.8
1.3 DERIVED UNITS
Area:
1 ft2
10.76 ft2
= length ´ breadth (L2):
= 0.0929 m2
= 1 m2
Force:
1 dyne
= mass ´ acceleration (mL/t 2):
= 1 g cm/s2
(Force applied on a mass of 1 g, which gives an acceleration of 1 cm/s2)
1 Newton (N) = 1 kg m/s2
= (1000 g) (100 cm)/s2
1 N
= 105 g cm / s2 = 105 dynes
Work/energy:
= 1 kg m/s2
1 erg is the amount of work done on a mass of 1 g when it is
displaced by 1 cm by applying a force of 1 dyne.
1 erg
1 Joule
1 Joule
= [1 dyne] ´ [1 cm]
= [1 g cm/s2] ´ [1 cm]
= 1 g cm2/s2
= (1 N) ´ (1 m)
= 105 g cm/s2 ´ 100 cm
= 107 g cm2/s2
= 107 erg
Heat Unit:
1 Btu
1 cal
1 J/s
= 0.252 kcal = 252 cal
= 4.18 J
=1W
UNITS AND DIMENSIONS
3
1.4 DEFINITIONS
System. This refers to a substance or group of substances under consideration, e.g. storage tank, water in a tank, hydrogen stored in cylinder, etc.
Process. Changes taking place within the system is called process, e.g.
burning of fuel, or reaction between two substances like hydrogen and
oxygen to form water.
Isolated system. Boundaries of the system are limited by a mass of
material, and its energy content is completely detached from all other matter
and energy. In an isolated system, the mass of the system remains constant,
regardless of the changes taking place within the system.
Extensive property. It is a state of system, which depends on the
mass under consideration, e.g. volume.
Intensive property. This state of a system is independent of mass. An
example of this property is temperature.
WORKED EXAMPLES
1.1
The superficial mass velocity is found to be 200 lb/h.ft2. Find its
equivalent in kg/s.m2
G (Mass velocity) = (200) lb/h.ft2
1
1
Ê 1
ˆ
= (200) ¥ Á
kg˜ ¥
¥
m2
Ë 2.205 ¯ (3600 s) (0.0929)
= 0.2712 kg/s.m2
1.2
Convert the heat transfer coefficient of value 100 Btu/h.ft2.°F into
W/m2 °C
È (0.252 kcal) ˘
2
h (Heat transfer coefficient) = (100) Í
˙ (0.0929 m )
Î (3600 s) ˚
(1.8 °C) [1 degree variation in Farh.
scale is equivalent to 1.8 times the
variation in celsius scale]
= 4.186 ¥ 10–2 kcal/s.m2 °C
= 4.186 ¥ 10–2 ¥ 103 ¥ 4.18 W/m2 °C
= 174.98 W/m2 °C
1.3
The rate of heat loss per unit area is given by (0.5) [(DT)1.25/(D)0.25]
Btu/h ft2 for a process, where, DT is in °F and D is in ft. Convert this
relation to estimate the heat flux in terms of kcal/h. m2 using DT in °C
and D in m.
4
PROCESS CALCULATIONS
We know that,
9
C1 + 32 = F1
5
9
C2 + 32 = F2
5
Therefore,
1.8 [DC] = (DF)
q
A
0.5
( 'T )1.25
( D)0.25
( 'T °F)1.25
q
, Btu/h ft2 = 0.5
A
( D ft)0.25
We know that,
1 Btu = 0.252 kcal
1 ft2 = 0.0929 m2
1 ft = 0.3048 m
For DT °F = 1.8 DT °C
Btu/h ft2
Ë [ 'T °F)1.25 Û
= 0.5 Ì
0.25 Ü
ÌÍ (D ft)
ÜÝ
= (0.5)
(1.8 'T °C)1.25
( D m/0.3048)0.25
= (0.7746)
Btu/h ft2
(a)
(b)
( 'T °C)1.25
(D m)0.25
= (0.252 kcal)/h (0.0929 m2)
= 2.713 kcal/h m2
(c)
From (b) and (c) we get the expression for heat flux in units of
kcal/h m2 with temperature difference in Celsius and diameter in metre
as:
Ë (0.7746)( 'T °C)1.25 Û
Heat flux, kcal/h m2 = Ì
Ü ´ 2.713
( D „ m)0.25
ÌÍ
ÜÝ
=
(2.101)( 'T °C)1.25
( D „ m)0.25
Now let us check the conversion with the following data:
D = 0.2 ft, i.e. D¢ = 0.06096 m
DT = 18 °F, i.e. DT = 10 °C
(d)
UNITS AND DIMENSIONS
From Eq. (a), heat flux is = 0.5
(18)1.25
(0.2)
0.25
Also, from Eq. (d), heat flux = 2.101
5
= 27.72 Btu/h ft2 = 75.2 kcal/h m2
(10)1.25
(0.06096)0.25
= 75.2 kcal/h m2
Both the values agree.
1.4
If Cp of SO2 is 10 cal/g mole K, what is the value in FPS units?
The Cp value is the same in all units, i.e. 10 Btu/lb mole °R.
1.5
1.6
Iron metal weighs 500 lb and occupies a volume of 29.25 litres. Find
the density in kg/m3.
Basis: 500 lb of Iron = 500/2.2 = 227.27 kg
29.25 lit = 29.25 ´ 10–3 m3
227.27
Density =
´ 10–3 = 7770 kg/m3
29.25
Etching operation follows the relation d = 16.2 – 16.2e–0.021t, where t is
in s. and d is in microns. Convert this equation to evaluate d in mm
with t in min.
d = 16.2 [1 – e–0.021t]
Let d¢ be in mm and t¢ be in min. (d = d¢ ´ 103 and t = t¢ ´ 60)
Then, d = d¢ × 103 = 16.2 [1 – e–0.021t × 60]
d¢ = 0.0162 [1 – e–1.26t¢]
1.7
The density of fluid is given by r = 70.5 exp (8.27 × 10–7). Convert
this equation to calculate the density in kg/m3 with pressure in N/m2.
1000 kg/m3 = 62.43 lb/ft3
14.7 psi = 1.0133 × 105 N/m2
1 kg/m3 = 62.43 × 10–3 lb/ft3
Let, r¢ be in kg/m3 and p¢ be in N/m2.
Then, (r¢ × 62.43 × 10–3) = 70.5 × exp (8.27 × 10–7 × p¢ × 14.7/1.0133 × 105)
r¢ (kg/m3) = 1.129 × 103 × exp [119.97 × 10–12 × p¢ (N/m2)]
1.8
Vapour pressure of benzene in the range of 7.5 °C to 104 °C is given
by log10 (p) = 6.9057 – 1211/(T + 220.8), where T is in °C and p is in
torr 1 torr = 133.3 N/m2. Convert it to SI units.
Let p¢ be in N/m2and T¢ be in K.
1211
È p Ø
Then, log É
= 6.9057 –
Ê 133.3 ÙÚ
(T „ 273) 220.8
1211
log p¢ – log 133.3 = 6.90305 –
T „ 52.2
log (p¢) = 9.0305 – 1211 .
T „ 52.2
6
PROCESS CALCULATIONS
EXERCISES
1.1
Convert the following quantities:
(a) 42 ft2/h to cm2/s
(b) 25 psig to psia
(c) 100 Btu to hp-h
(d) 30 N/m2 to lbf/ft2
(e) 100 Btu/h ft2 °F to cal/s cm2 °C
(f) 1000 kcal/h m°C to W/m K
1.2
The heat transfer coefficient for a stream to another is given by
h = 16.6 Cp G 0.8/D0.2
where h = Heat transfer coefficient in Btu/(h)(ft)2(°F)
D = Flow diameter, inches
G = Mass velocity, lb/(s)(ft)2
Cp = Specific heat, Btu/(lb) (°F)
Convert this equation to express the heat transfer coefficient in kcal/
(h)(m)2(°C)
With D = Flow diameter in m, G = Mass velocity in kg/(s)(m)2
and Cp = Specific heat, kcal/(kg) (°C)
1.3
Mass flow through a nozzle as a function of gas pressure and
temperature is given by m = 0.0549 p/(T)0.5 where m is in lb/min, p is
in psia and T is in °R, where T(°R) = T °F + 460. Obtain an
expression for the mass flow rate in kg/s with p in atmospheres (atm)
and T in K.
1.4
The flow past a triangular notch weir can be calculated by using the
following empirical formula:
q = [0.31 h2.5/g0.5] tan F
where q = Volumetric flow rate, ft3/s
h = Weir head, ft
g = Local acceleration due to gravity, ft/s2
F = Angle of V-notch with horizontal plane
1.5
In the case of liquids, the local heat transfer coefficient, for long tubes
and using bulk-temperature properties, is expressed by the empirical
equation,
h = 0.023 G0.8 k0.67 Cp0.5/(D0.2 m0.47)
where G = Mass velocity of liquids, lb/ft2.s
k = Thermal conductivity, Btu/ft.h.°F
Cp = Specific heat, Btu/lb °F
D = Diameter of tube, ft
m = Viscosity of liquid, lb/ft.s
Convert the empirical equation to SI units.
Mass Relations
2.1
2
MASS RELATIONS IN CHEMICAL REACTION
In stoichiometric calculations, the mass relations between reactants and
products of a chemical reaction are considered and are based on the atomic
weight of each element involved in the reaction.
For the following reactions the material balance is established as
indicated below:
(i)
CaCO3 ® CaO + CO2
(2.1)
[40 + 12 + 3 ´ 16] ® [40 + 16] + [12 + 32]
100
®
56
+
44
(ii)
3Fe + 4H2O ® Fe3O4 + 4H2
(2.2)
(3 ´ 55.84) + 4(2 ´ 1 + 16) ® (55.84 ´ 3 + 4 ´ 16) + 4(2 ´ 1)
167.52 +
72
®
231.52
+
8
239.52 ® 239.52
Based on the reactions given by Eqs. (2.1) and (2.2) we conclude that
when 100 parts by weight of CaCO3 reacts, 56 parts by weight of CaO and
44 parts by weight of CO2 are formed. Similarly, when 167.52 parts by
weight of iron reacts with 72 parts by weight of steam (water), we get
231.52 parts by weight of magnetite and 8 parts by weight of hydrogen.
Thus the total weight of reactants is always equal to the total weight of
products.
Such computations will help one to estimate the quantity of reactants
needed to obtain a specified amount of product.
gram atom (or g atom) = Mass in grams/Atomic weight
katom (or kg atom)
= Mass in kg/Atomic weight
gram mole (or g mole) = Mass in grams/Molecular weight
kmole (or kg mole)
= Mass in kg/Molecular weight
7
8
PROCESS CALCULATIONS
The conclusions based on reactions (2.1) and (2.2) on material balance
can be expressed in other forms too, as per the definitions given above:
1 kmole of CaCO3 gives 1 kmole of CaO and 1 kmole of CO2
Similarly, 3 kmoles of iron reacts with 4 kmoles of steam (water) to
yield 1 kmole of oxide and 4 kmoles of hydrogen. When such balances (on
molar basis) are made, the number of moles on the reactants side need not
be equal to the total numbers of moles on the product side.
One atom of oxygen weighs
= 16 grams O
One atom of hydrogen weighs
= 1 gram H
One molecule of oxygen weighs = 32 grams O
One molecule of hydrogen weighs = 2 grams H
In other words,
16 grams of oxygen
= 1 g atom of oxygen
32 pounds of oxygen
= 1 lb mole of oxygen
2 g atoms of oxygen
= 1 g mole of oxygen
1 gram of hydrogen
= 1 g atom of hydrogen
2 kg of hydrogen
= 1 kmole of hydrogen
2 kg atoms of hydrogen = 1 kmole of hydrogen
\
\
g atom or lb atom
g mole or lb mole
= Mass in grams or pounds/Atomic weight
= Mass in grams or pounds/Molecular weight
2.2 CONSERVATION OF MASS
The law of conservation of mass states that mass can neither be created nor
be destroyed. It is the basic principle adopted in solving the material balance
problems in chemical process calculations, whether a chemical reaction is
involved or not. However, while applying the law of conservation of mass,
one should not apply it for the conservation of molecules. We frequently
come across chemical reactions in which the total number of moles on the
reactant side is not equal to the total number of moles on the product side.
For example:
Na2CO3 + Ca(OH)2 ® CaCO3 + 2NaOH
The total number of moles on the reactant side is 2 and on the product
side 3. Here the mass balance is ensured but not the mole balance. Now
consider the reaction:
2Cr2O3 + 3CS2 ® 2Cr2S3 + 3CO2
Here the total number of moles both on the reactant side and the
product side is 5. Hence both the conservation of mass and the conservation
of moles are observed.
MASS RELATIONS
2.3
9
AVOGADRO’S HYPOTHESIS
1 g mole of any gaseous substance at NTP occupies 22,414 cc or
22.414 litres and 1 lb mole of the same substance occupies 359 ft3at NTP.
1 kmole of any gaseous substance occupies 22.414 m3 at NTP.
Note: NTP = Normal temperature (273 K) and pressure (1 atmosphere) are also
referred to at times as standard conditions (SC).
2.4
LIMITING REACTANT AND EXCESS REACTANT
For most of the chemical reactions the reactants will not be used in
stoichiometric proportion or quantities. One of the reactants will be present
in excess and remain unreacted even when the other reactant has completely
reacted. The reactant thus present in excess is termed excess reactant and
the other reactant which is present in a lesser quantity and cannot react with
whole of the other reactant (excess reactant) is called limiting reactant. All
calculations involved in estimating the quantity of product and conversion
are always based on the limiting reactant. The amount by which any
reactant is present in excess to that required to combine with the limiting
reactant is usually expressed as percentage excess. The percentage excess of
any reactant is defined as the percentage ratio of the excess to that
theoretically required by the stoichiometric equation for combining with the
limiting reactant. A limiting reactant is the one, which will not be present in
the product, whereas the excess reactant is the one, which will always be
present in the product.
Let us consider that 18 kg of carbon is burnt with 32 kg of oxygen. As
per stoichiometry
C + O2 Æ CO2
i.e. 12 kg of carbon will burn with 32 kg of oxygen to form 44 kg of CO2.
Hence, for 18 kg of carbon to react fully we should have 48 kg of
oxygen. Since 32 kg of oxygen alone is available, it is called the limiting
reactant and carbon is called the excess reactant. For 32 kg of oxygen to
react fully, it is sufficient to have 12 kg of carbon. However 6 kg of carbon
is present in excess.
Hence % excess of carbon is =
2.5
6
¥ 100 = 50%.
12
CONVERSION AND YIELD
These terms are used for a chemical reaction where the reactants give out
new compounds or products.
10
PROCESS CALCULATIONS
Conversion is the ratio of the amount of material actually converted to
that present initially, whereas the yield is the amount of desired product
actually formed compared to that which can be formed theoretically.
Conversion for a reaction is based on the limiting reactant whereas the
yield is based on the product formed.
2.6
COMPOSITION OF MIXTURES AND
SOLUTIONS
Different methods are available for expressing composition of mixtures of
gases, liquids and solids.
Conventionally the composition of solids is either expressed on weight
basis or mole basis. Let us consider a binary system comprising components
A and B.
W = Total weight of the system.
WA, WB = Weight of components A and B respectively.
MA, MB = Molecular weight of components A and B respectively,
if they are compounds.
AA, AB = Atomic weight of components A and B respectively,
if they are elements.
V = Volume of the system.
VA and VB = Pure component volume of components A and B
respectively.
2.6.1
Weight Percent
This is defined as the ratio of weight of a particular component to the total
weight of the system in every 100 part, i.e.
Weight % of A =
WA
¥ 100
W
This method of expressing composition is generally employed in solid
and liquid systems and not used in gaseous system. One major advantage of
weight percent is, its independence to changes in temperature and pressure.
The composition of a solid mixture is to be always taken as weight %
when nothing is mentioned above its units.
2.6.2
Volume Percent
The ratio of the volume of each component and the total volume of the
system, for each 100 part of the total volume is called volume percent
MASS RELATIONS
Volume % of A =
11
VA
´ 100
V
This method of expressing composition is employed always for gases,
rarely for liquids and seldom for solids. The composition of a gas mixture is
to be taken as volume % when nothing is mentioning about its units.
The volume % is also equal to mole % for ideal gases but not for
liquids and solids. This is based on Avogadro’s law.
2.6.3
Mole Fraction and Mole Percent
These concepts are generally adopted in the case of a mixture containing
molecules of different species.
WA
MA
Mole fraction of A =
WA
WB
M A MB
Mole % of A = Mole fraction of A ´ 100
2.6.4
Atomic Fraction and Atomic Percent
This is adopted when a mixture contains two or more atoms.
WA
AA
Atomic fraction of A =
WA WB
AA
AB
Atomic % of A = Atomic fraction of A ´ 100
2.6.5 Composition of Liquid Systems
In the case of liquids we come across more number of methods of
expressing compositions of the liquid constituents.
(i)
(ii)
(iii)
(iv)
(v)
Weight ratio = Weight of solute/weight of solvent
Mole ratio
= g moles of solute/g moles of solvent
Molality
= g moles of solute/1 kg of solvent
Molarity
= Number of g moles of solute/1 litre of solution
Normality (N) = Number of gram equivalents of solute/1 litre of
solution
Hence, concentration in grams per litre = Normality (N) ´ Equivalent
weight of solute
For very dilute aqueous solutions, molality = molarity
12
PROCESS CALCULATIONS
2.7 DENSITY AND SPECIFIC GRAVITY
Density is defined as mass per unit volume and it varies with temperature.
Specific gravity is the ratio of density of a liquid to that of water. However,
in the case of gases, it is defined as the ratio of its density to that of air at
same conditions of temperature and pressure.
Over a narrow range of temperature, the variation in density of solids
is not high. However, in the case of liquids and gases the variation in
density is significant. Similarly, the densities vary significantly with
concentration also. This property of density and specific gravity varying
with concentration is very widely used both in industries and markets as an
index for finding the composition of a system comprising a specific solute
and a specific solvent.
Several scales are in use in which specific gravities are expressed in
terms of a degree, which are related to specific gravities and densities by
arbitrary mathematical definitions.
2.7.1 Baume’ (°Be’) Gravity Scale
For liquids lighter than water
Degrees Baume’ =
140
– 130
G
G is the specific gravity at
60
¦ 15
µ
°F § °C¶
60
¨ 15
·
Thus, water will have a gravity of 10° Be’ and this degree decreases
with increase in specific gravity.
For liquids heavier than water
145
G
In this scale the degree increases with increase in specific gravity.
Degrees Baume’ = 145 –
2.7.2 API Scale (American Petroleum Institute)
This scale is used for expressing gravities of petroleum products. This is
similar to Baume’ scale for liquids lighter than water.
Degrees API =
141.5
– 131.5
G
MASS RELATIONS
2.7.3
13
Twaddell Scale
This scale is used for liquids heavier than water.
Degrees Twaddell (Tw) = 200 (G – 1.0)
2.7.4 Brix Scale
This scale is used in sugar industry and 1 degree Brix is equal to 1% sugar
in solution.
Degree Brix =
400
– 400
G
WORKED EXAMPLES
2.1
Convert 5000 ppm into weight %.
5000 × 100
10 6
2.2
= 0.5%
The strength of H3PO4 was found to be 35% P2O5. Find the weight %
of the acid.
The acid can be split into
2H 3 PO 4 → P2 O 5 + 3H 2 O
(2 × 98)
142
(3 × 18)
196 units of the acid contains 142 units of the pentaoxide.
The weight % of pentaoxide is (142/196) which is 72.5% for pure acid.
When the strength of pentoxide is 35%, the weight % of acid is
100 ⎞
⎛
⎜ 35 × 72.5 ⎟ = 48.3%
⎝
⎠
2.3
What is the volume of 25 kg of chlorine at standard condition?
25 kg Cl2 =
Volume =
2.4
25
kmoles of Cl2
2 × 35.46
25 × 22.414
= 7.9 m3
2 × 35.46
How many grams of liquid propane will be formed by the liquefaction
of 500 litres. of the gas at NTP? Molecular weight of propane (C3H8)
is 44
1 g mole of any gas occupies 22.414 litres at NTP
14
PROCESS CALCULATIONS
500
= 22.31 g moles
22.414
22.31 g moles of propane weighs = 22.31 ´ 44 = 981.52 g.
500 litres of propane at NTP =
2.5
Find the volume of (a) 100 kg of hydrogen and (b) 100 lb of
hydrogen at standard conditions?
(a) 100 kg of H2 = 50 kmoles of hydrogen
volume occupied by 50 kmoles of hydrogen º 50 ´ 22.414
= 1120.7 m3
(b) 100 lb of H2 º 50 lb moles of hydrogen
Volume occupied by 50 lb moles of hydrogen º 50 ´ 359 = 17950 ft3
2.6
A solution of naphthalene in benzene contains 25 mole % Naphthalene. Express the composition in weight %.
Basis: 100 g moles of solution
Component
Molecular
weight
Weight,
g mole
Actual
weight, g
128
78
25
75
25 ´ 128 = 3200
75 ´ 78 = 5850
3200 ´ 100/9050 = 35.35
5850 ´ 100/9050 = 64.65
9050
100.00
Naphthalene C10H8
Benzene C6H6
Total
2.7
Composition in
weight percent
What is the weight of one litre of methane CH4 at standard conditions?
22.414 litres of any gas at NTP is equivalent to 1 g mole of that gas
\
\
2.8
1
= 0.0446 g mole
22.414
Weight of one litre methane = 0.0446 ´ 16 = 0.714 g
1 litre of methane º 1 ´
A compound whose molecular weight is 103 analyses C : 81.5, H : 4.9
and N : 13.6 by weight. What is the formula?
Basis: 100 g of substance
Element
Atomic
weight
Weight, g
Weight,
g atom
Rounding of
atoms
Weight of
each element
Carbon
Hydrogen
Nitrogen
12
1
14
81.5
4.9
13.6
81.5/12 = 6.8
4.9/1 = 4.9
13.6/14 = 0.9
7
5
1
84
5
14
Total
103
Hence the formula obtained after rounding is correct.
So the molecular formula is C7H5N
15
MASS RELATIONS
2.9
An analysis of a glass sample yields the following data. Find the
mole %.
Na2O : 7.8%, MgO : 7.0%, ZnO : 9.7%, Al2O3 : 2.0%, B2O3 : 8.5%
and rest SiO2.
Basis: 100 g of glass sample
Component
Weight, g
Molecular weight
g mole
mole %
Na2O
MgO
ZnO
Al2O3
B2 O 3
SiO2
7.8
7.0
9.7
2.0
8.5
65.0
62.0
40.3
81.4
102.0
69.6
60.1
0.1258
0.1737
0.1192
0.0196
0.1221
1.0815
7.665
10.583
7.262
1.194
7.439
65.857
Total
100.0
—
1.6419
100.0
2.10 A gaseous mixture analyzing CH4 : 10%, C2H6 : 30% and rest H2 at
15 °C and 1.5 atm is flowing through an equipment at the rate of
2.5 m3/min. Find (a) the average molecular weight of the gas mixture,
(b) weight % and (c) the mass flow rate.
Basis: 100 g moles of the gaseous mixture.
Component
Weight,
g mole
Molecular
weight
Weight,
g
Weight %
CH4
C2H6
H2
10
30
60
16
30
2
160
900
120
13.56
76.27
10.17
Total
100
—
1180
100
The average molecular weight =
1180
= 11.8
100
⎛ 1.5 ⎞
Volumetric flow rate at standard conditions = 2.5 ¥ ⎜
⎟ ¥ (273/288)
⎝ 1 ⎠
= 3.555 m3/min
3.555
= 0.156 kmole
22.414
Mass flow rate = moles ¥ average molecular weight = 0.156 ¥ 11.8
= 1.84 kg/min
Moles of the gas =
2.11 In an evaporator a dilute solution of 4% NaOH is concentrated to 25%
NaOH. Calculate the evaporation of water per kg of feed.
16
PROCESS CALCULATIONS
Basis: 1 kg of feed.
NaOH present is 0.04 kg, which appears as 25% in the thick liquor
formed
0.04
= 0.16 kg
0.25
Weight of water evaporated = (1 – 0.16) = 0.84 kg
Weight of thick liquor formed =
Water evaporated per kg of feed = 0.84 kg
2.12 The average molecular weight of a flue gas sample is calculated by
two different engineers. One engineer used the correct molecular
weight of N2 as 28, while the other used an incorrect value of 14.
They got the average molecular weight as 30.08 and the incorrect one
as 18.74. Calculate the % volume of N2 in the flue gases. If the
remaining gases are CO2 and O2 calculate their composition also.
Basis: 100 g moles of flue gas
g mole
I Engineer
II Engineer
N2
x
28x
14x
CO2
y
44y
44y
O2
z
32z
32z
Total
100
3008
1874
Component
x + y + z = 100
(i)
28x + 44y + 32z = 3008
(ii)
14x + 44y + 32z = 1874
(iii)
Solving Eqs. (i), (ii) and (iii), we get
x = Moles of nitrogen = 81%
y = Moles of carbon dioxide = 11%
z = Moles of oxygen = 8%
2.13 An aqueous solution contains 40% of Na2CO3 by weight. Express the
composition in mole percent.
Basis: 100 g of solution
Component
grams
Molecular
weight
g mole
Composition in
mole %
Na2CO3
40
106
40/106 = 0.377
0.377 ´ 100/3.71 = 10.16
Water
60
18
60/18 = 3.333
3.333 ´ 100/3.71 = 89.84
3.710
100.00
Total
MASS RELATIONS
17
2.14 What is the weight of iron and water required for the production of
100 kg of hydrogen?
3Fe + 4H 2 O →
Fe 3 O 4
+ 4H 2 ↑
(4 ×18)
(4 × 2 ×1)
(3 × 55.84)
(3 × 55.84) + (4 ×16)
72
8
167.52
231.52
239.52
239.52
Method 1 (Based on absolute mass)
167.52 kg of Fe is required for producing 8 kg of H2
\
For producing 100 kg of H2 (by stoichiometry)
100 t 167.52
8
= 2094 kg
Iron (Fe) required =
Similarly, for getting 100 kg of H2 the amount of steam (H2O)
required is
100 t 72
= 900 kg
8
The total weight of reactants is
=
2094 kg Fe and 900 kg H2O = 2994 kg
231.52
´ 2094 = 2894 kg
The weight of Fe3O4 formed is =
167.52
\ The total weight of products is (100 kg H2 + 2894 kg Fe3O4)
= 2994 kg
The total weight of reactants is (2094 kg Fe + 900 kg H2O) = 2994 kg
Method 2 (Based on moles)
100 kg of H2
= 50 kmoles
4 kmoles H2 comes from
= 3 katoms of Fe (by stoichiometry)
50 kmoles of H2 comes from
=
Weight of 37.5 katoms Fe
= (37.5 ´ 55.84) = 2094 kg of iron
50 kmoles H2 from
=
Weight of 50 kmoles H2O
Total weight of reactants
= (50 ´ 18) = 900 kg H2O
37.5
=
kmoles
3
¦ 37.5 µ
= §
¶ ´ 231.52 = 2894 kg
¨ 3 ·
= 2994 kg
Total weight of products
= 2994 kg
Moles of Fe3O4 formed is
Weight of Fe3O4 formed is
50 µ
¦
§3 t 4 ¶
¨
·
¦ 50
µ
§ 4 t 4¶
¨
·
= 37.5 katoms Fe
kmoles of water
18
PROCESS CALCULATIONS
2.15 How much super phosphate fertilizer can be made from one ton of
calcium phosphate 93.5% pure?
Atomic weights are: Ca : 40, P : 31, O : 16, S : 32
Ca3(PO4)2 + 2H2SO4 Æ CaH4(PO4)2 + 2CaSO4
310
(2 ¥ 98)
(2 ¥ 136)
234
506
506
One ton of raw calcium phosphate contains 0.935 tons of pure calcium
phosphate
0.935
= 0.70577 tonne
\ Weight of super phosphate formed is = 234 ¥
310
2.16 SO2 is produced by the reaction between copper and sulphuric acid.
How much Cu must be used to get 10 kg of SO2?
Cu + 2H 2 SO 4 → CuSO 4 + SO 2 + 2H 2 O
64
63.54
64 kg of sulphur dioxide is obtained from 63.54 kg of copper.
10 kg of sulphur dioxide will be obtained from 9.93 kg of copper.
2.17 How much potassium chlorate must be taken to produce the same
amount of oxygen that will be produced by 2.3 g of mercuric oxide?
2KClO3
Æ
2HgO
Æ
(2 ¥ 122.46)
244.92
(2 ¥ 216.6)
2KCl
+
3O2
(2 ¥ 74.46)
148.92
2Hg
(6 ¥ 16)
96
+
O2
(2 ¥ 200.6)
(2 ¥ 16)
433.2 g HgO gives 32 g of oxygen
2.3 g of HgO will give (32 ¥ 2.3/433.2) = 0.1698 g of O2
0.1698 g O2 is obtained from (244.92 ¥ 0.1698)/96 = 0.4332 g of KClO3.
2.18 Ammonium phosphomolybdate is made up of the radicals NH3, H2O,
P2O5 and MoO3. What is % composition of the molecule with respect
to these radicals?
The formula of ammonium phosphomolybdate is
(NH4)3PO4 ◊ 12MoO3 ◊ 3H2O
First let us form the final product from the radicals:
3NH3 + 4.5H2O + 12MoO3 + ½P2O5 Æ (NH4)3PO4 12MoO3 · 3H2O
(3 ¥ 17 = 51) (4.5 ¥ 18 = 81) (12 ¥ 144 = 1728) (½ ¥ 142 = 71)
% of NH3
=
51 ¥
100
1931
=
2.64
% of H2O
=
81 ¥
100
1931
=
4.19
1931
MASS RELATIONS
% of MoO3 = 1728 ´
100
1931
71 ´
100
1931
Total
% of P2O5 =
=
89.49
=
3.68
19
= 100.00
2.19 How many grams of salt are required to make 2500 g of salt cake?
How much Glauber’s salt can be obtained from this?
The molecular formula of Glauber’s salt is Na2SO4 × 10H2O
(142 + 180 = 322)
2NaCl
(2 × 58.46 =116.92)
+ H 2 SO 4 → Na 2 SO 4 + 2HCl
98
142
(2 × 36.46)
Thus, 142 g of Na2SO4 is obtained from 116.92 g NaCl.
2500 g of salt cake is obtained from 116.92 ´ 2500/142 = 2058.45 g
NaCl
Hence, salt needed is 2058.45 g
Glauber’s salt (Na2SO4 × 10H2O) obtained is 2500 ´ 322/142 = 5669 g
2.20 (a) How many grams of K2Cr2O7 are equivalent to 5 g KMnO4?
(b) How many grams of KMnO4 are equivalent to 5 g K2Cr2O7?
2KMnO4 + 8H2SO4 + 10FeSO4 ® 5Fe 2(SO4)3 + K2SO4 + 2MnSO4
+ 8H2O
K2Cr2O7 + 7H2SO4 + 6FeSO4 ® 3Fe2(SO4)3 + K2SO4 + (Cr2SO4)3
+ 7H2O
2KMnO4
gives
5Fe2(SO4)3
(2 ´ 158 = 316)
(5 ´ 400 = 2000)
K2Cr2O7 gives 3Fe2(SO4)3
(294)
(3 ´ 400 = 1200)
\
294 g K2Cr2O7 is equivalent to
\
3 g K2Cr2O7
º
Similarly, 5 g KMnO4 º
1200 t 316
= 189.6 g KMnO4
2000
3 t 189.6
= 1.935 g KMnO4
294
5 t 294
= 7.75 g K2Cr2O7
189.6
5t3
= 7.75 g K2Cr2O7
1.935
2.21 If 45 g of iron react with H2SO4, how many litres of hydrogen are
liberated at standard condition?
Alternatively,
20
PROCESS CALCULATIONS
There are two possible reactions in this case:
(a) Case I
Fe + H 2 SO 4 → FeSO 4 + H 2
(55.85)
(i)
(2)
The weight of hydrogen formed by reaction (i) is = 45 ¥
2
= 1.611 g,
55.85
1.611
= 0.806 g mole
2
0.806 g mole ∫ 0.806 ¥ 22.414 = 18.06 litres
i.e.
(b) Case II
2Fe + 3H 2 SO 4 → Fe 2 (SO 4 )3 + 3H 2
(111.7)
(ii)
(6)
The moles of hydrogen formed by reaction (ii) is 45 ¥
6
= 2.418 g
111.7
2.418 g H2 = 1.209 g moles of hydrogen ∫ 1.209 ¥ 22.414 ∫ 27.1
litres
2.22 A natural gas has the following composition by volume CH4 : 83.5%,
C2H6 : 12.5%, and N2 : 4%. Calculate the following:
(a) composition in mole % (b) composition in weight % (c) average
molecular weight (AVMWT) (d) density at standard condition (kg/m3)
Basis: 100 kmoles of gas mixture
Component
Molecular
weight
mole %
Weight, kg
Weight %
CH4
16
83.5
83.5 ¥ 16 = 1336
1336 ¥ 100/1823 = 73.29
C 2H 6
30
12.5
12.5 ¥ 30 = 375
375 ¥ 100/1823 = 20.57
N2
28
4.0
4.0 ¥ 28 = 112
112 ¥ 100/1823 = 6.14
Total
1823
100.0
(c) Average molecular weight =
1823
100
= 18.23
Volume at standard condition = 00 ¥ 22.414 = 2241.4 m3
(d) Density of gas at standard condition =
1823
2241.4
= 0.813 kg/m3
21
MASS RELATIONS
2.23 Convert 54.75 g/litre of HCl into molarity.
Molarity = g moles/litre of solution
54.75
= 1.5
36.45
2.24 A solution of NaCl in water contains 230 g of NaCl per litre at 20 °C.
The density of the solution at this temperature is 1.148 g/cc. Find the
composition in (a) weight % (b) volume % of water (c) mole %
(d) atomic % (e) molality and (f) g NaCl/g water.
=
Basis: (a) 1 litre of solution has a weight of 1148 g
Component
Molecular
weight
Weight, g
Weight, %
g mole
NaCl
58.5
230
20.03
230/58.5 = 3.93 3.93/54.93 = 7.15
Water
18
918
79.97
918/18 = 51.00
51/54.93 = 92.85
Total
1148
100.00
54.93
100.00
mole %
(b) Volume % of water: 918 g is present in 1 litre of solution, i.e.
918 cc water is present in 1000 cc. of solution (density of water is
1 g/cc)
Volume % = 91.8%
(d)
Element
g atoms
Atomic %
Na
Cl
H
O
3.93
3.93
102.00
51.00
2.443
2.443
63.409
31.705
Total
160.86
100.000
(All are based on the molecular formula)
(e) Molality = g moles of solute in 1 kg of solvent (3.93 ´ 1000/918)
or, 3.93 g moles of NaCl is present in 918 g of water
(i.e.) 4.28 g moles/1000 g of solvent
Molality = 4.28
Molarity = Moles of solute per litre of the solution = 3.93
(f) g NaCl/g water =
230
= 0.252
918
22
PROCESS CALCULATIONS
2.25 A benzene solution of anthracene contains 10% by weight of the
solute. Find the composition in terms of (a) molality (b) mole fraction.
Basis: 100 g of solution
Component
Molecular
weight
Weight, g
Weight,
g mole
mole
fraction
Anthracene
178
10
(10/178)
0.046
0.0562
Benzene
78
90
(90/78)
0.954
1.1538
1.2100
1.00
Total
Molality = g moles of anthracene in 1000 g benzene
=
0.0562
90
¥ 1000 = 0.624
2.26 Calculate the weight of NaCl that should be placed in a 1 litre
volumetric flask to prepare a solution of 1.8 molality. Density of this
solution is 1.06 g/cc
Molality = g moles of NaCl/1000 g of water
= 1.8
or, 1.8 g moles NaCl = (1.8 ¥ 58.46) = 105.228 g
Component
Weight, g
Weight %
NaCl
105.228
9.52
H2O
1000.000
90.48
1105.228
100.00
Density of this solution = 1.06 g/cc
Volume of this solution, i.e. mass/density =
1105.228
= 1042.67 cc
1.06
or 1042.67 cc of this solution contains 105.228 g of NaCl
\
105.228
1042.67
= 100.92 g of NaCl.
1000 cc of this solution will have = 1000 ¥
NaCl needed = 100.92 g
MASS RELATIONS
23
2.27 For the operation of a refrigeration plant it is desired to prepare a
solution of 20% by weight of NaCl solution.
(a) Find the weight of salt that should be added to one gallon of
water at 30°C?
(b) What is the volume of this solution?
Basis: 100 lb of solution
It will have 20 lb NaCl and 80 lb water
80 lb water = 1.28 ft3 (since the density of water is 62.47 lb/ft3)
We know that 1 ft3 = 7.48 gallons
Therefore, 1.28 ft3 = 9.57 gallons.
20 µ
¶
¨ 9.57 ·
(a) Weight of salt per gallon of water = ¦§
= 2.09 lb.
(b) Specific gravity of NaCl solution at 30 °C = 1.14
\
Density of solution = 1.14 ´ 62.4 = 71.14 lb/ft 3
62.47
= 8.35 lb.
7.48
Total weight of solution = weight of water + weight of salt
Weight of 1 gallon of water =
= 8.35 + 2.09 = 10.44 lb.
10.44
= 0.147 ft3
71.14
= 1.1 gallons.
2.28 (a) A solution has 100° Tw gravity. What is its specific gravity and
°Be’?
Hence, volume of the above solution =
(b) An oil has a specific gravity of 0.79. Find °API and °Be’
(a) 100
= 200 (G – 1) \
°Be’ = 145 –
G = 1.5
145
145 µ
= 145 – ¦§
¶ = 48.3 °Be’
G
¨ 1.5 ·
¦ 141.5 µ
141.5 µ
¶ – 131.5 = 47.6
¶ – 131.5 = §
¨ 0.79 ·
G
¨
·
(b) °API = ¦§
140 µ
¶
¨ 0.79 ·
°Be’ = ¦§
– 130 = 47.2
2.29 An aqueous solution contains 15% ethyl alcohol by volume. Express
the composition in weight % and mole %. Density of ethyl alcohol
and water are 790 kg/m3 and 1000 kg/m3 respectively.
24
PROCESS CALCULATIONS
Basis: 1 m3 of solution.
Compound Molecular Volume, Density,
weight
m3
kg/m3
Weight,
kg
Number
of moles
Weight
%
Ethanol
Water
Total
118.5
850
968.5
2.576
47.222
49.798
12.235
5.173
87.765 94.827
100
100
46
18
0.15
0.85
1.00
790
1000
mole
%
2.30 The quality of urea is expressed in terms of nitrogen content. If the
nitrogen content in the sample is only 40%, estimate the purity of
sample in terms of urea content.
The molecular weight of urea (NH2CONH2) is 60 and that of N2 is 28.
Basis: 100 kg of sample
60 kg of urea has 28 kg of N2
100 kg of urea will have = 28 ×
100
= 46.67 kg of N2
60
(Theoretically)
The given sample has 40% N2
100
= 85.71%
46.67
2.31 If the nitrogen content in ammonium nitrate sample is 28%, estimate
the purity of ammonium nitrate.
Hence, the % purity is = 40 ×
Molecular weight of ammonium nitrate, NH4NO3 = 80
% Nitrogen in pure ammonium nitrate = 28 ×
100
= 35%
80
The % of nitrogen in the sample is 28
¦ 28 µ
¶
¨ 35 ·
Hence, the purity of ammonium nitrate is §
× 100 = 80%
2.32 Nitrobenzene is produced by reacting nitrating mixture with benzene.
The nitrating mixture contains 31.5% HNO3, 60% H2SO4 and 8.5%
H2O. A charge contains 663 kg of benzene and 1700 kg of nitrating
mixture which sent into the reactor. If the reaction is 95%, then
calculate the amount of nitrobenzene and spent acid produced.
The reaction is
C6H6 + HNO3 ® C6H5NO2 + H2O
Feed, C6H6 : 663/78 = 8.5 kmole
: 31.5% of 1700 kg = 535.5 kg = 8.5 kmoles
HNO3
: 60 % of 1700 kg = 1020 kg = 10.408 kmoles
H2SO4
: 8.5 % of 1700 kg = 144.5 kg = 8.028 kmoles
H2O
MASS RELATIONS
25
Reaction is 95% complete
Hence, HNO3 unreacted : 0.05 × 8.5 = 0.425 kmole
: 0.05 × 8.5 = 0.425 kmole
C6H6 unreacted
: 10.408 kmoles
H2SO4 unreacted
: 8.028 kmoles
H2O unreacted
: 8.5 × 0.95 = 8.075 kmoles
H2O formed
Nitrobenzene formed
: 8.5 × 0.95 = 8.075 kmoles
Component
Weight,
kmole
Molecular
weight
Weight,
kg
Weight,
%
HNO3
0.425
63
26.775
1.133
H2SO4
10.408
98
1020.000
43.165
H2O
(8.075 + 8.028) = 16.103
18
289.850
12.266
Nitrobenzene
8.075
123
993.225
42.032
C6H6
0.425
78
33.150
1.403
2363.00
100%
Total
Nitrobenzene produced = 993.225 kg
Spent acid = 26.775 + 1020.000 + 289.85 = 1336.63 kg
2.33 A sample of caustic soda flake contains 74.6% Na2O by weight.
Estimate the purity of flakes.
Reaction is as follows:
2NaOH ® Na2O + H2O
Amount of Na2O in pure flakes = 62 × 100/80 = 77.5%
% Purity = 0.746/0.775 × 100 = 96.26%
2.34 Two kg of CaCO3 and MgCO3 was heated to a constant weight of
1.1 kg. Calculate the % amount of CaCO3 and MgCO3 in reacting
mixture.
Reaction is as follows:
CaCO3 ® CaO + CO2
(100)
(56)
(44)
MgCO3 ® MgO + CO2
(84)
(40)
(44)
Let, x be the amount of CaCO3.
Therefore, (2 – x) be the weight of MgCO3
100 kg of CaCO3 gives 56 kg of CaO
26
PROCESS CALCULATIONS
56 x
= 0.56x kg of CaO.
100
Similarly, 84 kg of MgCO3 gives 40 kg of MgO
Therefore, x kg of CaCO3 gives
© 40 ¸
¹
« 84 º
Therefore, (2 – x) kg of MgCO3 gives ª
× (2 – x) kg of MgO
The weight of product left behind is 1.1 kg, i.e. weight of MgO +
CaO left behind
0.56x + (0.4672)(2 – x) = 1.1
0.0838x = 1.1 – 0.96524
Therefore, x = 1.761 kg
Component
Weight, kg
Weight, %
CaCO3
MgCO3
Total
1.761
0.239
2.000
88.05
11.95
100.00
2.35 The composition of NPK fertilizer is expressed in terms of N2, P2O5
and K2O each of about 15 weight %. Anhydrous ammonia, 100%
phosphoric acid and 100% KCl are mixed to get 1 ton of fertilizer.
Estimate the amount of filler in the NPK fertilizer.
Basis: 1000 kg of fertilizer
Reactions are:
2NH3
® N2
+ 3H2
(28)
(6)
(34)
2H3PO4
® P2O5 + 3H2O
(196)
(142)
(54)
2KCl + H2O ® K2O + 2HCl
(149)
(18)
(94)
(73)
N2, K2O and P2O5 are each equivalent to 15 weight % = 150 kg each
Ammonia reacted = 34 ×
150
28
H3PO4 needed
150
= 207.04 kg
142
= 196 ×
= 182.14 kg
= 149 × 150 = 237.77 kg
94
The amount of inert material/filler = 1000 – 626.95 = 373.05 kg
KCl needed
2.36 A solution whose specific gravity is 1 contains 35% A by weight and
the rest is B. If the specific gravity of A is 0.7, find the specific
gravity of B.
MASS RELATIONS
27
Basis: 1000 kg of solution
Weight of A: 350 kg
Weight of B: 650 kg
Volume of solution = 1 m3 (since density is 1000 kg/m3 due to
specific gravity being unity)
Mass/volume = density
Assuming ideal behaviour
350
700
650
SB
=1
rB = 1300 kg/m3
Therefore, specific gravity of B = 1.3
2.37 An aqueous solution contains 47% of A on volume basis. If the density of A is 1250 kg/m3, express the composition of A in weight %.
Basis: 1 m3 of solution
Volume of A in solution = 1 × 0.47 = 0.47 m3
Weight of A = 0.47 × 1250 = 587.5 kg
Volume of water = (1 – 0.47) = 0.53 m3
Therefore, the weight of water = 0.53 m3 × 1000 = 530 kg
Hence, weight % of A =
587.5
= 52.57%
530 587.5
2.38 An aqueous solution contains 43 g of K2CO3 in 100 g of water. The
density of solution is 1.3 g/cc. Find the composition in molarity and
molality.
Basis: 100 g of solvent
Weight of K2CO3 = 43 g
Weight of solution = 143 g
Density of solution = 1.3 g/cc
Volume of solution = 143 = 110 cc
1.3
Moles of solute = Weight/Molecular weight =
43
= 312 g moles
138
Molarity = g mole/volume of solution in lit =
0.312
= 2.833M
0.11
Molality = g mole/kg of solvent =
0.312
= 3.12 g moles/kg solvent.
0.1
28
PROCESS CALCULATIONS
2.39 A gaseous mixture contains ethylene: 30.6%, benzene: 24.5%, O2:
1.3%, ethane: 25%, N2: 3.1% and methane: 15.5% in volume basis.
Estimate the composition in mole %, weight %, average molecular
weight and density.
Basis: 100 kmoles of mixture
Compound
mole %
Molecular weight
Weight, kg
Weight %
C2H4
C6H6
O2
CH4
C2H6
N2
30.6
24.5
1.3
15.5
25
3.1
28
78
32
16
30
28
856.8
1911.0
41.6
248.0
750.0
86.8
22.00
49.07
1.07
6.37
19.26
2.23
3894.2
100.00
Total
Density = Weight/Volume =
3894.2
× 22.414 = 1.737 kg/m3 = 1.737 g/l.
100
2.40 A compound has a composition of 9.76% Mg, 13.01% S, 26.01%
O2 and 57.22% H2O by weight. Find the molecular formula of this
compound.
Basis: 100 g of compound
Compound
Weight,
g
Atomic weight
or Molecular
weight
Number
of moles
Converting to
whole numbers
dividing by 0.41
Mg
S
O
H2O
9.76
13.01
26.01
57.22
24
32
16
18
0.410
0.410
1.615
2.8738
1
1
3.94
6.92
Therefore, molecular formula of the compound = MgSO4.7H2O
2.41 A substance on analysis gave 1.978 g of Ag, 0.293 g of S and 0.587 g
of O2. Find the molecular formula of the compound.
Compound
Weight,
g
Atomic weight
or Molecular
weight
Number
of moles
Converting to
whole numbers
dividing by
9.156 × 10–3
Ag
S
O2
1.978
0.293
0.587
108
32
16
0.0183
9.156 × 10–3
0.0367
2
1
4.04
Molecular formula = Ag2SO4
MASS RELATIONS
29
2.42 Two engineers are estimating the average molecular weight of gas
containing oxygen and another gas. One uses the molecular weight as
32 and finds the average molecular weight as 39.8 and the other uses
the atomic weight of oxygen as 16 and finds the average molecular
weight as 33.4. Estimate the composition of the gas mixture.
By using the atomic weight of oxygen as 16, the value is 33.4 and by
using the molecular weight of oxygen, the value is 39.8.
Let x be the mole fraction of oxygen in the mixture and the molecular
weight of the other gas be M
39.8 = (x) × (32) + (1 – x) × (M)
33.4 = (x) × (16) + (1 – x) × (M)
Solving, we get x = 0.4
i.e the fraction of oxygen in the mixture is 0.4
2.43 A mixture of methane and ethane has an average molecular weight of
21.6. Find the composition.
Let the mole fraction of methane be X
21.6 = (Molecular weight of CH4)(X) + (Molecular weight of C2H6)
(1 – X)
21.6 = 16 × X + 30 × (1 – X)
Solving, we get X = 0.6
2.44 A mixture of FeO and Fe3O4 was heated in air and is found to gain
5% in mass. Find the composition of initial mixture.
Reactions involved are:
2FeO + 0.5 O2 ® Fe2O3
2Fe3O4 + 0.5 O2 ® 3 Fe2O3
Basis: 100 kg of feed mixture
Let X be the weight of FeO in the mixture
From 144 kg of FeO, Fe2O3 formed is 160 kg
Therefore, from X kg of FeO, Fe2O3 formed is 160 ×
X
144
From 232 kg of Fe3O4, Fe2O3 formed is 480 kg
(480)
464
Since 5% gain in mass is observed, the weight of final product is 105 kg, i.e.
Therefore, from (100 – X) of Fe3O4, Fe2O3 formed is (100 – X) ×
160 t X
(480)
100 X t
105
144
464
Solving, X, the weight of FeO = 20.25 kg
Fe3O4 = (100 – 83.45) = 79.75 kg
30
PROCESS CALCULATIONS
2.45 A sample of lime stone has 54.5% CaO. Find the weight % of lime
stone.
Basis: 100 kg of lime stone
100 kg of CaCO3 will have 56% CaO
If the CaO is 54.5%, then % of CaCO3 in the sample is
54.5 × 100/50 = 97.32%
2.46 Express the composition of magnesite in mole %.
Compound
Weight %
MgCO3
SiO2
H2O
81
14
5
Compound
Weight %
Molecular weight
moles
mole %
MgCO3
SiO2
H2O
81
14
5
84
60
18
0.964
0.233
0.278
65.34
15.82
18.83
2.47 The concentration of H3PO4 is expressed in terms of P2O5 content. If
35% P2O5 is reported, find the composition of H3PO4 by weight.
P2O5 + 3H2O ® 2H3PO4
(142)
(54)
(98)
i.e. 142 kg of P2O5 º 196 kg of H3PO4
Therefore, 35 kg of P2O5 º 35 ×
196
= 48.3 H3PO4
142
i.e. H3PO4 is 48.3%
2.48 Ten kg of PbS and 3 kg of oxygen react to yield 6 kg of Pb and 1 kg
of PbO2 according to the reaction shown below:
PbS + O2 ® Pb + SO2
PbS + 2O2 ® PbO2 + SO2
(1)
(2)
Estimate (i) unreacted PbS, (ii) % excess oxygen supplied, (iii) total
SO2 formed, and (iv) the % conversion of PbS to Pb.
PbS + O2 ® Pb + SO2
(239.2)
(32)
(207.2)
(1)
(64)
PbS + 2O2 ® PbO2 + SO2
(239.2)
(64)
(239.2)
(2)
(64)
207.2 kg of Pb comes from 239.2 kg of PbS [from stoichiometry Eq. (1)]
MASS RELATIONS
31
6
= 6.927 kg of PbS
207.2
239.2 kg of PbO2 comes from 239.2 kg of PbS
Therefore, 6 kg of Pb comes from 239.2 ×
Therefore, 1 kg of PbO2 comes from 1 kg of PbS
Therefore, total PbS reacted [from Eqs. (1) and (2)]
= 6.927 + 1 = 7.927 kg
Unreacted PbS = 10 – 7.927 = 2.073 kg
O2 required for this process
From Reaction 1:
32 kg of oxygen is needed to produce 207.2 kg of Pb
Therefore, to produce 6 kg of PbO2, oxygen needed = 6 ×
32
= 0.927 kg
207.2
From Reaction 2:
239.2 kg of PbO2 requires 64 kg of oxygen
Therefore, to produce 1 kg of PbO2, oxygen required is 268 kg
Therefore, total oxygen used = 0.927 + 0.268 = 1.195 kg
© (3 1.195) ¸
¹ × 100 = 151%
« 1.195 º
Percentage excess O2 supplied = ª
Amount of SO2 formed
If 207.2 kg of Pb is formed, SO2 formed is 64 kg
6
= 1.853 kg
207.2
If 239.2 kg of PbO2 is formed, SO2 formed is 64 kg
If 6 kg of Pb is formed, SO2 formed is 64 ×
64
= 0.268 kg
239.2
Total SO2 formed = 1.853 + 0.268 = 2.121 kg
If 1 kg of PbO2 is formed, SO2 formed is
% conversion of PbS fed to Pb = Mass of PbS converted to Pb/Total
mass of PbS
100
= 69.27%
10
2.49 The composition of a liquid mixture containing A, B and C is
peculiarly given as 11 kg of A, 0.5 kmole of B and 10 wt of % C.
The molecular weights of A, B and C are 40, 50 and 60 respectively
and their densities are 0.75 g/cc, 0.8 g/cc and 0.9 g/cc respectively.
Express the composition in weight %, mole %. Also give its average
molecular weight and density assuming ideal behaviours.
= 6.927 ×
32
PROCESS CALCULATIONS
Let the weight of mixture be W kg
Weight of A = 11 kg
Weight of C (10%) = 0.1W kg
Weight of B = W – 11 – 0.1W = 0.5 kmole
= 0.5 × 50 = 25 kg
i.e. weight of B = W – 11 – 0.1W = 25 kg
0.9W = 36 kg
W = 40 kg
i.e. total weight of mixture is 40 kg.
Component
Weight, Weight, Molecular
kg
%
weight
A
B
C
11
25
4
27.5
62.5
10.0
40
50
60
Total
40
100.00
Average density = Mass/Volume =
moles,
kmole
mole
%
Density, Volume,
kg/m3
m3
0.275
0.500
0.067
32.66
59.38
7.96
0.842
100.00
750
800
900
0.0147
0.0313
0.0044
0.0504
40
= 793.65 kg/m3
0.0504
Average molecular weight = Weight/Total moles =
40
= 47.5
0.842
(Check: Average molecular weight
= 40 × 0.3266 + 50 × 0.5938 + 60 × 0.0796 = 47.5)
EXERCISES
2.1
How many g moles are equivalent to 1.0 kg of hydrogen?
2.2
How many kilograms of charcoal is required to reduce 3 kg of arsenic
trioxide?
As2O3 + 3C Æ 3CO + 2As
2.3
Oxygen is prepared according to the following equation:
2KClO3 Æ 2KCl + 3O2. What is the yield of oxygen when 9.12 g
of potassium chlorate is decomposed? How many grams of potassium
chlorate must be decomposed to get 5 g of oxygen?
2.4
An aqueous solution of sodium chloride contains 28 g of NaCl per
100 cc of solution at 293 K. Express the composition in (a) percentage
NaCl by weight (b) mole fraction of NaCl and (c) molality. Density of
solution is 1.17 g/cc.
MASS RELATIONS
33
2.5
An aqueous solution has 20% sodium carbonate by weight. Express
the composition by mole ratio and mole percent.
2.6
A solution of caustic soda in water contains 20% NaOH by weight.
The density of the solution is 1196 kg/m3. Find the molarity,
normality and molality of the solution.
2.7
A saturated solution of salicylic acid in methanol contains 64 kg
salicylic acid per 100 kg methanol at 298 K. Find the composition by
weight % and volume %.
2.8
A solution of sodium chloride in water contains 270 g per litre at
323 K. The density of this solution is 1.16 g/cc. Estimate the
composition by weight %, volume %, mole %, atomic %, molality and
kg of salt per kg of water.
2.9
A mixture of gases has the following composition by weight at 298 K
and 740 mm Hg.
Chlorine: 60%, Bromine: 25% and Nitrogen: 15%.
Express the composition by mole % and determine the average
molecular weight.
2.10 Wine making involves a series of very complex reactions most of
which are performed by microorganisms. The initial concentration of
sugar determines the final alcohol content and sweetness of the wine.
The general convention is to adjust the specific gravity of the starting
stock to achieve a desired quality of wine. The starting solution has a
specific gravity of 1.075 and contains 12.7 weight % of sugar. If all
the sugar is assumed to be C12H22O11, determine
(a) kg sugar/kg H2O
(b) kg solution/m3 solution
(c) g sugar/litre solution
2.11 The synthesis of ammonia proceeds according to the following reaction
N2 + 3H2 ® 2NH3
In a given plant, 4202 lb of nitrogen and 1046 lb of hydrogen are fed
to the synthesis reactor per hour. Production of pure ammonia from
this reactor is 3060 lb/h.
(a) What is the limiting reactant?
(b) What is the percent excess reactant?
(c) What is the percent conversion obtained (based on the limiting
reactant)?
2.12 How many grams of chromic sulphide will be formed from 0.718 g of
chromic oxide according to the following equation?
2Cr2O3 + 3CS2 ® 2Cr2S3 + 3CO2
34
PROCESS CALCULATIONS
2.13 How many kilograms of silver nitrate are there in 55.0 g mole silver
nitrate?
2.14 Phosphoric acid is used in the manufacture of fertilizers and as a
flavouring agent in drinks. For a given 10 weight % phosphoric acid
solution of specific gravity 1.10, determine:
(a) the mole fraction composition of this mixture.
(b) the volume of this solution, which would contain 1 g mole H3PO4.
2.15 Hydrogen gas in the laboratory can be prepared by the reaction of
sulphuric acid with zinc metal
H2SO4 (l) + Zn(s) ® ZnSO4(s) + H2 (g)
How many grams of sulphuric acid solution (97%) must act on an
excess of zinc to produce 12.0 m3/h of hydrogen at standard
conditions. Assume all the acid used reacts completely.
2.16 Sulphur dioxide may be produced by the reaction
Cu + 2 H2SO4 ® CuSO4 + 2H2O + SO2.
Find how much copper and how much 94% sulphuric acid must be
used to obtain 32 kg of SO2.
2.17 Aluminium sulphate is produced by reacting crushed bauxite ore with
sulphuric acid as shown below:
Al2O3 + 3 H2SO4 ® Al2 (SO4)3 + 3 H2O
Bauxite ore contains 55.4% by weight Al2O3, the reminder being
impurities. The sulphuric acid contains 77.7% H2SO4, the rest being
water. To produce crude aluminium sulphate containing 1798 kg of
pure Al2(SO4)3, 1080 kg of bauxite ore and 2510 kg of sulphuric acid
solution are used. Find (a) the excess reactant, (b) % of excess
reactant consumed, and (c) degree of completion of the reaction.
2.18 600 kg of sodium chloride is mixed with 200 kg of KCl. Find the
composition in weight % and mole %.
2.19 What is the weight of iron and water required to produce 100 kg of
hydrogen.
2.20 Cracked gas from petroleum refinery has the following composition
by volume:
Methane: 42%, ethane: 13%, ethylene: 25%, propane: 6%, propylene:
9%, and rest n-butane. Find: (a) average molecular weight of mixture,
(b) Composition by weight, and (c) specific gravity of the gas
mixture.
2.21 A gas contains methane: 45% and carbon dioxide: 45% and rest
nitrogen. Express (i) the weight %, (ii) average molecular weight, and
(iii) density of the gas at NTP.
3
Ideal Gases
3.1
RELATION BETWEEN MASS AND VOLUME
FOR GASEOUS SUBSTANCES
3.1.1 Standard Conditions
1 atm. pressure or 760 mm Hg or 29.92 inches of Hg and 0 °C or 32 °F
By Avogadro’s Hypothesis,
1 g mole of any gas under standard conditions will occupy 22.414 litres
1 lb mole of any gas under standard conditions will occupy 359 cu.ft.
T(K) = T °C + 273.16
T(°R) = T °F + 459.69
3.1.2 Ideal Gas Law
The ideal gas law states that,
PV = nRT
P = Pressure of gas
V = Volume of n moles of gas
n = Number of moles of gas
R = Gas constant
T = Absolute temperature
Using the ideal gas law (PV = nRT) and the above information one can always
determine the weight of a gas if the volume is known and vice-versa.
Normal Temperature and Pressure/Standard Conditions
Parameters
Pressure
Molar volume
Absolute
temperature
Gas constant
English
Metric
1 atm
359 ft3/lb mole
2
1.033 kgf/cm
22.414 m3/kmole
1.01325 bar
22.414 m3/kmole
491.69 oR
0.73 atm ft3/lb mole oR
273.16 K
0.085 kgf m3/kmole K
273.16 K
0.083 Bar m3/kmole K
35
SI
36
PROCESS CALCULATIONS
Pressure
1 atm = 1.033 kgf/cm2 = 1.01325 bar
1 atm = 14.67 psia = 105 N/m2 = 760 mm Hg
1 atm = 760 Torr = 29.92 inches of Hg = 76 cm Hg
PV
R =
= 82.06 atm.cc/g mole K = 0.73 atm ft3/lb mole °R
nT
R = 10.73 lbf ft3/in2 lb mole °R
Avogadro’s Number: 6.023 ´ 1023 molecules per g mole
2.73 ´ 1026 molecules per lb mole
6.023 ´ 1026 molecules per kg mole
Different units are used to express pressure like atmosphere, mm of
Hg, psia, kg/cm2, bar, N/m2, and Pa. Similarly, volume is expressed in cm3,
m3, litre, ft3 and gallon. The temperature is expressed in °C, °F, K and °R.
However, the temperature used in the application of Ideal gas law is in terms
of K or °R.
Thus, the gas constant is a dimensional quantity. The following table
gives the gas constant in different units.
Temperature
R
R
R
K
K
K
K
Pressure
Volume
Gas constant
‘Rg’
Units of gas
constant
in3
ft3
ft3
m3
m3
cm3
cm3
18.51
10.73
0.73
8314
0.08206
82.06
6239.79
in3psia/lb mole R
ft3psia/lb mole R
ft3atm/lb mole R
m3Pa/kmole K
m3atm/kmole K
cm3atm/g mole K
(cm3cm Hg)g mole K
psia
psia
atmospheres
Pa
atmospheres
atmospheres
cm Hg
Rg = 8.314 J/(g mole) (K)
= 1545 ft lb/lb mole °R
3.2 GASEOUS MIXTURE
3.2.1 Partial Pressure (PP)
The partial pressure of a component gas that is present in a mixture of gases
is the pressure that would be exerted by that component gas if it alone were
present in the same volume and at the same temperature as the mixture.
3.2.2 Pure Component Volume (PCV)
The PCV of a component gas that is present in a mixture of gases is the
volume that would be occupied by that component gas if it alone were
present at the same pressure and temperature as the mixture.
IDEAL GASES
37
Components
A
B
C
Sum of the quantities
Partial pressure
Number of moles
Pure component volume
pA
nA
VA
pB
nB
VB
pC
nC
VC
P
n
V
3.2.3 Dalton’s Law
The total pressure (P) exerted by a gaseous mixture in a definite volume is
equal to the sum of partial pressures.
pA + pB + p C = P
where pA, pB, pC, … represent partial pressure of components A, B, C, … .
3.2.4
Amagat’s Law (or) Leduc’s Law
The total volume (V ) occupied by a gaseous mixture is equal to the sum of
the pure component volumes
VA + V B + V C = V
VA, VB, VC, … stand for pure component volume of components A, B,
C, …
Where ideal gas law is applicable;
(a)
n RT
n RT
nA RT
; pB = B
; pC = C
V
V
V
Adding all the partial pressures of A, B and C, we have,
pA =
Ë RT Û
P = pA + p B + p C = Ì
Ü ´ (nA + nB + nC)
ÍV Ý
nA
RT
V
×
×
;
V
RT (n A + nB + nC )
pressure fraction = mole fraction.
Dividing, pA /P =
(b)
PVA = nART; PVB = nBRT; PVC = nCRT
Adding P(VA + VB + VC) = RT (nA + nB + nC) = nRT = PV
Dividing,
or,
n A RTP
VA
= NA
=
P(n A + nB + nC ) RT
V
VA = NA × V
38
3.3
PROCESS CALCULATIONS
AVERAGE MOLECULAR WEIGHT
The weight of unit mole of the mixture is called average molecular weight,
which is also equal to total weight of the gas mixture divided by the total
number of moles in the mixture. This is applicable only for gaseous
mixtures and not for solid or liquid mixtures. For example, air contains 79%
nitrogen and 21% oxygen by volume.
Basis: 100 kmole
Number of moles of nitrogen = 79
and those of oxygen
= 21
Weight of a component = Number of moles ¥ respective molecular
weight
\ Weight of nitrogen = 79 ¥ 28 = 2212 kg
Weight of oxygen = 21 ¥ 32 = 672 kg
2884 kg
\ The weight of 1 kmole = 2884/100 = 28.84 kg
Hence, the average molecular weight of air = 28.84
3.4
DENSITY OF MIXTURE
Density is defined as the weight of a mixture per unit volume and is
independent of temperature. As the volume of liquids and gases is a strong
function of temperature, density also varies significantly with temperature
for a specified composition. However, in the case of solids the variation of
density with temperature is not very significant.
WORKED EXAMPLES
3.1
Calculate the volume of 15 kg of Chlorine at a pressure of 0.9 bar and
293 K.
Basis: 15 kg Cl2 = 15/71.0 = 0.2113 kmole
Its volume is 0.2113 ¥ 22.414 ¥
3.2
1.0
293
¥
= 5.643 m3
0.9
273
Calculate the volume occupied by 6 lb of chlorine at 743 mm Hg and 70 °F
Basis: 6 lb of Cl2 ∫ 6/71 = 0.0845 lb mole of chlorine
Volume at standard condition = (0.0845 ¥ 359) = 30.34 ft3
Volume at given condition
ÊPV ˆ ÊT ˆ
= Á 0 0˜ ¥Á 1˜
Ë T0 ¯ Ë P1 ¯
IDEAL GASES
39
760 Ø È 530 Ø
È
= É 30.34 –
Ù –É
Ù
Ê
743 Ú Ê 492 Ú
= 33.4 ft3
3.3
Calculate the weight of 200 cu.ft. of water vapour at 15.5 mm Hg and
23 °C
Basis: 200 ft3 of gas at given condition
È PV Ø È T Ø
Volume at standard condition = É 1 1 Ù – É 0 Ù
Ê P0 Ú Ê T1 Ú
15.5 Ø 273
È
= É 200 –
= 3.76 ft3
ٖ
Ê
760 Ú 296
3.4
Moles of water
È 3.76 Ø
= É
= 0.01047 lb mole
Ê 359 ÙÚ
Weight of water
= (0.01047 ´ 18) = 0.18846 lb
It is desired to compress 30 lb of CO2 to a volume of 20 ft3 at 30 oC.
Find the pressure of the gas stored (required)?
0.6818
È 30 Ø
Basis: 30 lb CO2 = É Ù = 0.6818 lb mole =
2.2046
Ê 44 Ú
= 0.3102 kmole
Volume at standard condition
= 0.3102 ´ 22.414 = 6.9528 m3
Volume at given condition
= 20 ´ 0.02832 = 0.5664 m3
ÈPV Ø ÈT Ø
Pressure at the given condition “P1” = É 0 0 Ù – É 1 Ù
Ê V1 Ú Ê T0 Ú
6.9528 Ø È 303 Ø
È
= É1 –
Ù –É
Ù
Ê
0.5664 Ú Ê 273 Ú
= 13.62 atm
3.5
Calculate the maximum temperature to which 10 lb of nitrogen enclosed
in a 30 ft3 chamber may be heated without exceeding 100 psi pressure.
Basis: 10 lb of N2 = 10/28 = 0.357 lb mole
Volume at standard condition = 0.357 ´ 359 = 128.21 ft3
\
Temperature ‘T1’
ÈT V Ø È P Ø
= É 0 1Ù – É 1 Ù
Ê V0 Ú Ê P0 Ú
30 Ø È 100 Ø
= ÉÈ 273 –
Ù –É
Ù
Ê
128.21Ú Ê 14.67 Ú
= 435.4 K = 162.4 °C
40
3.6
PROCESS CALCULATIONS
When heated to 100 °C and 720 mm Hg, 17.2 g of N2O4 gas occupies
a volume of 11,450 cc. Assuming that the ideal gas law applies,
calculate the percentage dissociation of N2O4 to NO2?
N 2 O 4 → 2NO 2
(92)
(2 × 46)
N2O4 present initially = 17.2/92 = 0.187 g mole;
Let ‘x’ g mole of N2O4 dissociate,
then, ‘2x’ g moles of NO2 is formed.
Total moles after dissociation = (0.187 – x + 2x) = 0.187 + x
Given condition (1)
Standard condition (0)
Pressure
720 mm Hg
760 mm Hg
Temperature
373 K
273 K
Volume
11,450 cc
?
Parameter
È PV Ø È T Ø
Volume at standard condition = É 1 1 Ù – É 0 Ù
Ê P0 Ú Ê T1 Ú
720 Ø È 273 Ø
È
= É11450 –
Ù –É
Ù
Ê
760 Ú Ê 373 Ú
= 7939.2 cc
Therefore, no. of g moles remaining =
\
\
7939.2
= 0.354
22, 414
(0.187 + x) = 0.354
x = 0.167
È 0.167 Ø
Percentage dissociation = É
´ 100 = 89.42%
Ê 0.187 ÙÚ
3.7
Calculate the average molecular weight of a gas having the following
composition by volume. CO2: 13.1%, O2: 7.7% and N2: 79.2%
Basis: 1 g mole of the gas
Component
Volume % =
mole %
CO2
O2
N2
13.1
7.7
79.2
Total
100.0
Molecular
weight
g mole
Weight, g
44
32
28
0.131
0.077
0.792
0.131 ´ 44 = 5.764
0.077 ´ 32 = 2.464
0.792 ´ 28 = 22.176
1.000
30.404
Average molecular weight is 30.404.
IDEAL GASES
3.8
41
Calculate the density in lb/ft3 at 29.0 inches of Hg and 30 °C for a
mixture of hydrogen and oxygen that contains 11.1% of hydrogen by
weight.
Basis: 1 lb of gas mixture
Component
Weight %
Molecular weight
lb mole
Hydrogen
Oxygen
0.111
0.889
2
32
0.111/2 = 0.0555
0.889/32 = 0.0278
Total
0.0833 lb moles
Temperature = 30 °C = 86° F = 546° R
Volume of the gas at the given condition = 0.0833 ´ 359 ´ (29.92/29)
´ (546/492) = 34.24 ft3
Density = (1/34.24) = 0.0292 lb/ft3
3.9
Calculate the density in g/litre at 70 °F and 741 mm Hg of air
Basis: 1 g mole of air
Component
Volume % = mole %
Molecular weight
Weight, g
Oxygen
Nitrogen
0.21
0.79
32
28
6.72
22.12
Total
28.84 g
È 530 Ø È 760 Ø
–
= 24.8 litres
Volume of air = 1 ´ 22.414 ´ É
Ê 492 ÙÚ ÉÊ 741 ÙÚ
È 28.84 Ø
Density of air = É
= 1.162 g/litre
Ê 24.8 ÙÚ
3.10 In 1000 ft3 of a mixture of hydrogen, nitrogen and carbon-dioxide at
250 °F, the partial pressures are 0.26, 0.32 and 1.31 atm. Assuming
‘Ideal Gas’ behaviour, find the following:
(a) lb moles of H2; (b) mole fraction and mole % H2; (c) pressure
fraction of H2 (d) partial volume of H2; (e) volume fraction and
volume % of H2; (f) weight of H2; (g) weight fraction and weight %
of H2; (h) average molecular weight; (i) density of gas mixture;
(j) density at standard condition
Also, show that volume % = pressure % = mole %
Basis: 1000 ft3 of gas mixture
(a) Partial pressure of hydrogen = 0.26 atm V = 1000 ft3, T = 710 °R
42
PROCESS CALCULATIONS
0.26 492
–
1.00 710
= 180.169 ft3
Volume of H2 at standard condition = 1000 ´
È 180.169 Ø
lb moles of H2 = É
= 0.502 lb moles
Ê 359 ÙÚ
(b) Total pressure = (0.26 + 0.32 + 1.31) = 1.89 atm
È 1000 Ø È 1.89 Ø È 492 Ø
Total moles = É
= 3.648 lb moles
–
–
Ê 359 ÚÙ ÊÉ 1.00 ÚÙ ÊÉ 710 ÚÙ
mole fraction of hydrogen =
0.502
= 0.138
3.648
È 0.26 Ø
= 0.138
(c) Pressure fraction of hydrogen = É
Ê 1.89 ÙÚ
(d) Partial volume of hydrogen is the volume occupied by 0.502 lb
moles of it at 1.89 atm and 710 °R
È 1 Ø È 710 Ø
Volume of H2 = 0.502 ´ 359 ´ É
= 138 ft3
–
Ê 1.89 ÚÙ ÊÉ 492 ÚÙ
138
= 0.138
1000
Thus volume % = pressure % = mole %
(e) Volume fraction of hydrogen =
(f) Weight of hydrogen = 0.502 ´ 2 = 1.004 lb
(g) Basis 100 lb moles of gas mixture
Mole fraction of nitrogen =
0.32
= 0.169
189
Mole fraction of carbon dioxide =
Mole fraction of hydrogen =
Component
Molecular
weight
mole %
1.31
= 0.693
1.89
0.26
= 0.138
1.89
lb mole
Weight, lb
Weight %
Hydrogen
2
Nitrogen
28
Carbon dioxide 44
13.8
16.9
69.3
13.8 13.8 ´ 2 = 27.6
16.9 16.9 ´ 28 = 473.2
69.3 69.3 ´ 44 = 3049.2
0.78
13.33
85.89
Total
100.0
100.0
100.00
3550.0
IDEAL GASES
43
È 3550 Ø
(h) Average molecular weight = É
= 35.5
Ê 100 ÙÚ
È 492 Ø È 1.89 Ø
–
(i) 1000 ft3 of gas at given condition º 1000 ´ É
Ê 710 ÙÚ ÉÊ 1.0 ÙÚ
= 1309.7 ft3 at NTP condition
Number of moles =
1309.7
= 3.648 lb moles º 3.648 ´ 35.5
359
= 129.55 lb.
Density at given condition = ÈÉ 129.55 ØÙ = 0.12955 lb/ft3
Ê 1000 Ú
( j) Density at standard condition:
È 1.89 Ø È 492 Ø
–
Volume at standard condition = 1000 ´ É
Ê 1.00 ÙÚ ÉÊ 710 ÙÚ
\
= 1309.7 ft3
\
È 129.55 Ø
Density = É
= 0.09892 lb/ft3
Ê 1309.7 ÙÚ
3.11 A certain gaseous hydrocarbon is known to contain less than 5 carbon
atoms. This compound is burnt with exactly the volume of oxygen
required for complete combustion. The volume of reactants (all gases)
is 600 ml and the volume of products (all gases) under the same
condition is 700 ml. What is the compound?
Basis: 1 mole of hydrocarbon.
Let the hydrocarbon be Cx Hy
y
y
CxHy + ÈÉ x ØÙ O2 ® xCO2 + ÈÉ ØÙ H2O
Ê
Ê 2Ú
4Ú
Moles:
1
yØ
È
ÉÊ x ÙÚ
4
®
yØ
È
Total moles of reactants = É1 x Ù
Ê
4Ú
yØ
È
Total moles of products = É x Ù
Ê
2Ú
Reactants (1 + x + y /4) 600 6
=
=
=
Products
( x + y /2)
700 7
È 7y Ø
7 + 7x + É Ù = 6x + 3y
Ê 4Ú
x
y
2
44
PROCESS CALCULATIONS
È 7y
Ø
3 y Ù = –7
(7x – 6x) + É
Ê 4
Ú
5y
= –7
4
Since the hydrocarbon has carbon atoms less than 5, we have x < 5;
solving above equation assuming carbon atoms as 1, 2, up to 5 we get
the values of y as indicated below:
x–
x = 1;
–
5y
= –8;
4
y=
32
5
x = 2;
–
5y
= –9;
4
y=
36
5
x = 3;
–
5y
= –10;
4
y=
40
=8
5
5y
44
= –11
y=
4
5
The value of y is to be an integer.
x=4
–
Hence, the hydrocarbon is C3H8 : (Propane)
The combustion reaction is C3H8 + 5O2 ® 3CO2 + 4H2O
3.12 Combustion gases having the following molal composition are passed
into an evaporator at 200 °C and 743 mm Hg (N2 : 79.2%, O2 : 7.2%,
CO2 : 13.6 %) Water is added to the stream as vapour and the gases
leave at 85 °C and 740 mm Hg with the following composition
N2 : 48.3%, O2 : 4.4% and H2O : 39%. Calculate (a) volume of gases
leaving evaporator per 100 litres of gas entering and (b) weight of
water added per 100 litres of gas entering.
(N2, O2, CO2) ®
Evaporator ® (N2, O2, CO2) + H2O
­
Water
Basis: 100 g moles gas entering
È 473 Ø È 760 Ø
= 3972.31 litres
Volume = 100 ´ 22.414 ´ É
–
Ê 273 ÙÚ ÉÊ 743 ÙÚ
This 100 g moles of entering gas º 61% of gases leaving.
\
g moles of gases leaving = 100/0.61 = 164 g moles.
Water added = 164 – 100 = 64 g moles = 1152 g.
760 Ø È 358 Ø
Volume of gas leaving = 164 ´ 22.414 ´ ÈÉ
–
Ê 740 ÙÚ ÉÊ 273 ÙÚ
= 4950.7 litres
IDEAL GASES
(a)
45
Volume of gas leaving 4950.0 t 100
100 litres gas-entering
3972.31
= 124.6 litres
(b)
Volume of water added 1152.0 t 100
29 g.
100 litres gas-entering
3972.31
3.13 How many g moles of nitrogen will occupy 1000 m3 at 112 ´ 103
N/m2 and 400 K
PV = nRT, where R = 8.314 J/g mole K = 8.314 (Pa) (m3)/g mole K
n=
PV
112 – 103 – 1000
=
= 33,678 g moles
RT
8.314 – 400
Alternatively,
P1 = 1.01325 ´ 105 N/m2
T1 = 273 K
since
P1V1 P2V2
=
T1
T2
V1 (at NTP) =
112 t 10 3 t 1000
273
= 754.4 m3
t
5
400
1.01325 t 10
1000 g moles of gas occupies 22.414 m3 at NTP,
1000 Ø
È
Hence, 754.4 m3 contains É 754.4 –
Ù = 33,658 g moles of nitrogen
Ê
22.414 Ú
(Error is due to the fact that T1 is taken as 273 K and not as 273.16 K
on taking T1 as 273.16 K the g moles of nitrogen will be 33,677)
3.14 In the manufacture of hydrochloric acid, a gas is obtained that
contains 25% HCl and 75% air by volume. This gas is passed through
an absorption system in which 98% of the HCl is removed. The gas
enters the system at 48.8 °C and 743 mm Hg and leaves at 26.7 °C
and 738 mm Hg. Applying the pure component volume method,
calculate:
(a) The volume of gas leaving per 1000 litres entering the absorption
column.
(b) The weight of HCl removed per 100 litres entering.
Basis: In 100 litres of entering gas volume of air = 75 litres
Pure component volume of HCl = 25 litres
Pure component volume of HCl absorbed = (25 ´ 0.98) = 24.5 litres
46
PROCESS CALCULATIONS
Pure component volume of HCl remaining = 0.5 litres
Volume of gas leaving = 75 + 0.5 = 75.5 litres
(a) Parameter
Entering condition
Leaving condition
Pressure
743 mm Hg
738 mm Hg
Temperature
48.8 °C
26.7 °C
Volume
75.5 litres
?
Volume of (entering) gas at leaving condition
= 75.5 ´
(b) Composition:
743 299.7
= 70.8 lit.
–
738 321.8
Component
Litre
Volume % (or) mole %
HCl
0.5
0.66
Air
75.0
99.34
Total:
75.5
100.00
Volume of HCl absorbed at standard condition
= 24.5 ´
743 492
= 20.3 litres
–
760 580
20.3
´ 36.5 = 33.057 g.
22.414
3.15 Absorbing chlorine in milk of lime produces calcium hypochlorite. A
gas produced by the Deacon process enters the absorption apparatus at
740 mm Hg and 75 °F. The partial pressure of Cl2 is 59 mm Hg and
the remainder being inert gas. The gas leaves at 80 °F and 743 mm Hg
with Cl2 having a partial pressure of 0.5 mm Hg. Calculate, by
applying the partial pressure method:
Weight of HCl absorbed =
(a) volume of gas leaving per 100 litres entering
(b) weight of Cl2 absorbed.
Basis: 100 litres of gases entering
Partial pressure of inert gas entering = 740 – 59 = 681 mm Hg
Partial pressure of inert gas leaving = 743 – 0.5 = 742.5 mm Hg
Volume of inert gases entering = 100 litres (681 mm Hg)
681
299.7
–
742.5
297
= 92.5 litres
Volume of inert gas leaving = 100 ´
IDEAL GASES
47
(a) Volume of gas leaving = 92.5 litres (743 mm Hg)
Volumes of Cl2 entering and leaving are 100 litres and 92.5 litres
Entering condition:
Parameter
Given condition (1)
Standard condition (0)
Pressure
Temperature
Volume
59 mm Hg
297 K
100 litres
760 mm Hg
273 K
?
Volume at standard condition of Cl2 entering
59 273
= 7.14 litres
–
760 297
Volume at standard condition of Cl2 leaving
= 100 ´
0.5
273
= 0.055 litre
–
760 299.7
Volume at standard condition of Cl2 absorbed
= 7.14 – 0.055 = 7.085 litres
= 92.5 ´
7.285
´ 71 = 22.45 g.
22.414
3.16 Nitric acid is produced by the oxidation of ammonia with air. In the
first step of the process, ammonia and air are mixed together and
passed over a catalyst at 700 °C. The following reaction takes place.
4NH3 + 5O2 ® 6H2O + 4NO. The gases from this process are passed
into towers where they are cooled and the oxidation is completed
according to the reactions:
(b) Weight of Cl2 absorbed =
2NO + O2 ® 2NO2
3NO2 + H2O ® 2HNO3 + NO
The NO liberated is re-oxidized in part and forms more nitric acid in
successive repetitions of the above reactions. The ammonia and the air
enter the process at 20 °C and 755 mm Hg. The air is present in such
proportion that the oxygen will be 20% in excess of that required
for complete oxidation of ammonia to nitric acid. The gases leave
the catalyst at 700 °C and 743 mm Hg. Given the overall reaction:
2NO + 1.5O2 + H2O ® 2HNO3, calculate the following:
(a) The volume of air to be used per 100 litres of NH3 entering the process
(b) The composition of gases entering the catalyzer
(c) The composition of gases leaving (assuming the reaction in
catalyzer is 85%)
48
PROCESS CALCULATIONS
(d) The volume of gases leaving the catalyzer for 100 litres ammonia
entering
(e) Weight of acid produced per 100 litres NH3, assuming 90% of the
nitric oxide entering the tower is oxidized to acid.
NH3
Air
Exit gas
Catalyzer
Absorber
HNO3
Basis: 1 g mole of NH3
overall reaction is NH3 + 2O2 ® HNO3 + H2O
O2 required is 2 g moles
But O2 supplied is 2 ´ 1.2 = 2.4 g moles
i.e. air supplied is 2.4 = 11.42 g moles
0.21
Thus, N2 = 9.02 g moles.
(a) Volume of air = 11.42 ´ 22.414 ´
293 760
–
= 276.4 litres
273 755
Volume of NH3 = 1 ´ 22.414 ´
293 760
= 24.2 litres
–
273 755
Volume of air/100 litres of NH3 =
276.4 – 100
= 1142.2 litres
24.2
(b)
Component
g mole
mole % = volume %
NH3
1.00
8.0
O2
2.40
19.3
N2
9.02
72.7
Total
12.42
100.0
(c) Gases leaving catalyzer are nitrogen, oxygen, ammonia, nitric
oxide and water:
N2 (all that enters) = 9.02 g moles
NH3 (85% conversion) = (1 – 0.85) = 0.15 g mole
5Ø
È
O2 consumed = É 0.85 – Ù = 1.06 g moles
Ê
4Ú
IDEAL GASES
49
O2 leaving = (2.4 – 1.06) = 1.34 g moles
NO formed = 0.85 g mole
6ˆ
Ê
H2O formed = Á 0.85 ¥ ˜ = 1.275 g moles
Ë
4¯
Component
N2
O2
NH3
NO
H2O
Total
g moles
9.02
1.34
0.15
0.85
1.275
12.635
mole % = volume % 71.40 10.60
1.20
6.70
10.10
100.000
(d) Volume of gases leaving the catalyzer
973 760
= 1031.8 litres
¥
273 243
This is the volume of gases leaving for 24.2 litres of ammonia
entering
= 12.635 ¥ 22.414 ¥
Therefore, for 100 litres of NH3 entering
= (1031.8 ¥ 100)/24.2 = 4264 litres of gas leaves
(e) NO entering the tower = 0.85 g mole
NO converted = 0.85 ¥ 0.9 = 0.765 g mole
HNO3 produced = 0.765 g mole = (0.765 ¥ 63) = 48 g
For 100 litres of NH3 weight of HNO3 produced
48 ¥ 100
= 199 g
24.2
3.17 1000 kg/h of an organic ester C19H36O2 is hydrogenated to C19H38O2
in a process. The company purchases its H2 in cylinders of 1 m3
capacity. The pressure in cylinder is initially 10 kg/cm2 (abs.) and
drops to 2 kg/cm2 (abs.) after use. The company works 24 h/day,
7 days a week. How many cylinders are needed per week?
=
Basis: Ester being processed in one week = (1000 ¥ 24 ¥ 7)
= 1,68,000 kg
C19H36O2 + H2 Æ C19H38O2
296
2
298
296 kg of ester reacts with 2 kg of H2
168000 kg of ester reacts with
2 ¥ 168000
= 1135 kg of hydrogen
296
1 kmole of any gas occupies 22.414 m3 at NTP
50
PROCESS CALCULATIONS
È 10 Ø È 1 Ø
H2 initially present in the cylinder (10 m3) = (1) ´ É Ù ´ É
Ù
Ê 1 Ú Ê 22.414 Ú
= 0.45 kmole
(by reducing to NTP condition)
È 1 Ø È 2Ø
= 0.09 kmole
H2 after use = É
–
Ê 22.414 ÙÚ ÉÊ 1 ÙÚ
\
H2 available from one cylinder = 0.36 kmole
1135 Ø
= 567.5 kmoles
H2 needed = 1135 kg = ÈÉ
Ê 2 ÙÚ
The number of cylinders required should be full number, rounded to
next highest value
\
È 567.5 Ø
= 1577
Cylinders required per week = É
Ê 0.36 ÙÚ
3.18 A mixture of toluene and air is passed through a cooler where some
toluene is condensed. 1000 ft3 of gases enter the cooler per hour at
100 °C and 100 mm Hg (gauge). The partial pressure of toluene is
300 mm Hg. 740 ft3 of gases leave cooler per hour at 40 mm Hg and
50 °C. Calculate the weight of toluene condensed per hour. Vapour
pressure of toluene at 50 °C = 90 mm Hg.
Basis: One hour
100 mm Hg (gauge) = 860 mm Hg (abs.)
300 Ø
È
3
Toluene entering = É1000 –
Ù = 348.8 ft
Ê
860 Ú
È 348.8 Ø È 860 Ø È 273 Ø
Weight = É
´ 92 = 74 lb.
–
–
Ê 359 ÙÚ ÉÊ 760 ÙÚ ÉÊ 373 ÙÚ
At exit, air is saturated with toluene.
È 90 Ø
Toluene leaving = 740 ´ É
= 90 ft3
Ê 740 ÙÚ
È 90 Ø È 740 Ø È 273 Ø
–
–
Weight = É
´ 92 = 19 lb.
Ê 359 ÙÚ ÉÊ 760 ÙÚ ÉÊ 323 ÙÚ
\
Weight of toluene condensed = 74 – 19 = 55 lb.
3.19 Air is dried from a partial pressure of 50 mm of water vapour to a
partial pressure of 10 mm. The temperature of entering air is 500 °F
and the pressure remains constant at 760 mm Hg. How much water is
removed per 1000 ft3 of entering air?
51
IDEAL GASES
Basis: 1000 ft3 of entering air.
È 1000 Ø È 492 Ø
= 1.43 lb moles
Moles of air entering = É
–
Ê 359 ÙÚ ÉÊ 960 ÙÚ
50 Ø
È
Moles of water in it = É1.43 –
Ù = 0.094 lb mole
Ê
760 Ú
Moles of dry air = (1.43 – 0.094) = 1.336 lb moles
10 Ø
È
Moles of water leaving = É1.336 –
Ù = 0.0178 lb mole
Ê
750 Ú
Water condensed = (0.094 – 0.0178) = 0.0762 lb mole = 1.37 lb
3.20 Chimney gas has the following composition:
CO2 : 9.5%, CO : 0.2%, O2 : 9.6% and N2 : 80.7%. Using ideal gas
law, calculate:
(a) its weight percentage
(b) volume occupied by 0.5 kg of gas at 30 °C and 760 mm Hg.
(c) density of the gas in kg/m3 at condition of (b)
(d) specific gravity of the gas mixture.
(Density of air may be taken as 1.3 g/cc)
Basis: 100 kmoles of chimney gas
(a)
Component Mol. weight
CO2
CO
O2
N2
44
28
32
28
Total
(b) 0.5 kg of gas =
Weight, kmole
Weight, kg
Weight %
9.5
0.2
9.6
80.7
(9.5 ´ 44) = 418.0
(0.2 ´ 28) =
5.6
(9.6 ´ 32) = 307.2
(80.7 ´ 28) = 2259.6
13.978
0.187
10.273
75.562
100.0
2990.4
100.000
0.5
= 0.01672 kmole
29.904
Volume at 30 °C, 760 mm Hg = 0.01672 ´ ÈÉ 303 ØÙ ´ 22.414
Ê 273 Ú
3
= 0.416 m
È 0.5 Ø
(c) Density = É
= 1.202 kg/m3
Ê 0.416 ÙÚ
(d) Specific gravity = density of gas/density of air =
1.202
= 0.925
1.300
52
PROCESS CALCULATIONS
3.21 A producer gas has the following composition CO2 : 4.4%, CO : 23%,
O2 : 2.6% and N2 : 70%. Calculate the following:
(a) volume of air at 25 °C and 750 mm Hg required for the
combustion of 100 m3 of gas at the same condition if 25% excess
air is used
(b) the composition and volume of gases leaving the burner at 350 °C
and 750 mm Hg per 100 m3 of gas burnt.
Basis: 100 kmoles of producer gas.
CO = 23 kmoles
O2 needed = 23/2 = 11.5 kmole; O2 in feed = 2.6 kmoles
O2 actually needed = (11.5 – 2.6) = 8.9 kmoles
8.9 – 1.25
= 52.98 kmoles
0.21
O2 supplied = 52.98 ´ 0.21 = 11.125 kmoles
N2 in air = 52.98 ´ 0.79 = 41.85 kmoles
Air supplied =
È 298 Ø È 760 Ø
–
Volume of feed = 100 ´ 22.414 ´ É
Ê 273 ÙÚ ÉÊ 750 ÙÚ
= 2479.28 m3
È 298 Ø È 760 Ø
–
Volume of air = 52.98 ´ 22.414 ´ É
= 1313.43 m3
Ê 273 ÚÙ ÊÉ 750 ÚÙ
(a) Volume of air/100 m3 of feed =
1313.43 – 100
= 52.98 m3
2479.28
(b) (assuming complete combustion)
CO2 leaving the burner = (4.4 + 23) = 27.400 kmoles
O2 leaving = (8.9 ´ 0.25)
= 2.225 kmoles
= 111.850 kmoles
N2 leaving = (70 + 41.85)
141.475 kmoles
Composition of gases leaving (volume % = mole %):
CO2 : 19.37%, O2 : 1.57% and N2 : 79.06%
È 623 Ø È 760 Ø
Volume of gases leaving = 141.475 ´ 22.414 ´ É
–
Ê 273 ÙÚ ÉÊ 750 ÙÚ
= 7332.92 m3
Volume of gases leaving/100 m3 of feed =
7332.92 – 100
= 295.76 m3
2479.28
53
IDEAL GASES
3.22 Natural gas has the following composition: CH4 : 94.1%, C2H6 : 3%
and N2 : 2.9%. This gas is piped from the well at 80°F and 80 psi.
Calculate the following.
(a) Partial pressure of N2
(b) Pure component volume of N2 per 100 ft3 of the gas
(c) Density
Basis: 100 ft3 of the natural gas
80 – 2.9
= 2.32 psia.
100
(b) Pure component volume of N2 = 0.029 ´ 100 = 2.9 ft3
(a) Partial pressure of N2 =
80 Ø È 492 Ø
(c) Volume of gas at NTP = 100 ´ ÈÉ
= 497 ft3
–
Ê 14.67 ÙÚ ÉÊ 540 ÙÚ
No. of moles of gas =
497
= 1.38 lb moles.
359
lb mole
Weight, lb
Component
mole %
CH4
C2H6
N2
94.1
3.0
2.9
1.38 ´ 0.941 = 1.298
1.38 ´ 0.03 = 0.042
1.38 ´ 0.029 = 0.040
(1.298 ´ 16) = 20.777
(0.042 ´ 30) = 1.260
(0.040 ´ 28) = 1.120
Total
100.0
1.380
23.157
23.157 Ø
Density of gas = ÈÉ
= 0.23157 lb/ft3
Ê 100 ÙÚ
È 23.157 Ø
Density at standard condition = É
= 0.0466 lb/ft3
Ê 497 ÙÚ
3.23 Compare pressures given by the ideal gas and van der Waals equation
for 1 mole of CO2 occupying a volume of (381 ´ 10–6) m3 at 40 °C
RT
= 6.831 ´ 106 N/m2
nV
where, R = 8.314 N.m/g mole K (J/g mole K); T = 313 K;
(a) Ideal gas law P =
V = 381 ´ 10–6 m3
a Ø
È
(b) van der Waals equation É P 2 Ù (V – b) = nRT
Ê
V Ú
a = 0.3646; b = 4.28 ´ 10–5 rest same as above
Ë RT Û È a Ø
P = Ì
ÜÉ 2Ù
Í (V b) Ý Ê V Ú
54
PROCESS CALCULATIONS
P =
©
¸ ©
8.314 t 313
0.3646 ¸
ª
6
5 ¹ ª
6 2 ¹
4.28 t 10 º « (381 t 10 ) º
« 381 t 10
= 5.2 ´ 10–6 N/m2
3.24 It is desired to market O2 in small cylinders having volumes of 0.5 ft3
and exactly containing 1 lb of gas at 120 °F. What is the pressure?
Basis: 1 lb of O2 = 1/32 = 0.031 lb mole; R = 0.73 ft3 atm/lb mole °R
T = 120 + 460 = 580 °R
nRT
580
= 0.031 ´ 0.73 ´
= 26.5 atm
V
0.5
3.25 A tire is inflated to 35 psig at 0 °F. To what temperature it can be
heated up to a pressure of 50 psig, volume remaining same?
P=
P1 = 35 + 14.67 = 49.67 psia; P2 = 50 + 14.67 = 64.67 psia.
T1 = 460 °R; V1 = V2; n1 = n2; T2 = ?
( P1V1 )
T1
\
( P2V2 )
; T2
T2
( P2T1 )
P1
(64.67 – 460)
49.67
600 °R
Temperature = 140 °F
3.26 Calculate densities of C2H6 and air at NTP.
At NTP 1 g mole occupies 22,414 cc.
Density of C2H6 =
1 – 30
= 1.3 ´ 10–3 g/cc
22414
1 – 28.84
= 1.29 ´ 10–3 g/cc
22414
3.27 Acetylene gas is produced according to the reaction
Density of air =
CaC2 + 2H2O ® C2H2 + Ca(OH)2
Calculate the number of hours of service that can be got from 1 lb of
carbide in lamp burning 2 ft3 of gas/hour at 75 °F and 743 mm Hg.
1
= 0.0156 lb mole.
64
CaC2 + 2H2O ® C2H2 + Ca(OH)2
Basis: 1 lb of CaC2 =
64
36
26
74
È 535 Ø È 760 Ø
Volume of acetylene got = 0.0156 ´ 359 ´ É
= 6.24 ft3
–
Ê 492 ÙÚ ÉÊ 743 ÙÚ
È 6.24 Ø
Time of burning = É
= 3.12 h.
Ê 2 ÙÚ
IDEAL GASES
55
A similar problem has been solved in MKS system also:
Data: 1 kg CaC2; 25 °C; 740 mm Hg; for a lamp burning gas 40 litres h.
Basis: 1 kg CaC2 =
1
= 0.0156 kmole.
64
298 Ø È 760 Ø
= 392.6 litres.
Volume of C2H2 = 0.0156 ´ 22414 ´ ÈÉ
–
Ê 273 ÚÙ ÊÉ 740 ÚÙ
392.6 Ø
Time of burning = ÈÉ
= 9.8 h.
Ê 40 ÙÚ
3.28 By electrolyzing a mixed brine a mixture of gases is obtained at the
cathode having the following composition by weight: Cl2 : 67%,
Br2 : 28%, O2 : 5%. Calculate:
(a) composition by volume
(b) density at 25 °C and 740 mm Hg
(c) specific gravity of the gas mixture.
Basis: 100 g of gas mixture
(a)
Component
Molecular
weight
Weight, g
Weight,
g mole
Volume %
= mole %
Cl2
Br2
O2
71
160
32
67
28
5
(67/71) = 0.945
(28/160) = 0.175
(5/32) = 0.156
74.0
13.7
12.3
Total
100
1.276
100
È 298 Ø È 760 Ø
(b) Volume of gases = 1.276 ´ 22.414 ´ É
= 32 litres
–
Ê 273 ÚÙ ÊÉ 740 ÚÙ
È 100 Ø
= 3.12 g/litre
Density = É
Ê 32 ÙÚ
(c) Density of air = 1.293 g/litre (at standard condition)
Density of air at 25 °C and 740 mm Hg
È 273 Ø È 740 Ø
= 1.293 ´ É
= 1.15 g/litre
–
Ê 298 ÚÙ ÊÉ 760 ÚÙ
Alternatively, volume of air at 25 °C and 740 mm Hg
È 760 Ø È 298 Ø
= 22.414 ´ É
= 25.13 litres
–
Ê 740 ÚÙ ÊÉ 273 ÚÙ
56
PROCESS CALCULATIONS
28.84
= 1.15 g/litre
25.13
3.12
Specific gravity of gas mixture =
= 2.7
1.15
3.29 A mixture of NH3 and air at 730 mm Hg and 30 °C contains 5.1 NH3
by volume. This gas is passed at a rate of 100 ft3/min. through an
absorption tower in which NH3 is removed. The gases leave the tower
at 725 mm Hg and 20 °C having 0.05% NH3 by volume. Calculate:
Density of air =
(a) the rate of flow of gases leaving the tower and
(b) weight of NH3 absorbed.
Basis: 100 ft3 of entering gases.
Volume of NH3 = 5.1 ft3 and of air = 94.9 ft3 (730 mm Hg, 30 °C)
È 730 Ø È 293 Ø
Volume of air at exit condition = 94.9 ´ É
= 92.4 ft3
–
Ê 725 ÙÚ ÉÊ 303 ÙÚ
92.4 ft3 of air º 99.95% of exit gas.
(a) \
92.4 Ø
Volume of exit gas = ÈÉ
= 92.447 ft3/min.
Ê 0.9995 ÙÚ
Volume of NH3 in exit = 0.047 ft3
È 725 Ø È 303 Ø
Volume of exit NH3 at inlet condition = 0.047 ´ É
–
Ê 730 ÚÙ ÊÉ 293 ÚÙ
= 0.0482 ft3
(b) Volume of NH3 absorbed = (5.1 – 0.0482) = 5.0518 ft3
È 730 Ø È 273 Ø
Volume of NH3 absorbed at NTP = 5.0518 ´ É
–
Ê 760 ÚÙ ÊÉ 303 ÚÙ
= 4.372 ft3
4.372
= 0.012 lb mole = 0.207020 lb
359
3.30 1000 ft3 of moist air at 740 mm Hg and 30 °C contains water vapour
in such proportions that its partial pressure is 22 mm Hg. Without the
total pressure being changed, the temperature is reduced to 15 °C and
some water condenses. After that the partial pressure of water is 12.7
mm Hg. Using partial pressure method, find the following:
Moles of NH3 =
(a) Volume of gas after cooling and
(b) Weight of water condensed.
Basis: 1000 ft3 of moist air at 740 mm Hg and 30 °C
Partial pressure of water = 22 mm Hg
IDEAL GASES
57
Partial pressure of air = (740 – 22) = 718 mm Hg
Partial pressure of air after cooling = (740 – 12.7) = 727.3 mm Hg
È 718 Ø È 288 Ø
= 938 ft3
(a) Volume of air after cooling = 1000 ´ É
–
Ê 727.3 ÙÚ ÉÊ 303 ÙÚ
After cooling volume of gases = Volume of water vapour + Dry air
(740 mm, 15 °C)
(12.7 mm, 15 °C)
È 12.7 Ø È 303 Ø
Volume of air leaving at inlet condition = 938 ´ É
–
Ê 22 ÙÚ ÉÊ 288 ÙÚ
= 570 ft3
Volume of water vapour condensed = 1000 – 570 = 430 ft3
430 Ø È 22 Ø È 273 Ø
(b) Water condensed = ÈÉ
´ 18 = 0.562 lb
–
–
Ê 359 ÙÚ ÉÊ 760 ÙÚ ÉÊ 303 ÙÚ
3.31 A producer gas has the following composition by volume CO : 23%,
CO2 : 4.4%, O2 : 2.6% and N2 : 70%
(a) Calculate the ft3 of gas at 70 °F and 750 mm Hg per lb of carbon
present.
(b) Calculate the volume of air required for the combustion of 100 ft3
of the gas if 20% excess air is used.
(c) Calculate the volumetric composition of gases leaving assuming
complete combustion.
(d) Calculate the volume of gases leaving at 600 °F and 750 mm Hg
per 100 ft3 gas burnt.
Basis: 100 lb moles of gas
i.e. carbon in the feed = 23 + 4.4 = 27.4 atoms
530 Ø È 760 Ø
(a) Volume of gases = 100 ´ 359 ´ ÉÈ
= 39,200 ft3
–
Ê 492 ÙÚ ÉÊ 750 ÙÚ
Weight of carbon present = (27.4 ´ 12) = 328.8 lb.
È 39200 Ø
= 119.2 ft3/lb.
(Volume of gas/lb of carbon) = É
Ê 328.8 ÙÚ
23
(b) O2 required for the combustion of CO = ÈÉ ØÙ = 11.5 lb moles
Ê 2Ú
O2 available in feed = 2.6 lb moles
Theoretical O2 required = (11.5 – 2.6) = 8.9 lb moles
O2 supplied = (8.9 ´ 1.2) = 10.68 lb moles
58
PROCESS CALCULATIONS
È 100 Ø
Air supplied = 10.68 ´ É
= 50.85 lb moles
Ê 21 ÙÚ
È 530 Ø È 760 Ø
Volume of air = 50.85 ´ 359 ´ É
= 19,930 ft3
–
Ê 492 ÙÚ ÉÊ 750 ÙÚ
(70 oF, 750 mm Hg)
È 530 Ø È 760 Ø
= 39,200 ft3
Volume of feed = 100 ´ 359 ´ É
–
Ê 492 ÙÚ ÉÊ 750 ÙÚ
19930 – 100
= 50.85 ft3
39200
(c) CO2 leaving = (23 + 4.4)
= 27.40 lb moles
= 1.78 lb moles
O2 remaining = (10.68 – 8.9)
N2 leaving (from air) = (50.85 – 10.68) = 40.17 lb moles
= 70.00 lb moles
N2 entering along with feed
= 110.17 lb moles
N2 Total
Total
= 139.35 lb moles
Volume of air/100 ft3 of feed =
Composition of
CO2
O2
N2
Volume %
19.66
1.28
79.06
(d) Volume of gases leaving
È 760 Ø È 1060 Ø
= 1,09,218 ft3
= 139.35 ´ 359 ´ É
–
Ê 750 ÙÚ ÉÊ 492 ÙÚ
Volume of gases/100 ft3 of feed =
1,09,218 – 100
= 278.6 ft3
39,200
3.32 A furnace is to be designed to burn coke at the rate of 200 lb/h having
a composition C : 89.1% and ash : 10.9%. The grate efficiency of the
furnace is such that 90% of the carbon present in the coke charged is
burnt. Air supplied is 30% in excess of that required for complete
combustion. It may be assumed that 97% of the carbon burnt is
oxidized to carbon dioxide and the rest to carbon monoxide.
(a) Calculate the composition of the flue gases.
(b) If the flue gases leave the furnace at 550 °F and 743 mm Hg,
calculate the rate of flow of gases in ft3/min.
Basis: 100 lb of coke.
C + O2 ® CO2
C : 89.1 lb. Ash : 10.9 lb.
Carbon burnt = 89.1 ´ 0.9 = 80.19 lb
IDEAL GASES
59
Carbon burnt to CO2 = 80.19 ´ 0.97 = 77.78 lb = 6.48 lb moles
Carbon burnt to CO = 80.19 ´ 0.03 = 2.41 lb = 0.2 lb mole
È 32 Ø
È 308.88 Ø
O2 supplied = 89.1 ´ É Ù ´ 1.3 = 308.88 lb = É
Ê 12 Ú
Ê 32 ÙÚ
= 9.65 lb moles
È 100 Ø
Air supplied = 9.65 ´ É
= 45.96 lb moles
Ê 21 ÙÚ
N2 in air = (45.96 – 9.65) = 36.31 lb moles
È 6.48 0.2 Ø
O2 required = É
ÙÚ = 6.58 lb moles
Ê
2
Excess O2 leaving = (9.65 – 6.58) = 3.07 lb moles
(a)
Component
lb mole
mole %
CO2
6.48
14.07
CO
0.20
0.43
O2
3.07
6.67
N2
36.31
78.83
Total
46.06
100.00
È 760 Ø È 1010 Ø
(b) Volume of flue gases = 46.06 ´ 359 É
= 34,722 ft3
–
Ê 743 ÙÚ ÉÊ 492 ÙÚ
Hence, for 200 lb/h of coke charge, volume of flue gases is
= (34,722 ´ 2) = 69,444 ft3/h
i.e. Volumetric flow rate of flue gases = 1157.4 ft3/min.
3.33 In the fixation of nitrogen by the arc process, air is passed through a
magnetically flattened electric arc. Some of the nitrogen is oxidized to
NO, which on cooling, oxidizes to NO2. Of the NO2 formed,
66% will be associated to N2O4 at 26 °C. The gases are then
passed into water washed absorption towers where nitric acid is
formed.
3NO2 + H2O ® 2HNO3 + NO
NO liberated in this reaction will be reoxidized in cooler.
In the operation of such a plant it is possible to produce gases from
the arc furnace in which the NO is 2% by volume while hot. The
gases are cooled to 26 °C at 750 mm Hg before entering the
absorption column.
60
PROCESS CALCULATIONS
(a) Calculate the composition of hot gases leaving the furnace
assuming air is at NTP.
(b) Calculate the partial pressure of NO2 and N2O4 in the gas entering
the absorption apparatus.
(c) Calculate the weight of acid formed per 1000 litres of gas
entering the absorption system if the combustion to HNO3 of the
combined nitrogen in the furnace gases is 85% complete.
H 2O
Air
N2,O2
Arc
NO
2%
Cooler
NO2
26 °C,
750 mm
Chamber
N 2O 4
Abs.
Col.
HNO3
Basis: 100 g moles of air contain N2 : 79 g moles and O2 : 21 g moles.
(a) Reaction in arc furnace is given as
N2 + O2 ® 2NO
which means ‘x’ mole of N2 react to give 2x mole of NO.
Total moles of gas leaving = (79 – x) + (21 – x) + 2x = 100 (by
stoichiometry)
N2
O2
NO
when x = 1, we have,
% NO = 2 =
N2
O2
NO
¯
¯
¯
78
20
2
2x
100
\x=1
(b) Reaction in cooler NO + ½O2 ® NO2
2 g moles of NO gives 2 g moles of NO2
O2 remaining = (20 – 1) = 19 g moles
Reaction in chamber is 2NO2 ® N2O4
66% of NO2 remains as N2O4
Moles of NO2 associated to N2O4 = (2 ´ 0.66) = 1.32
\
N2O4 formed = 0.66 g mole
NO2 remaining = 2 – 1.32 = 0.68 g mole
NO2
Total moles:
N 2O 4
O2
N2
Total
0.68 + 0.66 + 19 + 78 = 98.34
IDEAL GASES
Partial pressure of NO2 =
61
0.68 – 750
= 5.188 mm Hg.
98.34
0.66 – 750
= 5.030 mm Hg.
98.34
(c) Combined N2: in NO2 – 0.68 g mole
Partial pressure of N2O4 =
in N2O4 – (0.66 ´ 2) = 1.32 g moles/2 g moles
3NO2 + H2O ® 2HNO3 + NO
Thus,
Total moles of gas in 1000 litres of entering gas
=
PV
RT
1000
Û
È 750 Ø Ë
= É
–
Ê 760 ÙÚ ÌÍ (0.082)(299) ÜÝ
= 40.25 g moles
(0.68)
= 0.278 g mole
98.34
3 moles of NO2 yields 2 moles of HNO3.
Moles of NO2 = (40.25) ´
È 2Ø
\ 0.278 kmole of NO2 yields 0.278 ´ É Ù ´ 0.85 = 0.158 kmole of HNO3
Ê 3Ú
(85% conversion)
\
Weight of HNO3 formed = 0.158 ´ 63 = 9.95 g.
3.34 The gas leaving a gasoline stabilizer has the following analysis by volume
C3H8 : 8%, CH4 : 78%, C2H6 : 10% and C4H10 : 4%
This gas leaving at 90 oF and 16 psia at the rate of 70,000 ft3/h is fed
to gas reforming plant where the following reaction takes place.
CnH2n + 2 + nH2O ® nCO + (2n + 1)H2
CO + H2O ® CO2 + H2
(1)
(2)
Reaction (1) is 95% complete and Reaction (2) is 90% complete. Find
(a) Average molecular weight of the gas leaving stabilizer;
(b) weight of gas fed to reforming plant (lb/h) (c) weight of H2
leaving (lb/h) and (d) composition of gases leaving (weight %)
Basis: One hour = 70,000 ft3 of gas.
È 492 Ø È 16 Ø
Volume of gas at NTP = 70,000 ´ É
= 68,295 ft3
–
Ê 550 ÙÚ ÉÊ 14.67 ÙÚ
È 68,295 Ø
Moles of gas = É
= 190.24 lb moles
Ê 359 ÙÚ
62
PROCESS CALCULATIONS
Components
C3H8
CH4
C2H6
C4H10
Total
Volume %
lb mole
Molecular weight
Weight, lb
8
15.22
44
669.68
78
148.39
16
2374.24
10
19.02
30
570.6
4
7.61
58
441.38
100
190.24
4055.9
(a) Average molecular weight of the gas leaving stabilizer
=
4055.9
= 21.32
190.24
(b) C 3 H 8 + 3H 2 O → 3CO + 7H 2
44
54
84
14
CH 4 + H 2 O → CO + 3H 2
16
18
28
6
C 2 H 6 + 2H 2 O → 2CO + 5H 2
30
36
56
10
C 4 H10 + 4H 2 O → 4CO + 9H 2
58
72
112
18
Weight of water produced from C3H8 = (15.22 ´ 3 ´ 18) = 821.88 lb
Weight of water produced from CH4
= (148.39 ´ 1 ´ 18) = 2,671.02 lb
Weight of water produced from C2H6 = (19.02 ´ 2 ´ 18) = 684.72 lb
Weight of water produced from C4H10 = (7.61 ´ 4 ´ 18)
= 547.92 lb
Total
= 4,725.54 lb
CO + H2O ® CO2 + H2
Total CO formed = (15.22 ´ 3) + 148.39 + (19.02 ´ 2) + (7.61 ´ 4)
= 262.53 lb moles.
H2O needed for forming CO = (262.53 ´ 18) = 4725.54 lb.
Weight of gas fed to reformer = [4055.9 + (4725.54 ´ 2)]
= 13,506.98 lb/h.
(c) Hydrogen formed:
From Reaction (1) (hydrocarbons undergo 95% conversion), we
have
(15.22 ´ 0.95 ´ 7) + (148.39 ´ 0.95 ´ 3) + (19.02 ´ 0.95 ´ 5) +
(7.61 ´ 0.95 ´ 9) = 679.535 lb moles
Total CO formed = (262.53 ´ 0.95) = 249.40 lb moles
Hydrogen from CO (90% conversion) = (249.40 ´ 0.90)
= 224.46 lb moles
\ Total hydrogen leaving reformer = (679.535 + 224.46) ´ 2
= 1808 lb/h
63
IDEAL GASES
(d) Gases leaving unreacted
= HC, H2O, CO, CO2, H2
H2
C3H8 15.22 ´ 0.05 ´ 44
CH4 148.39 ´ 0.05 ´ 16
C2H6 19.0 ´ 0.05 ´ 30
C4H10 7.61 ´ 0.05 ´ 58
CO 249.4 ´ 0.1 ´ 28
CO2 249.4 ´ 0.9 ´ 44
H2O 262.53 ´ 0.05 ´ 18
=
=
=
=
=
=
=
=
[4,725.54 – (224.46 ´ 18)] =
H2O
1808.000 lb
33.484 lb
118.712 lb
28.530 lb
22.069 lb
698.320 lb
9876.240 lb
236.277 lb
(from Reaction 1)
685.260 lb
(from Reaction 2)
13,506.892 lb
Component
C 3H 8
CH4
C 2H 6
C4H10
CO
CO2
H 2O
H2
Weight, lb
33.484
118.712
28.53
22.069
698.32
9,876.24
921.537
Weight %
0.248
0.879
0.212
0.163
5.17
73.12
Total
1808 13,506.892
6.823 13.385
100.000
3.35 Analysis of a sewage gas sample from municipal sewage plant is
given. CH4 : 68%, CO2 : 30% and NH3 : 2%. 600 m3/h of this gas at
30 oC and 2 atmospheres is flowing through a pipe. Find (a) the
average molecular weight of the sewage gas and (b) the mass rate of
flow of gas in kg/h and (c) density of the gas.
Basis: 100 kmoles of the sewage gas.
(a) the average molecular weight can be calculated from the following
table.
Gas
Molecular weight
Weight, kmole
Weight, kg
CH4
16
68
68 ´ 16 = 1,088
CO2
44
30
30 ´ 44 = 1,320
NH3
17
2
2 ´ 17 =
Total
—
100
Average molecular weight =
34
2,442
2, 442
= 24.42
100
(b) Volumetric flow rate at standard condition
È 273 Ø È 2 Ø
= 600 ´ É
= 1081.2 m3/h
–
Ê 303 ÙÚ ÉÊ 1 ÙÚ
64
PROCESS CALCULATIONS
È 1081.2 Ø
Molar flow rate of gases = É
= 48.24 kmoles/h
Ê 22.414 ÙÚ
We know that 100 kmoles of this gas weighs = 2,442 kg
Therefore, the weight of 48.24 kmoles
2, 442 – 48.24
= 1,178 kg.
100
Hence, the mass flow rate of gas in kg/h = 1,178 kg/h
=
2442
= 2.26 kg/m3.
1081.2
3.36 In the process of manufacturing Cl2, HCl gas is oxidized with air as
follows:
(c) Density of the gas is
4HCl + O2 ® 2Cl2 + 2H2O
If the air used is 30% excess and oxidation is 80% complete, find the
composition of dry gases leaving.
Basis: 4 kmoles HCl gas.
O2 needed
= 1 kmole
O2 supplied
= 1.3 kmoles
N2 entering
79
= 1.3 ´ ÈÉ ØÙ = 4.89 kmoles
Ê 21 Ú
O2 remaining
= (1.3 – 0.8) = 0.5 kmole
HCl un-reacted = (4 ´ 0.2) = 0.8 kmole
Cl2 formed
= (2 ´ 0.8) = 1.6 kmoles
Composition of dry gases leaving is presented in the following table
Gas
mole
mole %
HCl
O2
N2
Cl2
0.80
0.50
4.89
1.60
10.27
6.42
62.77
20.54
Total
7.79
100.00
3.37 A mixture of NH3 and air at 730 mm Hg and 30 °C contains 5.1%
NH3. The gas is passed through an absorption tower at the rate of
100 m3/h where NH3 is removed. The gases leave the tower at
725 mm Hg and 20 °C having 0.05% NH3. Calculate (a) the rate of
flow of gas leaving the tower and (b) weight of NH3 absorbed in kg/h.
IDEAL GASES
0.05% NH3,
20 °C
725 mm Hg
Abs.
Tower
100 m3/h,
30 °C,
730 mm Hg
5.1% NH3
65
Basis: One hour of operation
È 100 Ø È 730 Ø È 273 Ø
Moles of gas entering per hour = É
–
–
Ê 22.414 ÙÚ ÉÊ 760 ÙÚ ÉÊ 303 ÙÚ
= 3.86 kmoles
NH3 entering = 3.86 ´ 0.051 = 0.197 kmole
Air entering = (3.86 – 0.197) = 3.663 kmoles
3.663 kmoles of air contains 0.05% NH3 while leaving the absorber
Therefore the total moles of gases leaving the absorber is
3.663
= 3.665 kmoles
0.9995
Hence, NH3 leaving the tower = 0.002 kmole.
=
(b) NH3 absorbed = (0.197 – 0.002) ´ 17 = 3.317 kg
Volumetric flow rate of exit gas
È 760 Ø È 293 Ø
= 3.665 ´ 22.414 ´ É
= 92.42 m3/h
–
Ê 725 ÙÚ ÉÊ 273 ÙÚ
3.38 The analysis of a flue gas, from a fuel gas containing no N2 has
CO2 : 4.62%, CO : 3.08%, O2 : 8.91% and N2 : 83.39%. Calculate
(a) kmole of dry air supplied per kmole of dry flue gas (b) % excess air
(c) analysis of the fuel gas which is a mixture of CH4 and C2H6.
Basis: 100 kmoles of dry flue gas.
83.39
= 105.56 kmoles (from nitrogen balance)
0.79
kmole of dry air/kmole of dry flue gas = 1.0556
(a) Air supplied =
CH4 ®
C 2H6 ®
Air
®
CO2
CO
O2
N2
Reactions are:
CH4 + 2O2
® CO2 + 2H2O
C2H6 + 31/2O2 ® 2CO2 + 3H2O
(b) O2 supplied = 105.56 ´ 0.21 = 22.17 kmoles
O2 consumed = (22.17 – 8.91) = 13.26 kmoles
O2 needed for conversion of CO to CO2 =
3.08
= 1.54 kmoles
2
66
PROCESS CALCULATIONS
\
O2 needed for complete combustion = 14.80 kmoles
Excess oxygen = (8.91 – 1.54) = 7.37 kmoles
È 7.37 Ø
´ 100 = 49.8%
% Excess air = É
Ê 14.80 ÙÚ
(c) Let CH4 be ‘x’ kmole and C2H6 be ‘y’ kmole
Making a CO2 balance, x + 2y = (4.62 + 3.08) = 7.70
O2 needed (by stoichiometry) 2x + 3.5y = 14.8
Solving for x and y, we get x = 5.3 and y = 1.2
Gas
Weight, kmole
mole %
CH4
C2 H 6
5.3
1.2
81.54
18.46
Total
6.5
100.00
3.39 A furnace is fired with coke containing 90% carbon and 10% ash. The
ash pit residue after being washed with water analyze 10% carbon;
40% ash and rest water. The flue gas analysis shows CO2 : 14%,
CO : 1%, O2 : 6.4% and the rest N2
Calculate the following:
(a) Volume of flue gas produced at 750 mm Hg and 250 °C per tonne
of coke charged.
(b) % Excess air used
(c) % Of carbon charged which is lost in the ash.
Basis: 100 kmoles of exit gas.
N2 = 100 – (14 + 1 + 6.4) = 78.6 kmoles
From the foregoing, air supplied = 78.6/0.79 = 99.5 kmoles; O2 in air
supplied is = 20.9 kmoles
Furnace
Flue gas
Coke (F)
Ash (P)
O2 reacted = 20.9 – 6.4 = 14.5 kmoles.
Let F be the coke supplied and P be the ash in the pit (in kg)
Total carbon reacted = Total carbon in flue gas = 15 katoms
Carbon balance: 0.9F = (15 ´ 12) + 0.1P
(1)
IDEAL GASES
Ash balance:
0.1F = 0.4P
67
(2)
(i.e.) F = 4P
Substituting for (1) from (2), we get
(0.9 ´ 4P) = 180 + 0.1P
(3.6P – 0.1P) = 180
180
= 51.43 kg (ash)
3.5
F = 205.72 kg (coke)
\ P=
2C + O2 ® 2CO
C + O2 ® CO2;
CO + ½O2 ® CO2
Carbon lost in ash: 10% = 51.43 × 0.1 = 5.143 kg (for 205.72 kg of
coke fed)
O2 needed theoretically:
O2 supplied: 20.9 kmoles
O2 needed (theoretically):
Total Carbon fed = 205.72 × 0.9
= 185.148 kg
= 15.429 katoms
O2 needed for conversion of C to CO2 is 15.429 kmoles (by stoichiometry)
Excess O2 = 20.9 – 15.429 = 5.471 kmoles
È 5.471Ø
´ 100 = 35.46 %
(b) % Excess air = É
Ê 15 ÙÚ
(a) Total carbon fed = (0.9 ´ 205.72) = 185.148 kg
205.72 kg of coke gives 100 kmoles of gas
È 100 Ø
1000 kg of coke gives 1000 ´ É
= 486.1 kmoles of the gas
Ê 205.72 ÙÚ
Aliter
For 1000 kg of coke fed, carbon in coke: 1000 × 0.9 = 900 kg
1000
= 25 kg
205.72
Carbon reacted = 900 – 25 = 875 kg = 72.916 katoms
Carbon lost in ash = 5.143 kg ×
\
15 katoms of carbon reacted º 100 kmoles of flue gas
72.916 katoms reacted º 486.1 kmoles flue gas
68
PROCESS CALCULATIONS
Volume of flue gas produced
È 760 Ø È 523 Ø
= 486.1 ´ 22.414 ´ É
–
Ê 750 ÙÚ ÉÊ 273 ÙÚ
= 21,151.2 m3/ton of coke.
È 51.43 – 0.1Ø
(c) Carbon lost = É
´ 100 = 2.78%
Ê 185.148 ÙÚ
3.40 Hot air is being used to dry a wet wallboard. The hot air enters the
drier at 768 mm Hg and 166.7 °C. The partial pressure of water
vapour in this air is 25 mm Hg. At the exit partial pressure of water is
100 mm Hg. For a total pressure of 760 mm Hg at 111.1 °C in this
process, calculate the volume of exit air per cubic metre of inlet gas.
Basis: 1 m3 of inlet air = 1000 litres = 106 cc
È 1 Ø È 768 Ø È 273 Ø
Moles of air entering = É
= 0.02799 kmole
–
–
Ê 22.414 ÙÚ ÉÊ 760 ÙÚ ÉÊ 439.7 ÙÚ
In the incoming stream, moles of water/mole of wet air
25
= 0.0326
768
In the outgoing stream, moles of water/mole of wet air
=
100
= 0.1316
768
moles of dry air entering = 0.02799 (1 – 0.0326)
= 0.02707 kmole
=
moles of wet air leaving = 0.02707 (1 + 0.1316)
= 0.03063 kmole
Volume of exit air
È 384.1Ø
= 0.03063 ´ 22.414 ´ É
Ê 273 ÙÚ
= 0.9659 m3
3.41 Applying Ideal gas law find out the maximum temperature which
20 kg of CO2 enclosed in 20 m3 chamber may be heated to with
pressure not exceeding 20 bar.
20
= 0.454 kmole
44
Volume at standard conditions: 0.454 ´ 22.414 = 10.19 m3
Basis: 20 kg of carbon dioxide =
P1 = 1 bar, V1 = 10.19 m3, T1 = 273 K
P2 = 20 bar, V2 = 20 m3, T2 = ?
IDEAL GASES
We know that
È P1V1 Ø
ÉÊ T ÙÚ
1
69
È P2V2 Ø
ÉÊ T ÙÚ
2
Thus T2 is found to be 10,716.3 K = 10,433.3 °C
3.42 A telescopic gas holder contains 1000 m3 of gas saturated with water
vapour at 20 °C and a pressure of water 155 mm Hg above
atmosphere. The barometer reads 725 mm Hg. Find the weight of
water in the gas. Vapour pressure of water is 20 mm Hg.
Basis: 1000 m3 of the gas at the given conditions
155 mm of water = 11.39 mm Hg.
Given pressure = 725 + 11.39 = 736.39 mm Hg.
Volume at standard conditions
=
1000 – 273 – 736.39
= 902.79 m3.
293 – 760
20
725
= 1.111 kmole = 20 kg of water
Amount of water =
22.414
3.43 A gaseous mixture has the following three components X, Y, Z and the
composition is as expressed below. Find the molecular weight of Y
component.
902.79 –
Component
mole %
Weight %
Molecular weight
Weight
X
35
–
85
35 × 85
Y
–
20.0
?
40 × M
Z
25
–
60
25 × 60
Total
4475 + 40M
Basis: 100 moles of mixture
Let M be the molecular weight of component Y
Therefore, mole % of Y = 40%
Weight, % of Y = 20 = (40 × M) ×
100
(35 – 85) (40 – M ) (25 – 60)
89500 + 800 M = 4000 M
Therefore, M = 27.97
3.44 One hundred m3 of mixture of N2, CO2 and H2 in the ratio of 4 : 3 : 1
is at 150 °C and 2 atm pressure. Find the mole fraction, weight % of
each, average molecular weight, and total weight of the mixture.
È 4Ø
Partial pressure of N2 = É Ù × 2 = 1 atm
Ê 8Ú
70
PROCESS CALCULATIONS
Ê 3ˆ
Partial pressure of CO2 = Á ˜ × 2 = 0.75 atm
Ë 8¯
Ê 1ˆ
Partial pressure of H2 = Á ˜ × 2 = 0.25 atm
Ë 8¯
v ˘ È T0 ˘ È 1 ˘
È
Number of moles of N2 = Í pN 2 ¥ ˙ ¥ Í ˙ ¥ Í
T ˚ Î P0 ˚ Î 22.414 ˙˚
Î
È 100 ˘ È 273 ˘ Ê 1 ˆ
= Í1 ¥
¥
¥Á
˜ = 2.88 kmoles
423 ˙˚ ÍÎ 1 ˙˚ Ë 22.414 ¯
Î
v ˘ È T0 ˘ È 1 ˘
È
Number of moles CO2 = Í pCO2 ¥ ˙ ¥ Í ˙ ¥ Í
T ˚ Î P0 ˚ Î 22.414 ˙˚
Î
È 0.75 ¥ 100 ˘ È 273 ˘ Ê 1 ˆ
= Í
˜ = 2.16 kmoles
˙¥Í
˙¥Á
Î 423 ˚ Î 1 ˚ Ë 22.414 ¯
È 0.25 ¥ 100 ¥ 273 ˘
Number of moles of H2 = Í
˙ = 0.72 kmole
Î 473 ¥ 1 ¥ 22.414 ˚
Compound
Weight,
kmoles
Molecular
weight
Weight,
kg
Weight
%
mole
%
N2
2.88
28
80.676
45.531
0.500
CO2
2.16
44
95.080
53.66
0.375
H2
0.72
2
1.441
0.81
0.125
Total
5.76
177.197
00.00
1.000
Average molecular weight = 2 × (0.125) + 28 × (0.5) + 44 × (0.375)
= 30.75
Total weight = 177.197 kg
EXERCISES
3.1
A liquefied mixture of n-butane, n-pentane and n-hexane has the
following composition in percent.
n-C4H10 : 50
n-C5H12 : 30
n-C6H14 : 20
Calculate the weight fraction, mole fraction and mole percent of each
component and also the average molecular weight of the mixture.
IDEAL GASES
71
3.2
A steel tank having a capacity of 25 m3 holds carbon dioxide at 30 °C
and 1.6 atm. Calculate the weight of the carbon dioxide in grams.
3.3
A steel container has a volume of 200 m3. It is filled with nitrogen at
22 °C and at atmospheric pressure. If the container valve is opened
and the container heated to 200 °C, calculate the fraction of the
nitrogen which leaves the container.
3.4
Natural gas has the following composition in volumetric percent:
CH4 : 80%, C2H6 : 15% and N2 : 5%
Calculate (a) composition in mole %, (b) composition in weight %
(c) Average molecular weight, and (d) density at standard condition.
3.5
A typical flue from a chimney is found to contain the following
composition by weight: Oxygen : 16%, Carbon monoxide : 4%,
Carbon dioxide : 17% and rest nitrogen. Calculate average molecular
weight and density of the gas at NTP.
3.6
A mixture of gases analyzing 20% methane, 50% ethane and rest
hydrogen by volume at a temperature of 283 K and a pressure of
5 atmosphere, flows through a pipe line at the rate of 60 m3/h. The
internal pipe diameter is 50 mm. Express the concentration in
kmole/m3, velocity in pipeline and density of the gas mixture.
3.7
Air contains 79% nitrogen and 21% oxygen by volume. Estimate its
density at 20 °C and 741 mm Hg pressure.
3.8
One kilogram of benzene is stored at a temperature of 50 °C and a
pressure of 600 atmospheres. Calculate the volume.
3.9
Find the maximum temperature to which 20 kg of CO2 enclosed in
20 m3 chamber may be heated without exceeding a pressure of
20 bars.
3.10 A gas contains 81.8% carbon and 18.2% hydrogen by weight. If
369 ml of the gas at 22 °C and 748 mm Hg weighs 0.66 g, what is
the formula of the gas?
3.11 A gas analyzes 60% methane and 40% ethylene by volume. It is
desired to store 12.3 kg of this gas mixture in a cylinder having a
capacity of 5·14 ´ 10–2 m3 at a maximum temperature of 45 °C.
Calculate the pressure inside the cylinder by assuming that the mixture
obeys the ideal gas laws.
3.12 The flue gas of a burner at 800 °C and a pressure 2.5 atm has the
following composition by weight.
Nitrogen : 65%
CO2
: 15%
H2O
: 12%
72
PROCESS CALCULATIONS
O2
: 7%
CO
: 1%
Find (a) composition by volume
(b) the average density of the flue gas
(c) mole fraction of the components
3.13 How many kilogram of liquid propane will be formed by liquefaction
of 6 m3 of the gas at 500 kPa and 300 K?
3.14 The following is the analysis of a mixture of gases by weight:
chlorine : 65%, bromine : 27% and rest oxygen. Calculate the
composition by volume %, mole % and the average molecular weight.
3.15 A natural gas having CH4 : 94%, C2H6 : 3% and N2 : 3% is piped from
the well at 298 K and 3 atm pressure. Find (a) partial pressure of N2,
(b) volume of N2 per 100 m3 of gas and (c) density of the gas.
3.16 A gas flowing at 1000 litres/s has the following composition:
CH4 : 10%, C2H6 : 30% and H2 : 60% at 303 K and 2000 mm Hg
pressure. Calculate (a) The mole fraction of each component, (b) the
concentration of each component, g mole/cc, (c) partial pressure of
each component, (d) the molar density of the mixture, (e) mass flow
rate of the gas and (f) average molecular weight.
3.17 Two hundred eighty kg of nitrogen and 64.5 kg of hydrogen are
brought together and allowed to react at 550 °C and 300 atm. It is
found that there are 38 kmoles of gases present at equilibrium.
(a) What is the limiting reactant, (b) what is the % excess of excess
reactant, and (c) % conversion of hydrogen to ammonia.
3.18 A natural gas containing 90% methane, 5% ethane and 5% nitrogen is
piped from a well at 25 °C and 1 atm pressure. Assuming the validity
of ideal gas law, find:
(a) Partial pressure of nitrogen
(b) Volume of nitrogen/100 m3 of gas
(c) Density of mixture
(d) Average molecular weight of mixture
3.19 Acetylene gas is produced according to the reaction
CaC2 + 2 H2O ® C2H2 + Ca (OH)2
Calculate the number of hours of service that can be got from 2.5 kg
of carbide in air lamp burning 100 m3 of gas/hour at 25 °C and
760 mm Hg.
IDEAL GASES
73
3.20 The following is the analysis of a mixture of gases by weight:
Chlorine: 60%, bromine: 25% and rest nitrogen.
Calculate (a) the composition by mole %, (b) average molecular
weight, and (c) density at 298 K and 740 mm Hg.
3.21 In the manufacture of nitric acid, ammonia gas and air are mixed at
7 atm pressure and 650 °C and passed over a catalyst. The
composition of this gas mixture on weight basis is nitrogen: 70.5%,
oxygen 18.8%, water vapour: 1.2% and ammonia 9.5%. Assuming the
validity of the ideal gas law, find (a) composition in mole % and
(b) density in kg/m3.
3.22 Calculate the volume occupied by 30 g of chlorine at a pressure of
743 mm Hg and 21.1 °C.
3.23 Propane is liquefied for storage in cylinders. How many kilograms of
it will be formed by liquefying 500 litres of the gas at standard
conditions?
Vapour Pressure
4
Whenever we come across a liquid system, it is generally in contact with its
own vapour over the liquid surface. This vapour exerts a pressure like gases,
which is called vapour pressure. When the liquid is at its boiling condition
the vapour pressure will be equal to the surrounding pressure. When that
pressure becomes equal to the atmospheric pressure, the boiling point is
referred as normal boiling point (NBP). Whenever we have a binary system,
if the sum of their partial pressures equals the surrounding pressure then the
system will boil.
4.1
EFFECT OF TEMPERATURE ON VAPOUR
PRESSURE
Illustrated by the effect of temperature on vapour pressure is Clapeyron
equation
dp
l
=
dT T (VG - VL )
where, p represents vapour pressure, T : absolute temperature, l : heat of
vaporization at T, VG : volume of gas and VL : volume of liquid.
If the volume of liquid is neglected and applicability of the ideal gas
law assumed, the above reaction reduces to Clausius–Clapeyron equation.
dp l dT
=
p RT 2
1
-l
¥d
T
R
where R is gas law constant and l , molal latent heat of vaporization.
When the temperature does not vary over a wide range, it may be
assumed that the molal latent heat of vaporization is constant and the above
equation may be integrated between the limits p0 and p and T0 and T.
or
d(ln p) =
74
VAPOUR PRESSURE
75
Ê p ˆ l Ê 1 1ˆ
ln Á ˜ = Á - ˜
Ë p0 ¯ R Ë T0 T ¯
or
Ê p ˆ Ê l ˆÊ 1 1ˆ
log Á ˜ = Á
˜
Ë p0 ¯ Ë 2.303 R ¯ ÁË T0 T ˜¯
Effect of Temperature on V.P:
B
Antoine equation, log p* (mm Hg) = A –
(where, p* is the
T +C
vapour pressure and A, B, C are constants)
Limitations:
(i) Applicable only in the range of applicability of A, B, C.
(ii) Pressures below 10 bar.
4.2 HAUSBRAND CHART
Steam distillation takes place at point of intersection of the curves at
which
p A = p – pW
\ p = p A + pW
Total pressure = vapour pressure of component + vapour pressure of
water
p – pW
pA
Vapour
pressure
Temperature
WORKED EXAMPLES
4.1
The vapour pressure of ethyl ether at 0 °C is 185 mm Hg. Latent heat
of vaporization is 92.5 cal/g. Calculate vapour pressure at 20 °C and
35 °C.
Molecular weight of ethyl ether = 74.
l = (92.5 ¥ 74) = 6845 cal/g mole
R = 1.99 cal/g mole K
T0 = 273 K, T1 = 293 K, T2 = 308 K
76
PROCESS CALCULATIONS
At 20 °C,
È
ØÈ 1
6845
1 Ø
ÉÊ 2.303 – 1.99 ÚÙ ÉÊ 273 293 ÙÚ
È p Ø
log É
Ê 185 ÙÚ
Therefore, p = 437 mm Hg.
At 35 °C,
È
ØÈ 1
6845
1 Ø
ÉÊ 2.303 – 1.99 ÚÙ ÉÊ 273 308 ÙÚ
È p Ø
log É
Ê 185 ÙÚ
Hence, p = 773 mm Hg.
4.2
It is proposed to purify benzene from small amount of non-volatile
solutes by subjecting it to distillation with saturated steam under
atmospheric pressure of 745 mm Hg. Calculate the temperature at
which the distillation will proceed and the weight of steam
accompanying 1 g of benzene vapour.
Temperature, °C Vapour pressure of
benzene, mm Hg
Vapour pressure Total pressure,
of water, mm Hg
mm Hg
60
65
68
69
390
460
510
520
150
190
215
225
70
550
235
540
650
725
745
distillation
temperature,
69 °C
785
p – pW
pB
Vapour
pressure
Temperature
Basis: 1 g mole of mixed vapour.
g mole of C6 H 6
g mole of water
g C6 H 6
g water
g water
g benzene
Vapour pressure of C6 H 6
Vapour pressure of water
2.31 –
0.1
78
18
10.01
520
225
2.31
VAPOUR PRESSURE
4.3
77
It is decided to purify myristic acid (C13H27COOH) by steam
distillation under 740 mm Hg pressure. Calculate the temperature and
the weight of steam to be used per kg of acid.
At 99 °C: Vapour pressure of H2O = 740 mm Hg
Vapour pressure of acid = 0.032 mm Hg
\ The distillation temperature can be taken as 99 °C
Basis: 1 kmole of mixed vapour.
Moles of the acid È 0.032 Ø
: É
´ 1 = 4.3 ´ 10–5 kmoles = 0.0098 kg
Ê 740 ÙÚ
Mole of water
Moles of water = 1 kmole = 18 kg
Weight of steam
kg of acid
4.4
È 18 Ø
ÉÊ
Ù
0.0098 Ú
1840 kg
The acid given in example 4.3 is to be distilled at 200 °C by use
of superheated steam. It may be assumed that the relative saturation of
the steam with acid vapours will be 80% (a) Find the weight of steam
required per kg of acid distilled at 740 mm Hg. (b) Calculate the
weight of steam per kg of acid if 26 inches of Hg vacuum is
maintained.
(a)
Vapour pressure of acid at 200 °C = 14.5 mm Hg
Partial pressure of acid (80% saturation) = (14.5 ´ 0.8)
= 11.6 mm Hg
Basis: 1 kmole of mixed vapour
È 11.6 Ø
Weight of acid = É
´ 1 = 0.0157 kmole = 3.58 kg
Ê 740 ÙÚ
Weight of water = (1 – 0.0157) = 0.9843 kmole = 17.70 kg
Steam
È 17.7 Ø
= É
= 4.95 kg
Ê 3.58 ÙÚ
kg of acid
(b)
26 inches of Hg. vacuum = 740 – (26 ´ 25.4) = 80 mm Hg
È 11.6 Ø
Weight of acid = É
´ 1 = 0.145 kmole = 33.1 kg
Ê 80 ÙÚ
Weight of water (steam) = 0.855 kmole = 15.4 kg
Steam
15.4 Ø
= ÈÉ
= 0.465 kg
Ê 33.1 ÙÚ
kg of acid
Discussion on examples 4.3 and 4.4: When ordinary steam is used at
atmospheric pressure, 1840 kg of steam is needed. When superheated
steam at 200 °C is used, 4.95 kg is needed. Due to low pressure and
hence low boiling point under vacuum, quantity of steam needed is
0.465 kg only.
78
4.5
PROCESS CALCULATIONS
Calculate the total pressure and the composition of the vapours
in contact with a solution at 100 °C containing 35% Benzene,
40% Toluene and 25% Xylene by weight. At 100 °C, the vapour
pressures of benzene (molecular weight : 78) is 1340 mm Hg, toluene
(Molecular weight : 92) is 560 mm Hg and Xylene (molecular
weight : 106) is 210 mm Hg.
Basis: 100 kg of solution.
Composition of vapours and their partial pressure
Component
Mol.
wt
Vapour
pressure,
mm Hg
Weight
%
Weight,
kmole
Mole
fraction
in
liquid
phase
Partial
pressure,
mm Hg
Mole
fraction
in
vapour
phase
Benzene
(C6H6)
Toluene
(C7H8)
Xylene
(C8H10)
78
1340
35
35/78 = 0.449
0.401
0.401 ´ 1340 = 536
0.673
92
560
40
40/92 = 0.435
0.388
0.388 ´ 560 = 217
0.272
106
210
25
25/106 = 0.236
0.211
0.211 ´ 210 = 44
0.055
100
1.12
1.000
797
1.000
Total
Total pressure = 797 mm Hg.
4.6
An aqueous solution of NaNO3 having 10 g moles of salt/1 kg of
water boils at 108.7 °C at 760 mm Hg. Assume that the relative
vapour pressure of the solution is independent of temperature. Find
the vapour pressure of the solution at 30 °C and the boiling point
elevation.
Since the solution boils at 108.7 °C, the vapour pressure of solution
= 760 mm Hg.
Vapour pressure of water at 108.7 °C = 1030 mm Hg (from Steam
Tables)
k
Vapour pressure of solution
Vapour pressure of solvent
760
1030
0.74.
Vapour pressure of water at 30 °C = 31.8 mm Hg (from Tables)
Vapour pressure of solution at 30 °C = (31.8 ´ 0.74) = 23.5 mm Hg,
È from Cox chart Ø
Boiling point of water at 23.5 mm Hg = 24.8 °C É
Ê from Steam Tables ÙÚ
Boiling point elevation
= (Boiling point of solution – Boiling point of solvent)
= 30 – 24.8 = 5.2 °C
VAPOUR PRESSURE
4.7
79
The following table gives vapour pressure data:
Temperature °C
69
70
75
80
85
90
95
99.2
Hexane (A), mm Hg 760
780
915
1060
1225
1405
1577
1765
Heptane (B), mm Hg 295
302
348
426
498
588
675
760
Assuming Raoult’s law is valid, use the above data to calculate for
each of the above temperature the mole percent ‘x’ of hexane in the
liquid and the mole percent ‘y’ of hexane in vapour at 760 mm Hg.
p A = xA P A ;
p B = xB P B ;
(xA + xB) = 1
where PA, PB are vapour pressures of hexane (A) and heptane (B)
respectively.
pA = yAP;
pB = yBP;
(yA + yB) = 1
pA + pB = P; xB = (1 – xA )
(xAPA) + (1 – xA)PB = P
xA(PA – PB) = (P – PB)
xA 760 + (1 – xA)295 = 760
At 69 °C, (Boiling point of hexane)
xA(760 – 295) = (760 – 295); \ xA = 1
760
=1
760
At 70 °C xA (780 – 302) = (760 – 302); \ xA = 0.96
pA = (0.96 ´ 780) = 748.8; \ yA = 748.8/760 = 0.985
yA =
At 75 °C xA (915 – 348) = (760 – 348); \ xA = 0.73
pA = (0.73 ´ 915) = 668; \ yA = 668/760 = 0.880
At 80 °C xA (1060 – 426) = (760 – 426); \ xA = 0.53
pA = (0.53 ´ 1060) = 562; \ yA = 562/760 = 0.740
At 85 °C xA (1225 – 498) = (760 – 498); \ xA = 0.36
pA = (0.36 ´ 1225) = 441; \ yA = 441/760 = 0.58
At 90 °C xA (1405 – 588) = (760 – 588); \ xA = 0.21
pA = (0.21 ´ 1405) = 295; \ yA = 295/760 = 0.39
At 95 °C xA (1577 – 675) = (760 – 675); \ xA = 0.0945
pA = (0.0945 ´ 1577) = 149; \ yA = 149/760 = 0.196
At 99.2 °C Boiling point of heptane
xA (1765 – 760) = (760 – 760); \ xA = 0 = yA
4.8
A solution of methanol in water containing 0.158 mole fraction
alcohol boils at 84.1 °C (760 mm Hg). The resulting vapour contains
80
PROCESS CALCULATIONS
0.553 mole fraction of alcohol. How does the actual composition of
the vapour compare with composition calculated from Raoult’s Law?
Temperature, °C
Vapour pressure, mm Hg
80
1340
100
2624
To get vapour pressure at 84.1 °C, the equation
ln p = A – B/(T – 43)
can be used, ( p, vapour pressure in mm Hg and T, temperature in
Kelvin A and B are constants)
Solving A = 18.2 and B = 3420;
Vapour pressure at 84.1 ºC = 1520 mm Hg.
According to Raoult’s law; pA = xAPA = (0.158 ´ 1520) = 240;
240
yA =
= 0.316
760
Ë Actual value Calculated value Û
% Error = Ì
Ü ´ 100
Actual value
Í
Ý
È 0.553 - 0.316 Ø
= É
ÙÚ ´ 100 = 42.8%
Ê
0.553
4.9
Methane burns to form CO2 and water. If 1 lb mole is burnt with 10%
excess pure O2 and the resulting gas mixture is cooled and dried,
calculate (a) volume of dry exit gas at 70 °F and 750 mm Hg.
(b) Partial pressure of O2 in exit (c) Weight of water removed.
Basis: 1 lb mole of methane.
CH4 + 2O2 ® CO2 + 2H2O
O2 supplied = 2 ´ 1.1 = 2.2 lb moles
Gases leaving after drying: CO2 : 1.0 lb mole; O2 : 0.2 lb mole.
È 530 Ø È 760 Ø
(a) Volume of dry exit gas = 1.2 ´ 359 ´ É
–
Ê 492 ÙÚ ÉÊ 750 ÙÚ
= 470.26 ft3
0.2 – 750
= 125 mm Hg
1.2
(c) Water removed = 2 lb moles = 36 lb
(b) Partial pressure of O2 in exit =
4.10 Bottled liquid gas is sold. Determine (a) the pressure of the system;
(b) vapour composition. The composition in mole % in liquid phase is
given as follows:
n-Butane : 50%, Propane : 45%, and Ethane : 5%
VAPOUR PRESSURE
81
Vapour pressure (in bar) at 30 °C are n-butane: 3.4, propane: 10.8 and
ethane: 46.6
Gas
n-Butane
Propane
Ethane
Total
Mole %
Vapour pressure at 30 °C (bar)
Partial pressure (bar)
50
3.4
1.70
(3.4 ´ 0.5)
19.12
45
10.8
4.86
(10.8 ´ 0.45)
54.67
5
46.6
2.33
(46.6 ´ 0.05)
26.21
100
Vapour composition %
8.89
100
4.11 A solvent recovery system delivers a gas saturated with benzene
vapour which analyzes on a benzene free basis as follows:
CO : 15%, O2 : 4% and N2 : 81%. This gas is at 21.1 °C and 750 mm
Hg. It is compressed to 5 atm and cooled to 21.1 °C after compression.
How many kilograms of benzene are condensed by this process per
1000 m3 of original mixture? The vapour pressure of benzene at 21.1
°C is 75 mm Hg.
Basis: 1000 m3 of original mixture
È 1000 Ø È 750 Ø È 273 Ø
Moles of original mixture = É
–
–
Ê 22.414 ÙÚ ÉÊ 760 ÙÚ ÉÊ 294.1ÙÚ
= 40.87 kmoles
È 75 Ø
Benzene present originally = 40.87 ´ É
= 4.087 kmoles
Ê 750 ÙÚ
Moles of gas other than benzene = 40.87 – 4.087 = 36.783 kmoles
Vapour pressure of benzene = 75 mm Hg = 75/760 = 0.0987 atm;
(other gas is Tie element)
È 0.0987 Ø
Benzene after compression = 36.783 ´ É
Ê 5 0.0987 ÙÚ
= 0.741 kmole
Benzene condensed = (4.087 – 0.741) ´ 78 = 261 kg
4.12 N2 from a cylinder is bubbled through acetone at 840 mm Hg and
323 K at the rate of 0.012 m3/h. The N2 saturated with acetone
vapour, leaves at 760 mm Hg and 308 K at 0.023 m3/h. Find the
vapour pressure of acetone at 308 K.
Molar flow rate of N2 =
0.012
= 5 × 10–4 kmole/h
¦ 760 µ ¦ 323 µ
22.414 §
¶§
¶
¨ 840 · ¨ 273 ·
82
PROCESS CALCULATIONS
(N2 + CH3COCH3) =
0.023
= 9.09 × 10–4 kmole/h
¦ 1.013 µ ¦ 308 µ
22.414 §
¶§
¶
¨ 1.013 · ¨ 273 ·
Let y be the mole fraction of N2 in leaving stream.
Then, 9.09 × 10–4y = 5 × 10–4
Solving, we get y = 0.55
Thus, mole fraction of acetone = 0.45
(PT) × (y) = (Partial pressure of acetone) = (VP)acetone (y)acetone
(760)(0.45) = (VP) (1.0) = 342 mm Hg
4.13 Determine the pressure of the system and equilibrium VP at 30 oC.
Assuming all ethane is removed, estimate the pressure and vapour
phase composition components of the system at 30 oC.
Compound
Mole fraction, x, in liquid phase
n-butane
n-propane
Ethane
0.50
0.45
0.05
Compound
n-butane
n-propane
Ethane
Mole fraction, VP at 30 °C,
x, in liquid phase mm Hg
0.50
0.45
0.05
3.4
10.8
46.6
Total
y (Mole fraction) =
Partial pressure, Y, mole fraction
mm Hg
× 100
1.7
4.86
2.33
19.12
54.67
26.21
8.89
100.00
Partial pressure
Total pressure
È 1.7 Ø
= É
× 100 = 19.12 for n-butane
Ê 8.89 ÙÚ
54.67 for n-propane
26.21 for ethane.
When ethane is removed, total moles will be 0.95 kmole.
Compound
Mole fraction, VP at 30 °C,
x, in liquid phase mm Hg
n-butane 0.50/0.95 = 0.5263
n-propane
0.4737
Total
3.4
10.8
Partial pressure, Y, mole fraction
mm Hg
× 100
1.789
4.86
25.91
74.09
6.904
100.00
VAPOUR PRESSURE
83
4.14 Estimate the liquid phase composition of a mixture of benzene and
toluene at 80 oC when their gas phase compositions are 80% benzene
and 20% toluene. Vapour pressures of benzene and toluene are
1340 mm Hg and 560 mm Hg respectively at 80 oC.
Vapour pressure of benzene is 1340 mm Hg at 80 oC
Vapour pressure toluene is 560 mm Hg at 80 oC
At equilibrium, partial pressure of benzene
= (Mole fraction, xB, of benzene) × (1340)
= (Total pressure) × (Mole fraction in vapour phase)
= PT (0.8)
i.e. (xB) × (1340) = PT × (0.8)
Similarly for toluene,
xT × (560) = PT × (0.2)
PT × (0.8) + PT × (0.2) = PT
We also know that, xB + xT = 1
4 × (1 – xB) × (560) = 4 × PT × (0.2)
xB × (1340) = PT × (0.8)
4 (1 – xB) 560 = xB (1340)
2240 – 1340xB – 2240xB = 0
Solving, xB = 0.6
4.15 A certain quantity of an organic solvent (molecular weight 125 and
density 1.505 g/cc) is kept in an open flask and boiled long enough so
that the vapour fills the vapour space completely by displacing all the
air. The flask is closed, evacuated and the contents reach equilibrium
at 30 °C. Vapour pressure of solvent at 30 °C is 240 mm Hg. It is
observed that only 10 ml of liquid solvent is present finally.
(i) What is the pressure in the flask at equilibrium?
(ii) What is the total mass in grams of organic liquid in the flask?
(iii) What fraction of the organic liquid present in the flask is in
vapour phase at equilibrium?
Equilibrium vapour pressure = 240 mm Hg at 30 °C and the pressure
in the system is 240 mm Hg.
Volume of vapour space = 4000 – 10 = 3990 ml
T = 303 K ; P = 240 mm Hg
Moles at NTP =
3990 t 240 273
1
t
t
= 0.05065 g mole
303
760 22414
84
PROCESS CALCULATIONS
Molecular weight = 125
Weight = 0.05065 × 125 = 6.331 g
Mass of liquid left behind = 10 ml × 1.505 g/cc = 15.05 g
Total weight = 15.05 + 6.331 = 21.381 g
Mole fraction of vapour = (6.331/21.381) = 0.2961
4.16 Benzene and toluene form an ideal solution. If vapour pressure of
benzene is 70 mm Hg and that of toluene is 20 mm Hg and their mole
fractions in liquid phase are 0.7 and 0.3 respectively, calculate their
vapour phase composition.
Let A be benzene and B be toluene
Let PP and PT be partial pressure and total pressure respectively
PPA = (PT) (yA ) = xA PA
PPB = (PT)(yB) = xB PB
PT(yA + yB) = xA (PA) + (1 – xA) PB = PP
PT = PPA + PPB
PT = (0.7)(70) + (0.3)(20)
= 49 + 6 = 55 mm Hg
Hence, yA = (0.7)(70)/55 = 0.89
yB = (0.3)(20)/55 = 0.11
4.17 Vapour pressure of pure benzene and toluene are 960 mm Hg and
380 mm Hg respectively at 86.0 °C. Calculate the liquid phase and
vapour phase composition at that temperature.
Total pressure = 760 mm Hg
Let x and y be the liquid and vapour phase compositions respectively.
Suffix B and T denote benzene and toluene respectively
Then, by Raoult’s law,
PTot = (xB) (PB) + (xT) (PT)
760 = (xB) (960) + (1 – xB) 380
Solving,
xB = 0.655
xT = 0.345
yB =
PB x B 960 t 0.655
= 0.827
PTot
760
In the same way, yT =
PT xT 380 t 0.345
= 0.173
PTot
760
VAPOUR PRESSURE
85
4.18 Vapour pressure of benzene is 3 atm and that of toluene is 1.333 atm.
A liquid fuel containing 0.4 mole benzene and 0.6 mole toluene is
vaporized. Estimate the mole fraction of benzene in vapour phase.
Partial pressure of benzene = (3)(0.4)
= 1.2
Partial pressure of toluene = (1.333)(0.6) = 0.8
Total pressure
= 2.0 atm
ybenzene = 1.2/2.0 =0.6
EXERCISES
4.1
The vapour pressure of benzene can be calculated from the Antoine
equation.
ln (p*) = 15.9008 – [2788.51/(–52.36 + T)]
where p* is in mm Hg and T is in K. Determine the latent heat of
vaporization of benzene at its normal boiling point of 353.26 K, and
compare with the experimental value.
4.2
Prepare a Cox chart for ethyl acetate. The vapour pressure of ethyl
acetate is 200 mm Hg abs. at 42 °C and 5.0 atm at 126.0 °C. By
using the chart estimate the boiling point of ethyl acetate at 760 mm
Hg and compare with the experimental value (77.1 °C).
4.3
The following data gives the vapour pressure (VP) for benzene (A)–
toluene (B) system. Compute the vapour-liquid equilibrium data at a
total pressure of 760 mm Hg.
VPA, 760 811 882 957 1037 1123 1214 1310 1412 1530 1625 1756 —
mm Hg
VPB, —
mm Hg
4.4
4.5
314 345 378 414
452
494 538
585
635
689
747 760
The following data gives the vapour pressure data for a binary system.
Compute the VLE vapour–liquid equilibrium data at a total pressure of
760 mm Hg.
Vapour pressure of A, mm Hg
760
860
1002
1160
1262
Vapour pressure of B, mm Hg
433
498
588
698
760
It is desired to purify nitrobenzene by steam distillation under a
pressure of 760 mm Hg. Distillation takes place at 99 °C. Vapour
pressure of nitrobenzene is 20 mm Hg. Estimate the weight of
nitro benzene associated per kg of steam in distillate. Assuming the
vaporization efficiency to be 75%, estimate the weight of nitrobenzene per kg of steam in distillate.
86
PROCESS CALCULATIONS
4.6
Methyl alcohol and ethyl alcohol at 100 °C have vapour pressures of
2710 mm Hg and 1635 mm Hg respectively. Calculate the total
pressure and composition of the vapour in contact with a liquid
containing 30 weight % of methyl alcohol and 70 weight % of ethyl
alcohol at 100 °C.
4.7
A liquefied fuel has the following analysis: C2H6 : 2%, n-C3H8 : 40%,
i-C4H10 : 7%, n-C4H10 : 47% and C5H12 : 4%. It is stored in cylinders
for sale. (a) Calculate the total pressure in cylinder at 21 °C and the
composition of the vapour evolved. (b) Find the total pressure at 21 °C
if all C2H6 were removed.
Vapour pressure data (mm Hg):
C2H6 : 28500, n-C3H8 : 6525, i-C4H10 : 2250, n-C4H10 : 1560 and
C5H12 : 430.
4.8
Calculate the total pressure and the composition of the vapours and
liquid in molar quantities at 100 °C for a solution containing 45%
benzene, 40% toluene and 15% xylene by weight. At 100 °C, the
vapour pressures are benzene (C6H6) (molecular weight: 78) : 1340
mm Hg; Toluene (C7H8) (molecular weight: 92) : 560 mm Hg, xylene
(C7H8) (molecular weight: 106) : 210 mm Hg.
4.9
The partial pressure of actetaldehyde in a solution containing 0.245 kg
of CH3CHO in 20 kg of water is 190 mm Hg at 93.5 °C. Calculate
the partial pressure of CH3CHO over 0.15 molal solution in water.
4.10 Ethyl acetate at 30 °C exerts a vapour pressure of 110 mm Hg.
Calculate the composition of the saturated mixture of ethyl acetate and
air at a temperature of 30 °C and an absolute pressure of 900 mm Hg
pressure. Express the composition by (a) volume %, and (b) weight %.
5
Psychrometry
5.1 HUMIDITY
Humidification operation is a classical example of an interphase transfer of
mass and energy, when a gas and a pure liquid are brought into intimate
contact. The term humidification is used to designate a process where the
liquid is transferred to gas phase and dehumidification indicates a process
where the transfer is from the gas phase to the liquid phase. The matter
transferred between phases in both the cases is the substance constituting the
liquid phase, which either vaporizes or condenses indicating respectively the
humidification process or the dehumidification process.
5.2 DEFINITIONS
Dry bulb temperature (DBT): The temperature measured by a bare
thermometer or thermocouple is called the dry bulb temperature.
Wet bulb temperature (WBT): The temperature measured by a
thermometer or thermocouple with a wet wick covering the bulb, under
equilibrium condition, is called the wet bulb temperature.
Absolute humidity (nv): The substance that is transferred (vapour) is
designated by A and the main gas phase is designated by B.
It is defined as the moles of vapour carried by unit mole of vapour
free gas.
[ nv ]
Ë yA Û
Ì Ü
Í yB Ý
pA
pB
Ë pv Û
Ì
Ü
Í p pv Ý
moles of A
moles of B
(5.1)
When the quantities are expressed in mass, then it is called mass
absolute humidity (n¢v ) or Grosvenor humidity.
nv„
ËM Û
[ nv ] – Ì A Ü
Í MB Ý
Ë pv Û Ë M A Û
Ì
ܖÌ
Ü
Í p pv Ý Í M B Ý
87
mass of A
mass of B
(5.2)
88
PROCESS CALCULATIONS
Relative humidity or relative saturation ( yr%): It is normally expressed
in percentage. If pv is the partial pressure under a given condition and ps is
the vapour pressure at DBT of the mixture, then
Relative saturation = yr % =
pv
¥ 100
ps
(5.3)
Percentage saturation or percentage absolute humidity ( yP): It is
defined as the percentage of the ratio of humidity under given condition to
the humidity under the saturated condition.
Percentage saturation = yP % =
nv
¥ 100
ns
(5.4)
Dew point: This is the temperature at which a vapour-gas mixture
becomes saturated when cooled at constant total pressure out of contact with
a liquid. The moment the temperature is reduced below the dew point, the
vapour will condense as a liquid dew.
Humid heat: The humid heat CS is the heat required to raise the
temperature of unit mass of gas and its accompanying vapour by one degree
centigrade at constant pressure.
CS = CAn¢v + CB
(5.5)
where CA and CB are specific heats of vapour and gas respectively.
Enthalpy: The enthalpy H¢ of a vapour-gas mixture is the sum of the
enthalpies of the gas and the vapour content. For a gas at a DBT of tG, with
a humidity of n¢v , the enthalpy relative to the reference state ‘t0’ is,
H¢ = enthalpy of gas + enthalpy of vapour component.
= CB(tG – t0) + n¢v [CA(tG – tDP) + lDP + CA,L (tDP – t0)]
where
(5.6)
lDP = Latent heat of vaporization at dew point.
CA,L = Specific heat of component ‘A’ (vapour) in liquid phase.
The above Eq. (5.6) will reduce to Eq. (5.7) when tDP itself is taken as
the reference temperature, t0
H¢ = CB (tG – t0) + n¢v [CA(tG – t0) + l 0]
= CS (tG – t0) + n¢v l 0
(5.7)
Humid volume: The humid volume vH of a vapour-gas mixture is the
volume of unit mass of dry gas and its accompanying vapour at the
prevailing temperature and pressure. The expression for humid volume in
m3/kg is
PSYCHROMETRY
89
È Ê 1 ˆ Ê nv¢ ˆ ˘
105 ˘
È tG¢ + 273 ˘ È
22.414
1.013
+
¥
¥
¥
¥
vH = Í Á
Í
˙
˙
Í 273 ˙
˜ Á
˜
p ˙˚
Î
˚ ÍÎ
ÎË M B ¯ Ë M A ¯ ˚
È Ê 1 ˆ Ê nv¢ ˆ ˘ È tG¢ + 273 ˘
= (8315) Í Á
˙
˜ +Á
˜˙ ¥ Í
p
˚
ÎË MB ¯ Ë M A ¯ ˚ Î
(5.8)
where, p = total pressure N/m2
A typical psychrometric chart is shown at the end of the chapter.
Various properties of air-water system can be obtained from the chart.
Alternatively, the equations given can be used to determine the properties.
Saturated vapour: When a gas or gaseous mixture remains in contact
with a liquid surface it will acquire vapour from the liquid until the partial
pressure of the vapour in the gas mixture equals the vapour pressure of the
liquid at its existing temperature when the vapour concentration reaches this
equilibrium concentration, the gas is said to be ‘saturated’ with the vapour.
Partial saturation: If a gas contains a vapour in such proportions that its
partial pressure is less than the vapour pressure of the liquid at the existing
temperature, the mixture is partially saturated.
Relative saturation: The relative saturation of such a mixture may be
defined as the percentage ratio of the partial pressure of the vapour to
the vapour pressure of the liquid at the existing temperature. The relative
saturation is therefore a function of both the composition of the mixture and
its temperature as well as of the nature of the vapour.
Relative saturation also represents the following ratios: (a) The ratio of
the percentage of the vapour by volume to the percentage by volume that
would be present if the gas were saturated at the existing temperature and
total pressure, (b) the ratio of the weight of vapour per unit volume of
mixture to the weight per unit volume present at saturation at the existing
temperature and total pressure.
Percentage saturation: It is defined as the percentage ratio of the existing
weight of vapour per unit weight of vapour free gas to the weight of vapour
that would exist per unit weight of vapour free gas if the mixture were
saturated at the existing temperature and pressure.
This represents the ratio of the existing moles of vapour per mole of
vapour free gas to the moles of vapour that would be present per mole
of vapour free gas if the mixture were saturated at the existing temperature
and pressure.
p
Relative saturation % = yr = v ¥ 100
ps
90
PROCESS CALCULATIONS
where,
pv = Partial pressure of vapour
ps = Vapour pressure of pure liquid
Percentage saturation = yp =
nv
¥ 100
ns
where,
nv = moles of vapour per mole of vapour free gas actually present
ns = moles of vapour per mole of vapour free gas at saturation.
If p is the total pressure, then from Dalton’s law,
Ê pv ˆ
nv = Á
,
Ë p - pv ˜¯
and
Ê ps ˆ
ns = Á
Ë p - ps ˜¯
nv Ê pv ˆ Ê p - ps ˆ
=
ns ÁË ps ˜¯ ÁË p - pv ˜¯
Ê p - pv ˆ
yp = (yr) Á
Ë p - pv ˜¯
WORKED EXAMPLES
5.1
An air (B)-water (A) sample has a dry bulb temperature of 50 °C and
a wet bulb temperature of 35 °C. Estimate its properties at a total
pressure of 1 atm.
1 atm = 1.0133 ¥ 105 N/m2
Molecular weight of air = 28.84
(i) n¢v (chart) = 0.03 kg w.v/kg.d.a = 0.04833 kmole of w.v/kmole of d.a.
(ii) % humidity (chart) = 35%
(iii) % relative saturation = partial pressure/vapour pressure
Partial pressure under the given condition is given by
molal humidity (0.0483) =
0.0483 =
pA
p - pA
pA
1.0133 ¥ 105 - pA
Ê
105 ˆ
20.69 = Á1.0133 ¥
–1
p A ¯˜
Ë
pA = 4.672 ¥ 103 N/m2
where,
pA = Partial pressure of water
p = Total pressure
PSYCHROMETRY
91
Vapour pressure of water (Steam Tables) at 50 °C = 92.51 mm Hg
= 12.332 ´ 103 N/m2
103
´ 103 = 37.88%
12.332
(iv) Dew point = 31.5°C
(v) Humid heat = CS = CB + CA n¢v
(Eq. 5.5)
= 1.005 + 1.884 (0.03)
= 1.062 kJ/kg of dry air °C
(vi) Enthalpy
(For a reference temperature of 0 °C)
(Eq. 5.7)
(a) H¢ = CS (tG – t0) + n¢v l0
l0 = 2502 kJ/kg
H¢ = 1.062 (50 – 0) + (0.03) (2502) = 128.16 kJ/kg da
(b) Enthalpy of saturated air = 274 kJ/kg
Enthalpy of dry air = 50 kJ/kg
\ Enthalpy of wet air = 50 + (274 – 50)(0.35) = 128.4 kJ/kg
(vii) Humid volume vH
\
% R.H. = 4.672 ´
Ë È 1 Ø È nv„ Ø Û Ë tG„ 273 Û
(a) vH = (8315) Ì É
Ü
Ù É
ÙÜ – Ì
Ý
Í Ê M B Ú Ê M A Ú Ý Í pt
Û
Ë È 1 Ø È nv„ Ø Û Ë
(325)
= (8315) Ì É
É ÙÜ – Ì
Ü
Ù
5
Ê
Ú
Ê 18 Ú Ý Í 1.0133 – 10 Ý
Í 28.84
= 0.969 m3 mixture/kg of dry air
Alternatively,
(b) Specific volume of saturated air = 1.055 m3/kg of dry air
Specific volume of dry air = 0.91 m3/kg of dry air
By interpolation vH = 0.91 + (1.055 – 0.91)(0.35)
= 0.961 m3/kg of dry air
5.2
Ether at a temperature of 20 °C exerts a vapour pressure of 442 mm Hg.
Calculate the composition of a saturated mixture of nitrogen and ethyl
ether vapour at 20 °C and 745 mm Hg and express the same in the
following forms.
(a) Percentage composition by volume
(b) Composition by weight
(c) lb of vapour/ft3 of mixture
(d) lb of vapour/lb of vapour free gas
(e) lb moles of vapour/lb mole of vapour free gas
92
PROCESS CALCULATIONS
(a) Basis: 1ft3 of gas mixture
442 – 1
= 0.593 ft3 = 59.3 Volume %;
745
Nitrogen = 40.7% = 0.407 ft3
Percentage of ethyl ether =
(b) Basis: 1 lb mole of mixture
Molecular
weight
lb mole
Weight, lb
Weight %
Ether
74
0.593
(0.593 ´ 74) = 43.9
79.4
Nitrogen
28
0.407
(0.407 ´ 28) = 11.4
20.6
1.000
55.3
100.0
Component
Total
760 Ø È 293 Ø
(c) Volume of gas mixture = 1 ´ 359 ´ ÈÉ
= 393 ft3
–
Ê 745 ÙÚ ÉÊ 273 ÙÚ
\
È 43.9 Ø
lb of vapour/ft3 = É
= 0.112
Ê 393 ÙÚ
(d) lb of vapour/lb of vapour free gas =
43.9
= 3.85
11.4
0.593
= 1.457
0.407
A mixture of acetone vapour and nitrogen contains 14.8% acetone by
volume. Calculate yr and yp at 20 °C and 745 mm Hg
(e) lb mole vapour/lb moles vapour free gas =
5.3
Vapour pressure of acetone at 20 °C = 184.8 mm Hg
Partial pressure of acetone = (0.148 ´ 745) = 110 mm Hg
È 110 Ø
´ 100 = 59.7%
yr = relative saturation = É
Ê 184.8 ÙÚ
0.148 Ø
= 0.174
nv = ÈÉ
Ê 0.852 ÙÚ
Ë È 184.8 Ø Û
Ì ÉÊ 745 ÙÚ Ü
Ü
ns = Ì
Ì È 560.2 Ø Ü
Ì ÉÊ 745 ÙÚ Ü
Í
Ý
yp =
nv
ns
È 0.248 Ø
= 0.329
ÊÉ 0.752 ÚÙ
0.174
= 52.9%
0.329
Always yr > yp
PSYCHROMETRY
5.4
93
Moist air is found to contain 8.1 grains of water vapour per ft3 at
30 °C. Calculate the temperature to which it must be heated in order
that its relative saturation will be 15% (7000 grains = 1 lb)
Basis: 1 ft3 of water vapour–air mixture
8.1
= 1.16 ´ 10–3 lb º 6.42 ´ 10–5 lb moles
7000
Pure component volume of water at 30 °C
Amount of water =
È 303 Ø
= 6.42 ´ 10–5 ´ 359 ´ É
= 0.0256 ft3
Ê 273 ÙÚ
È 0.0256 Ø
Partial pressure of water vapour = 760 ´ É
= 19.4 mm Hg
Ê 1.00 ÙÚ
19.4
= 130 mm Hg
0.15
This corresponds to a temperature of 57 °C.
Vapour pressure of water vapour =
5.5
A stream of gas at 70 °F and 14.3 psi and 50% saturated with water
vapour is passed through a drying tower, where 90% of water vapour
is removed. Calculate the pound of water removed per 1000 ft3 of
entering gas. Vapour pressure of water at 70 °F = 0.36 psi.
Basis: 1000 ft3 of entering gas 70 °F and 14.3 psi
È 1000 Ø È 14.3 Ø È 492 Ø
lb mole of gas mixture = É
–
–
Ê 359 ÙÚ ÉÊ 14.67 ÙÚ ÉÊ 530 ÙÚ
= 2.52 lb moles
n
yp = 0.5 = v
ns
where nv = lb mole of water/lb mole of vapour free gas.
Ë 0.36 Û
nv = 0.5 ´ ns = 0.5 ´ Ì
Ü
Í 14.3 0.36 Ý
= 0.013 mole of water/mole of dry gas
È 0.013 Ø
= 0.032 lb mole
Amount of water entering = 2.52 ´ É
Ê 1.013 ÙÚ
Amount of water removed = 0.032 ´ 0.9 = 0.0288 lb mole
Weight of water removed = (0.0288 ´ 18) = 0.518 lb.
5.6
A mixture of benzene vapour and air contains 10.1% benzene by volume.
(a) Calculate the dew point of the mixture when at a temperature of
25 °C and 750 mm Hg.
94
PROCESS CALCULATIONS
(b) Calculate the dew point when the mixture is at a temperature
of 30 °C and 750 mm Hg.
(c) Calculate the dew point when the mixture is at a temperature of
30 °C and 700 mm Hg.
Basis: 1 mole of mixture.
(a) Partial pressure of benzene = 0.101 ´ 750 = 75.7 mm Hg
From the vapour pressure data for benzene, it is found that this
pressure corresponds to a temperature of 20 °C, the dew point.
(b) Partial pressure of benzene remains same, i.e. 75.7 mm Hg. Hence
dew point also remains same, i.e. 20 °C (only temperature of gas
mixture has changed.)
(c) Partial pressure of benzene = 0.101 ´ 700 = 70.7 mm Hg.
Dew point for this partial pressure is 18.7 °C. From these results,
it is seen that the dew point does not depend on the temperature
but does vary with total pressure.
5.7
It is proposed to recover acetone which is used as a solvent in an
extraction process by evaporation into a stream of nitrogen. The
nitrogen enters the evaporator at 30 °C containing acetone such that
its dew point is 10 °C. It leaves at a temperature of 25 °C with a dew
point of 20 °C. The atmospheric pressure is 750 mm Hg.
(a) Calculate the vapour concentration of the gases entering and
leaving the evaporator expressed in moles of vapour/mole of
vapour free gas.
(b) Calculate the moles of acetone evaporated per mole of the vapour
free gas passing through the evaporator.
(c) Calculate the weight of acetone evaporated per 1000 ft3 of gases
entering.
(d) Calculate the volume of gases leaving the evaporator per 1000 ft3
of gases entering.
Data: Vapour pressure of acetone:
At 10 °C = 116 mm Hg and at 20 °C = 185 mm Hg.
(a) Entering gases: partial pressure of acetone = 116 mm Hg
partial pressure of N2 = 634 mm Hg
116
mole of acetone/mole of N2 =
= 0.183
634
Leaving gases: partial pressure of acetone = 185 mm Hg
partial pressure N2 = 750 – 185 = 565 mm Hg
185
mole of acetone/mole of N2 =
= 0.328
565
PSYCHROMETRY
95
Basis: for (b), (c) and (d): 1 lb mole of N2 passing through evaporator.
(b) Moles of acetone evaporated = (0.328 – 0.183) = 0.145
(c) Total moles entering = (1 + 0.183) = 1.183 lb moles
Ê 760 ˆ Ê 303 ˆ
¥
Volume of entering gas = 1.183 ¥ 359 ¥ Á
= 477 ft3
Ë 750 ˜¯ ÁË 273 ˜¯
Weight of acetone evaporated = (0.145 ¥ 58) = 8.4 lb
8.4 ¥ 1000
= 17.6 lb
477
(d) Total moles of gas leaving = (1 + 0.328) = 1.328 lb moles
Acetone evaporated per 1000 ft3 =
Ê 760 ˆ Ê 303 ˆ
Volume of gases leaving = 1.328 ¥ 359 ¥ Á
¥
Ë 750 ˜¯ ÁË 273 ˜¯
= 526 ft3
Volume of gases leaving per 1000 ft3 =
5.8
526 ¥ 1000
= 1102 ft3
477
Air at a temperature of 20 °C and pressure 750 mm Hg has a relative
humidity of 80%
(a) Calculate the molal humidity of air.
(b) Calculate the humidity of this air if its temperature is reduced to
10 °C and its pressure increased to 35 psi, condensing out some
of the water.
(c) Calculate the weight of water condensed from 1000 ft3 of gas.
(d) Calculate the final volume of wet air leaving.
Data: Vapour pressure of H2O at 20 °C =17.5 mm Hg and at
10 °C = 9.2 mm Hg.
(a) Initial partial pressure of water = (17.5 ¥ 0.8) = 14 mm Hg
14
750 - 14
= 0.019 kmole of water vapour (w.v.)/kmole of dry air (d.a.)
Initial humidity =
(b) Final partial pressure of water = 9.2 mm Hg (saturated)
Final total pressure =
35 ¥ 760
= 1810 mm Hg
14.7
9.2
È
˘
Final humidity = Í
˙
Î 1810 - 9.2 ˚
= 0.0051 kmole of water vapour/kmole of dry air
3
Basis: 1000 ft of wet air, 20 °C and 750 mm Hg.
96
PROCESS CALCULATIONS
(c) Volume at standard condition and hence weight of mixture in lb mole
È 1000 Ø È 750 Ø È 273 Ø
ÉÊ
Ù –É
Ù –É
Ù = 2.56 lb moles
359 Ú Ê 760 Ú Ê 293 Ú
È 736 Ø
lb mole of dry air = 2.56 ´ É
= 2.51 lb moles
Ê 750 ÙÚ
lb mole of water vapour initially = (0.019 ´ 2.51)
= 0.0477 lb mole
lb mole of water finally = (0.0051 ´ 2.51) = 0.0127 lb mole
\ water condensed = 0.035 lb moles = 0.63 lb.
(d) Total moles finally present = (2.51 + 0.0127) lb moles
= 2.5227 lb moles
È 760 Ø È 283 Ø
–
= 395 ft3
Volume of wet air = 2.5227 ´ 359 ´ É
Ê 1810 ÙÚ ÉÊ 273 ÙÚ
5.9
A mixture of dry flue gases and acetone at a pressure of 750 mm Hg
has a dew point of 25 °C. It is proposed to condense 90% of the
acetone by cooling to 5 °C and compressing. Calculate the final
pressure in psi for acetone.
Vapour pressure at 25 °C = 229.2 mm Hg.
Vapour pressure at 5 °C = 89.1 mm Hg.
Basis: 1 lb mole of mixture of gases.
Partial pressure of acetone = 229.2 mm Hg; (equal to vapour pressure
as it is at dew point)
lb moles of acetone in it = 229.2/750 = 0.306 lb mole
Dry flue gases = (1 – 0.306) = 0.694 lb mole
Acetone finally present = 0.306 ´ 0.1 = 0.0306 lb mole
Final gas = 0.694 + 0.0306 = 0.7246 lb mole
lb mole of acetone in final gas/lb mole of final gas mixture
0.0306
= 0.0422
0.7246
Final pressure of acetone = 89.1 mm Hg
=
89.1
= 2110 mm Hg
0.0422
= 40.8 psi.
5.10 Air at a temperature of 20 oC and 750 mm Hg has a relative humidity
of 80%. Calculate,
\
Final pressure of gas mixture =
(i) The molal humidity of the air
PSYCHROMETRY
97
(ii) The molal humidity of this air if its temperature is reduced to
10 °C and pressure increased to 2000 mm Hg condensing out
some of the water and
(iii) Weight of water condensed from 1000 litre of the original wet air.
Vapour pressure of water at 20 °C = 17.5 mm Hg
Vapour pressure of water at 10 °C = 9.2 mm Hg
At 20 °C and 750 mm Hg,
p
p
yr = 0.8 = v = v
ps 17.5
where, partial pressure of H2O vapour = pv = 14 mm Hg.
(i) Molal humidity at 20 °C, 750 mm Hg =
pv
14
=
p - pv 750 - 14
= 0.019 g mole of water vapour/g mole of dry air
(ii) Molal humidity at 10 °C, 2000 mm Hg
(when saturated, vapour pressure = partial pressure, we have,
ps = pv = 9.2 mm Hg)
È
˘
ps
9.2
=Í
p - ps Î 2000 - 9.2 ˙˚
= 4.6 ¥ 10–3 g mole of water vapour/g mole of dry air
(iii) Basis: 1000 litres of original wet air
Ê 1000 ˆ Ê 750 ˆ Ê 273 ˆ
g mole of wet air = Á
¥
¥
Ë 22.414 ˜¯ ÁË 760 ˜¯ ÁË 293 ˜¯
= 41.02
Ê 1 ˆ
= 40.25
g mole of dry air = 41.02 ¥ Á
Ë 1.019 ˜¯
\ water condensed = (difference in humidity) ¥ (flow rate of dry air)
¥ (molecular weight of water)
= (0.019 – 4.6 ¥ 10–3) ¥ 40.25 ¥ 18 = 10.43 g
5.11 A material is to be dried from 16% moisture by weight (wet basis) to
0.5% by circulation of hot air. The fresh air contains 0.02 kg of
water/kg of dry air. Find the volume of fresh air required if 1000 kg/h
of dried material is to be produced. The exit humidity of air is 0.09 kg
of water/kg of dry air. The air enters at 301 K and at atmospheric
pressure.
98
PROCESS CALCULATIONS
301 K Air 0.02
16% wet solid
0.09
Drier
Dried solid 0.5%
Basis: 1000 kg/h product.
Dry solid = 995 kg
Moisture = 5 kg
\
feed =
995
= 1184.5 kg
0.84
Water in feed = 189.5 kg
Dry solid is the Tie Element
Water removed = 1184.5 – 1000 = 184.5 kg
1 kg of dry air removes = (0.09 – 0.02) = 0.07 kg H2O
\
Weight of dry air needed =
184.5
= 2636 kg/h
0.07
2636 Ø
Dry air needed = ÈÉ
= 91.40 kmoles
Ê 28.84 ÙÚ
2636 – 0.02
= 2.93 kmoles
18
Total moles of wet air = 94.33 kmoles
Water in entering air =
\
301 Ø
= 2331.17 m3
Volume of wet air = 94.33 ´ 22.414 ´ ÈÉ
Ê 273 ÙÚ
5.12 A tunnel drier is used to dry an organic paint. 1000 lb/h of feed
having 10% water is to be dried to 0.5%. Air is passed counter current
to the flow of paint. Air enters at 760 mm Hg. 140 °F and 10%
humidity while it leaves at 750 mm Hg 95 °F and 70% humidity.
What flow rate of air is needed?
Air 10%
Dry product 0.5%
Air 70%
Drier
1000 lb/h; 10% moisture
Basis: 1000 lb/h of feed
Dry material = 900 lb
Moisture = 100 lb
900
Product =
= 904.5 lb
0.995
PSYCHROMETRY
99
Water removed = (1000 – 904.5) = 95.5 lb/h
From humidity chart the following details are obtained:
Humidity of entering air = 0.025 lb mole water vapour/lb mole dry air
Humidity of leaving air 0.035 lb mole water vapour/lb mole dry air
1 lb mole of dry air removes 0.01 lb mole H2O = 0.18 lb
È 95.5 Ø
= 530.5 lb mole/h
Dry air needed = É
Ê 0.18 ÙÚ
\
Total air entering = 530.5 ´ 1.025 = 543.77 lb mole/h
È 600 Ø
Volumetric flow rate of air = (543.77 ´ 359) ´ É
Ê 492 ÙÚ
= 2,38,065 ft3/h.
5.13 Air at 100 oF and 20% relative humidity is forced through a cooling
tower. The air leaves at 85 °F and 70% relative humidity. Atmospheric
pressure is 29.48 inches of Hg. Calculate the following.
(a) The pounds of water evaporated per pound of dry air.
(b) Volume of air leaving per 1000 ft3 of entering air.
Vapour pressure of water at 100 °F = 1.9325 inches of Hg and at 85°F
= 0.8754 inches of Hg.
(a) Partial pressure of H2O in entering air = 1.9325 ´ 0.2
= 0.3865 inches of Hg
Partial pressure of H2O in exit air = 0.8754 ´ 0.7
= 0.6128 inches of Hg
(lb mole H2O/lb mole dry air) in entering air
=
0.3865
= 0.01328
29.48 0.3865
(lb mole H2O/lb mole dry air) in outgoing air
=
0.6128
= 0.02123
29.48 0.618
lb mole H2O evaporated/lb mole of dry air = 0.00795
È 18 Ø
lb H2O/lb of dry air = 0.00795 ´ É
= 5 ´ 10–3 lb = 0.005 lb
Ê 28.84 ÙÚ
È 1000 Ø È 43.2 Ø
(b) 1000 ft3 of entering air = É
= 2.447 lb moles
–
Ê 359 ÙÚ ÉÊ 560 ÙÚ
100
PROCESS CALCULATIONS
For 1.01328 lb moles of wet air entering, 1.02123 lb moles of wet
air leaves
\
for 2.447 lb moles of wet air entering
= 2.447 ´
1.02123
= 2.466 lb moles of wet air leaves
1.01328
È 535 Ø
= 962.8 ft3
Volume of air leaving = 2.466 ´ 359 ´ É
Ê 492 ÙÚ
5.14 Leather containing 100% of its own weight of water is dried by means
of air. The dew point of entering air is 40 °F while that of leaving air
is 55 °F. If 2000 lb of wet air is forced through the drier per hour,
how many lb of water is removed per hour. Total pressure 750 mm Hg.
Vapour pressure of water at 40 °F = 6.3 mm Hg
and at 55 °F = 11 mm Hg.
Basis: One lb mole of dry air
Water in entering air
=
6.3
750 6.3
= 0.00847 lb mole of water vapour/lb mole of dry air
= 0.00527 lb of water vapour/lb of dry air
11
Water in leaving air =
750 11
= 0.01488 lb mole of water vapour/lb mole of dry air
= 0.00926 lb of water vapour/lb of dry air
\
water vapour removed = 0.00926 – 0.00527
= 0.00399 lb of water vapour/lb of dry air
È 2000 Ø
Dry air entering per hour = É
= 1989.52 lb
Ê 1.00527 ÙÚ
Weight of water removed/h = 1989.52 ´ 0.00399 = 7.938 lb.
5.15 Acetone is used as a solvent in a certain process. Recovery of the
acetone is accomplished by evaporation into a stream of N2 followed
by cooling and compression of the final gas mixture. In the solvent
recovery unit 50 lb of acetone are removed per hour. The N2 is
admitted at 100 °F and 750 mm Hg partial pressure of acetone in
incoming nitrogen is 10 mm Hg. The nitrogen leaves at 85 °F, 740
mm Hg and 85% saturation.
PSYCHROMETRY
101
(a) How many ft3 of incoming gas must be admitted per hour to
obtain the required rate of evaporation of the acetone?
(b) How many ft3 of gases leave the unit per hour?
N2, Acetone 100 °F 750 mm Hg
partial pressure of acetone 10 mm Hg
Drier
N2, Acetone
85°F, 740 mm Hg
85% saturation
(Molecular weight of Acetone (CH3)2CO is 58. Vapour pressure of
acetone at 85 °F = 287 mm Hg)
Basis: 100 lb moles of gas mixture entering
Acetone removed =
50
= 0.862 lb mole
58
Ë 750 10 Û
N2 in entering mixture = Ì
Ü ´ 100 = 98.67 lb moles
Í 750 Ý
Acetone entering = 100 – 98.67 = 1.33 lb moles
1.33 Ø
lb mole of acetone/lb mole of N2 in entering stream = ÈÉ
Ê 98.67 ÙÚ
= 0.0135
Ë
Û
pv
yp = 0.85 = Ì
Ü
Í 740 pv Ý
Ë 287 Û
Ì 740 287 Ü
Í
Ý
where
yp (percentage saturation) = 85%
pv (partial pressure)
= 258 mm Hg
258
740 258
= 0.5385
lb mole of acetone/lb mole of N2 in leaving stream =
\ 1 lb mole of N2 removes: (0.5385 – 0.0135)
= 0.525 lb mole of acetone
0.862
= 1.642 lb mole
0.525
Total moles of gas entering = 1.642 ´ 1.0135 = 1.664 lb moles
lb mole N2 needed/h =
È 560 Ø È 760 Ø
–
(a) Volume of gases admitted = 1.664 ´ 359 ´ É
Ê 492 ÙÚ ÉÊ 750 ÙÚ
= 689.0 ft3
102
PROCESS CALCULATIONS
(b) Total moles of exit gas = 1.642 ¥ 1.5385 = 2.526 lb moles
Ê 545 ˆ Ê 760 ˆ
Volume of exit gases = 2.526 ¥ 359 ¥ Á
¥
Ë 492 ˜¯ ÁË 740 ˜¯
= 1031.6 ft3
5.16 By adsorption in silica gel you are able to remove all the weight
(0.93 kg) of water from moist air at 15 °C and 98.6 kPa. The same
air measures 1000 m3 at 20°C and 108 kPa when dry. What was
the relative humidity of the air?
Basis: 1000 m3 of dry air at the given conditions.
108
Ê
ˆ
¥ 293 ¥ 100˜ = 44.895
kmole of the dry air: Á 1000 ¥ 273 ¥
Ë
¯
22.414
0.93 ˆ
= 0.052 kmole
Water removed: 0.93 kg = ÊÁ
Ë 18 ˜¯
Total wet air = 44.895 + 0.052 = 44.947 kmoles
98.6 ˆ
Ê
pv = Partial pressure of water = Á 0.052 ¥
˜ = 0.114 kPa
Ë
44.947 ¯
ps = Vapour pressure of water = 15 mm Hg = 2 kPa
Q Relative humidity, yr =
pv
Ê 0.114 ˆ
, yr = Á
¥ 100 = 5.7%,
Ë 2 ˜¯
ps
5.17 At 297 K and 1 bar, the mixture of N2 and C6H6 has a relative
humidity of 60%. It is desired to recover 90% C6H6 present by
cooling to 283 K and compressing to a suitable pressure. What is the
pressure to which the mixture should be pressurised? Vapour pressures
of benzene at 283 K and 297 K are 6 kN/m2 and 12.2 kN/m2
respectively.
Relative humidity = 0.6 = Partial pressure/Vapour pressure
Partial pressure =
Humidity =
12.2 × 60 × 10 3
= 7.32 × 103 N/m2
100
7.32 × 10 3
101.3 × 10 − 7.32 × 10
3
3
×
78
= 0.217 kg of w.v./kg of d.a.
28
Final humidity = 0.10 × 0.217 = 0.0217 kg of w.v./kg of d.a.
0.0217 =
6 × 10 3
PT − 6 × 10
3
×
78
28
Therefore, PT, the total pressure = 7.642 × 105 N/m2
PSYCHROMETRY
103
5.18 Humid air at 75 °C, 1.1 bar and 30% relative humidity is fed at a rate
of 1000 m3/h. Determine the molar flow rates of water and dry air
entering the process, molar humidity, absolute humidity, percentage
humidity and dew point.
Vapor pressure at 75 °C = 289 mm Hg
Partial pressure of water vapour
= Vapour pressure of water × Relative humidity
= 86.7 mm Hg
Total pressure = 1.1 bar = 1.1 × 105 N/m2
1 atm = 760 mm Hg = 1.013 × 105 N/m2
Therefore, 1.1 bar = 825.36 mm Hg
Mole fraction of water vapour = (86.7/825.36) = 0.105
Volumetric flow rate = 1000 m3/h at 75 °C and 1.1 bar
1000 t 825 273
1
t
t
38 kmoles/h
348
760 22.414
Water flow rate = 38 × 0.105 = 3.99 kmoles/h
Dry air rate = 38 – 3.99 = 34.01 kmoles/h
86.7
= 0.117 kmole w.v./kmole d.a.
Humidity =
825.36 86.7
18
Mass humidity = 0.117 ×
= 0.073 kg w.v./kg d.a.
28.84
Percentage humidity = 0.117 ×
100
= 21.7%
289 /(825 289)
Dew point = 49 °C
5.19 A piece of wood entering furnace at 0.15 kg/s containing 60% moisture is
dried in a countercurrent dryer to give a fixed product containing 5%
moisture content. The air used is at 100 °C and is having water vapour at
a partial pressure of 103 N/m2. The air leaves the dryer at 70% RH and at
40 °C. Estimate the air required to remove the moisture. The vapour
pressure of water at 40 °C = 7.4 × 103 N/m2
Water in incoming wood = 0.15 × 0.6 = 0.09 kg/s
Weight of dried wood = 0.15 × 0.4 = 0.06 kg/s
Weight of water in dried wood is 5%
0.06
= 0.0632 kg/s
0.95
Weight of water removed = 0.15 – 0.0632 = 0.0868 kg/s
Weight of final product =
104
PROCESS CALCULATIONS
Humidity of entering air =
103
1.013 – 10 5 103
18
28.84
= 0.00623 kg w.v./kg d.a.
= 9.97 × 10–3 ×
Humidity of leaving air,
Relative humidity = Partial pressure/Vapour pressure
0.7 = Partial pressure/7.4 × 103
Partial pressure = 5.18 × 103 N/m2
Humidity = 5.18 ×
103
101.3 – 103 5.18 – 103
= 0.0336 kg w.v./kg d.a.
Increase in humidity = 0.0336 – 0.00662 = 0.02738 kg w.v./kg d.a.
Weight of water in kg to be removed
Water removed per kg of dry air
0.0868
=
= 3.17 kg/s
0.02738
Weight of wet air needed = 3.17 × (1.00622) = 3.1897 kg/s
Dry air needed =
Weight of wet air leaving = (3.17) × (1.0336) = 3.277 kg/s
5.20 Air at 740 mm Hg at 30 °C contains water vapour at a partial pressure
of 22 mm Hg. 1000 m3 of the air is cooled to 15 °C and the partial
pressure of water vapour is brought to 12.7 mm Hg. During this
process, water gets partially condensed. Estimate the amount of air at
new condition
Moles of humidified air at initial condition
Ë 1000 – 760 – 303 Û
= Ì
Ü = 39.14 kmoles
Í 22.414 – 740 – 273 Ý
22
= 0.02973
740
Mole fraction of water vapour in original air when cooled to 15 °C
Mole fraction of water vapour in original air (30 °C) =
12.7
= 0.017
740
Therefore, mole fraction of dry air = 0.983
=
Moles of wet air at 15 °C =
37.976
= 38.63 kmoles
0.983
PSYCHROMETRY
105
Volume of dry air at new condition (assuming ideal behaviour)
760 Û Ë 288 Û
Ë
3
= Ì38.63 – 22.414 –
Ü – Ì 740 Ü = 938.11 m /kg dry air
273
Í
Ý Í
Ý
Water removed is (39.14) × (0.02973) – (38.63 × 0.017)
= 0.507 kmole
= 0.547 × 18 kg
= 9.126 kg
5.21 Air with humidity of 0.004 kg of water vapour per kg of dry air is
bubbled through water at a rate of 0.1 kg/s. The air leaves with a
humidity of 0.014 kg of water vapour per kg of dry air. Estimate the
time needed to vaporize 0.1 m3 of water.
Weight of water evaporated = 0.1 × 1000 = 100 kg (since the density
is 1000 kg/m3)
Weight of water removed/kg of dry air = 0.014 – 0.004 = 0.01 kg
Water removed per second by using 0.1 kg of air
= 0.01 × 0.1 = 0.001 kg/s
Hence, time needed to evaporate 100 kg of water
=
100
= 100000 s
0.001
100000
= 27.7 h
3600
5.22 Air containing 0.01 kg of water vapour per kg of dry air is passed
through a bed of silica gel to get air containing 0.001 kg of water
vapour per kg of dry air for a specific application. The column has
2.1 kg of silica gel initially and after 5 h of operation, the weight of
silica gel is found to be 2.2 kg. Calculate the flow rate of air both at
inlet and outlet.
=
Water adsorbed by silica gel in 5 h = 2.2 – 2.1 = 0.1 kg
Water adsorbed/h =
2.2 2.1
= 0.02 kg/h
5
Making a balance for water:
Water in incoming air = Water in leaving air + Water adsorbed
Let G be the flow rate of dry air in kg/h.
(G) (0.01) = (G) (0.001) + 0.02
G (0.01 – 0.001) = 0.02
106
PROCESS CALCULATIONS
0.02
= 2.222 kg/h
0.009
i.e. the flow rate of dry air in kg/h = 2.222 kg/h
G =
Flow rate of wet air at inlet = 2.222[1 + 0.01] {since G = Gs (1 + x)}
= 2.244 kg/h
Flow rate of wet air at outlet = 2.222 (1 + 0.001)
= 2.224 kg/h
EXERCISES
5.1
Toluene–air mixture is available such that the partial pressure of the
vapour is 1.33 kPa. The total pressure is 99.3 kPa and the temperature
is 21 °C. Calculate (a) the relative humidity, (b) The moles of toluene
per mole of vapour free gas, (c) the weight of toluene per unit weight
of vapour free gas, (d) the percent saturation and (e) the percentage of
toluene by volume.
5.2
Air is available at a DBT of 50 °C and a wet bulb temperature of
30 °C. Estimate its humidity, % saturation and humid volume using
humidity chart.
5.3
A material is to be dried from 20% moisture by weight (wet basis) to
1.0% by circulation of hot air. The fresh air contains 0.02 kg of
water/kg of dry air. Find the volume of fresh air required if
1000 kg/h of dried material is to be produced. The exit humidity of air
is 0.09 kg water vapour/kg dry air. The air enters at 300 K and at
atmospheric pressure.
5.4
A mixture of benzene–air is saturated at 1 atm and 50 °C. The vapour
pressure of benzene is 275 mm Hg at 50 °C. Estimate the mass
humidity and molal humidity.
5.5
A mixture of benzene–air is available at 750 mm Hg and 60 °C. The
partial pressure of benzene is 150 mm Hg at 50 °C. Estimate the mass
humidity and molal humidity.
5.6
Air-water vapour mixture is available at a DBT of 35 °C and a
relative saturation of 70%. Estimate its humidity, dew point, humid
volume, adiabatic saturation temperature, humid heat and enthalpy.
5.7
Air at 50% relative humidity (RH) is to be used for a specific
operation. Air is available at a DBT of 35 °C and 27 °C. What should
be the final temperature to which it should be heated? It is found that
a DBT of 30 °C will be quite comfortable to the labourers. How will
you obtain this condition? Indicate the exact temperature to which it
PSYCHROMETRY
107
should be taken before bringing it to 30 °C and 50% RH. If it is
necessary to use 5000 m3/h of air at the final condition mentioned
above, estimate the quantity of air needed at its original condition.
5.8
Air at 80% saturation is to be maintained in a weaving room. Air at a
DBT of 41 °C and WBT of 24 °C is available. This air for weaving
room is obtained by saturating it initially and then heating it to 80%
saturation. Estimate the final temperature of the air and its humidity.
5.9
Air at a temperature of 60 °C and a pressure of 745 mm Hg and a
% humidity of 10 is supplied to a drier at a rate of 120 m3/h. Water is
evaporated in the drier at a rate of 25 kg/h. The air leaves at a
temperature of 35 °C and a pressure of 742 mm Hg. Calculate
(a) percentage humidity of air leaving the drier and (b) volume of wet
air leaving the drier per hour. (use vapour pressure data from Steam
Tables).
5.10 Air at a temperature of 30 °C and a pressure of 100 kPa has a relative
humidity of 80%. Calculate (a) the molal humidity of air, (b) molal
humidity of this air if its temperature is reduced to 15 °C and its
pressure increased to 200 kPa, condensing out some of the water,
(c) weight of water condensed from 100 m3 of the original wet air in
cooling to 15 °C and compressing to 200 kPa, and (d) the final
volume of the wet air of part (c)
Data: Vapour pressure of water at 30 °C = 4.24 kPa
Vapour pressure of water at 15 °C
= 1.70 kPa
5.11 Air at 85 °C and absolute humidity of 0.03 kg water vapour per kg
dry air, 1 std. atm is contacted with water at the adiabatic saturation
temperature and is thereby humidified and cooled to 70% saturation.
What are the final temperature and humidity of the air?
5.12 An air-water vapour sample has a dry bulb temperature (DBT) of
55 °C and an absolute humidity 0.030 kg water/kg dry air at 1 std.
atm pressure. Using humidity chart, calculate (a) the relative humidity,
if vapour pressure of water at 55 °C is 118 mm Hg and (b) the humid
volume in m3/kg dry air.
5.13 A drier is used to remove 100 kg of water per hour from the material
being dried. The available air has a humidity of 0.010 kg bone dry air,
and a temperature of 23.9 °C and is heated to 68.3 °C before entering
the drier. The air leaving the drier has a wet-bulb temperature of
37.8 °C and a dry-bulb temperature of 54.4 °C. Calculate (a) air
consumption rate, (b) humid volume of air before and after preheating,
(c) wet-bulb temperatures (WBT) of air before and after preheating,
and (d) dew point of the air leaving the drier.
108
PROCESS CALCULATIONS
5.14 The air supply for a drier has a dry-bulb temperature of 23 °C and a
wet-bulb temperature of 16 °C. It is heated to 85 °C by heating coils
and introduced into the drier. In the drier, it cools along the adiabatic
cooling line and leaves the drier fully saturated.
(a) What is its humidity?
(b) What is the dew point of the initial air?
(c) How much water will be evaporated per 100 cubic metre of
entering air?
(d) How much heat is needed to heat 100 cubic metre air to 85 °C?
(e) At what temperature does the air leave the drier?
5.15 A mixture of carbon disulphide vapour and air contains 23.3% carbon
disulphide (CS2) by volume. Calculate the relative saturation and
percent saturation of the mixture at 20 °C and 740 mm Hg pressure.
Vapour pressure of CS2 at 20 °C = 300 mm Hg
5.16 Air at 60 °C and 745 mm Hg having a percent humidity of 10% is
supplied to a drier at the rate of 1100 m3/h. Water at the rate of
20 kg/h is to be removed from the wet material. The air leaves the
drier at 35 °C and 742 mm Hg. Calculate (a) percent humidity of
leaving air and (b) volumetric flow rate of leaving air.
5.17 Air is available at a DBT of 310 K and a WBT of 305 K. Estimate
humidity, % saturation, dew point, humid volume and enthalpy using
humidity chart. Estimate humid volume and enthalpy using equations
and compare the results.
5.18 One thousand m3/min of methane saturated with water vapour at
1 atm, 49 °C is cooled to 10 °C, so that part of water vapour is
condensed. The mixture is then reheated to 24 °C at 1 atm. What is
(a) the volumetric flow rate of leaving mixture, and (b) amount of
water condensed?
Data: Vapour pressure of water at 49 °C, 24 °C and 8 °C are 100 mm
Hg, 27 mm Hg and 8 mm Hg respectively.
5.19 Air at a temperature of 30 °C and pressure 760 mm Hg has a relative
humidity of 60%, calculate:
(a) The absolute humidity of air
(b) The humidity of this air if its temperature is reduced to 20 °C and
the pressure increased to 2600 mm Hg, condensing out some of
the water.
(c) The weight of water condensed from 1000 m3 of the original
wet air. VP of water at 30 °C and 20 °C are 32 mm Hg and
17.5 mm Hg respectively.
PSYCHROMETRY
109
5.20 A mixture of air water vapour is present at 40 °C and 762 mm Hg.
The volume of the mixture is 5 litres and the partial pressure of water
vapour in the mixture is 27.7 mm Hg. The mixture is heated to 60 °C
at constant volume. Compute the following:
(a) Partial pressure of water vapour
(b) Partial pressure of dry air
(c) Total pressure, and
(d) Molal humidity.
5.21 A mixture of acetone vapour and nitrogen contains 15% acetone by
volume. Calculate molal humidity, partial pressure of acetone, relative
saturation and % saturation at 20 °C and 745 mm Hg. Vapour pressure
of acetone at 20 °C = 184.8 mm Hg.
5.22 Air at 60 °C and pressure of 745 mm Hg and a % humidity of 20
is supplied to a dryer at 1500 m3/h. Water is removed at a rate of
2.5 kg/h. Air leaves at 35 °C at 742 mm Hg, vapour pressure at 35 °C
= 43 mm Hg, and vapour pressure at 60 °C = 149.38 mm Hg
Estimate:
(a) Percentage humidity of air leaving
(b) Volumetric flow rate of wet air leaving
110
PROCESS CALCULATIONS
Psychrometric chart for air–water system at 760 mm Hg pressure.
6
Crystallization
Crystallization is a process in which the solid particles are formed from
a homogeneous phase. During crystallization, the crystals form when a
saturated solution gets cooled. The solution left behind after the separation
of crystals is known as mother liquor which will also be a saturated
solution. The mixture of crystals and mother liquor is known as magma.
A simple illustration of a crystallization process is shown below.
Water evaporated, E kg/h
xE
Feed, F kg/h
xF
Crystallizer
Mother liquor, M kg/h
xM
Crystal, C kg/h
xC
Feed rate of solution of concentration, xF
: F kg/h
Amount of water evaporated
(xE = 0 as salt in water evaporated will be zero)
: E kg/h
Amount of mother liquor of concentration xM
: M kg/h
(Mother liquor will always be a saturated solution)
Weight of crystal formed of concentration, xC
Total mass balance gives, F = C + M + E
Component balance gives, FxF = CxC + MxM + ExE
: C kg/h
(6.1)
(6.2)
Since we normally know F, E, xF xM xE we can find C and M by solving the
Eqs. (6.1) and (6.2). xE will always be zero as it is free from salt.
When we get anhydrous salt, xC is 1.0. Some times we get hydrated
salt during crystallization process. Under such circumstances xC will not be
1.0 but will be less than unity and this can be obtained by molecular weight
of anhydrous salt divided by the molecular weight of hydrated salt. The
above principles are explained through the following problems.
111
112
PROCESS CALCULATIONS
WORKED EXAMPLES
6.1
A solution of sodium chloride in water is saturated at a temperature of
15 °C. Calculate the weight of NaCl that can be dissolved by 100 kg
of this solution if it is heated to 65 °C.
Solubility of NaCl at 15 °C = 6.12 kg mole/1000 kg H2O
Solubility of NaCl at 65 °C = 6.37 kg mole/1000 kg H2O
Basis: 1000 kg of water
NaCl in saturated solution at 15 °C = (6.12 ¥ 58.5) = 358 kg
NaCl in saturated solution at 65 °C = (6.37 ¥ 58.5) = 373 kg
NaCl that may be dissolved further = (373 – 358) = 15 kg
Ê 1000 ˆ
Water in 100 kg of saturated solution at 15 °C = 100 ¥ Á
Ë 1358 ˜¯
= 73.6 kg
\ amount of additional NaCl that will be dissolved
15 ¥ 73.6
= 1.1 kg
1000
After a crystallization process, a solution of calcium chloride in water
contains 62 kg CaCl2 per 100 kg of water. Calculate the weight of this
solution necessary to dissolve 250 kg of CaCl2 ◊ 6H2O at 25 °C
(solubility = 7.38 kg mole CaCl2/1000 kg H2O)
=
6.2
Basis: x kg of water is present in original solution
CaCl2 ◊ 6H2O has the molecular weight = 111 + 108 = 219
250 kg of solution will have 126.7 kg of CaCl2 and 123.3 kg of 6H2O
Total CaCl2 entering the process = (0.62x + 126.7) kg
(1)
Water entering the process = (x + 123.3) kg
Solubility of CaCl2 at saturated condition for 1000 kg water
= 7.38 kmoles = 819.18 kg
\ CaCl2 leaving the process along with (x + 123.3) kg of water
Ê 819.18 ˆ
= Á
(x + 123.3) kg
Ë 1000 ˜¯
Equating (1) and (2), we get
(0.62x + 126.7) = 0.81918 (x + 123.3)
0.62x + 126.7 = 0.81918x + 101 – 0.19918x = – 25.7
\
x = 129.03 kg
(2)
CRYSTALLIZATION
113
100 kg of water is equivalent to 162 kg of solution
162
100
= 209 kg of solution
129.03 kg of water is equivalent to 129.03 ´
Original solution needed = 209 kg.
6.3
A solution of sodium nitrate in water at a temperature of 40 °C
contains 49% NaNO3 by weight.
(a) Calculate the percentage saturation of this solution
(b) Calculate the weight of NaNO3 that may be crystallized from
1000 kg of solution by reducing the temperature to 10 °C
(c) Calculate the percentage yield of the process.
Solubility of NaNO3 at 40 °C = 51.4% by weight.
Solubility of NaNO3 at 10 °C = 44.5% by weight.
Basis: 1000 kg of original solution
(a)
kg of NaNO3
kg of water
49
= 0.96
51
È kg of NaNO3 Ø
at 40 °C, É
Ê kg of water ÙÚ saturation
51.4
= 1.06
48.6
È 0.96 Ø
% saturation of given solution = É
´ 100 = 90.8%
Ê 1.06 ÙÚ
(b) Let a kg be NaNO3 crystallized out
NaNO3 balance = (1000 ´ 0.49) = (1000 – a) ´ (0.445) + a;
Solving, a = 81 kg
È 81 Ø
(c) % Yield = É
100 = 16.53%
Ê 490 ÙÚ
6.4
A solution of K2Cr2O7 in water contains 13% by weight. From
1000 kg of this solution is evaporated 640 kg of water. The remaining
solution is cooled to 20 °C. Calculate the amount and the percentage
yield of K2Cr2O7 crystals formed.
Solubility at 20 °C = 0.39 kmole/1000 kg H2O
Basis: 1000 kg of original solution
K2Cr2O7 = 130 kg;
Water = 870 kg
Water present finally = (870 – 640) = 230 kg
114
PROCESS CALCULATIONS
0.39 Ø
K2Cr2O7 present then = 230 ´ ÈÉ
´ 294 = 26.4 kg
Ê 1000 ÙÚ
\
K2Cr2O7 crystallized = 130 – 26.4 = 103.6 kg
103.6
= 79.7%
130
A solution of sodium nitrate in water contains 100 g of salt per 1000 g of
water. Calculate the amount of ice formed in cooling 1000 g of this
solution to –15 °C (solubility at –15 °C = 6.2 g mole/1000 g H2O)
% Yield =
6.5
Basis: 1000 g of original solution
È 100 Ø
= 91 g
NaNO3 present initially = 1000 ´ É
Ê 1100 ÙÚ
Water present initially = 909 g
1000 – 91
= 172.6 g
Water retained along with NaNO3 =
6.2 – 85
\ Ice formed = Original water – water retained along with NaNO3
= (909 – 172.6) = 736.4 g
6.6
1000 kg of a 30% aqueous solution of Na2CO3 is slowly cooled to
20 °C. During cooling, 10% water originally present evaporates. The
crystal is Na2CO3× 10H2O. If the solubility of anhydrous Na2CO3 at
20 °C is 21.5 kg/100 kg of water, what weight of salt crystallizes out?
Basis: 1000 kg of solution.
Salt = 300 kg
Water = 700 kg
Water evaporated = 700 ´ 0.1 = 70 kg
Let a kg of salt crystallize out. Let the weight of mother liquor be b kg
Molecular weight of Na2CO3 = 106 and that of Na2CO3 × 10H2O = 286
a + b = (1000 – 70) = 930 kg
106 Ø È
21.5 Ø
È
Na2CO3 balance = É a –
Ù Éb –
Ù = 300 kg
Ê
286 Ú Ê
121.5 Ú
Solving, a = 700 kg; b = 230 kg
Check (water balance):
Ë
È 180 Ø Û Ë
È 100 Ø Û
Ì 700 – ÉÊ 286 ÙÚ Ü Ì 230 – ÉÊ 121.5 ÙÚ Ü = 630 kg
Í
Ý Í
Ý
6.7
A batch of saturated Na2CO3 solution of 100 kg is to be prepared at
50 °C. The solubility is 4.48 g mole/1000 g H2O at 50 °C.
CRYSTALLIZATION
115
(i) If the monohydrate were available, how many kg of water would
be required to form the solution?
(ii) If the decahydrate is available how many kg of salt will be required?
Basis: 100 kg of saturated solution at 50 °C.
Solubility at 50 °C = (4.48 ´ 106) kg Na2CO3/1000 kg H2O
i.e. 474.88 kg Na2CO3/1474.88 kg solution.
100 – 474.88
1474.88
= 32.2 kg
In 100 kg of solution, Na2CO3 present =
water present = (100 – 32.2) = 67.8 kg
Molecular weight of Na2CO3 × H2O = 106 + 18 = 124 kg
Molecular weight of Na2CO3 × 10H2O = 106 + 180 = 286 kg
(i) Water needed for monohydrate = Actually needed-water from
hydrated salt
Ë 32.2 – 18 Û
= (67.8) – Ì
Ü
Í 106 Ý
= 62.33 kg
286
= 86.88 kg
106
Discussion: In case (ii), water available in the salt is
(ii) Weight of decahydrate needed = 32.2 ´
180
= 54.68 kg
106
The additional water needed hence is 13.12 kg, thus making the total
water to 13.12 + 54.68 = 67.8 kg (which corresponds to solubility data)
32.2 ´
6.8
10,000 kg/hr of 6% solution of salt in water is fed to an evaporator.
Saturated solution is produced and some salt crystallizes out. The hot
crystals with some adhering solution are centrifuged to remove some
of the solution. Then the crystals are dried to remove the rest of water.
During an hour test, 837.6 kg of concentrated solution was removed in
evaporator, 198.7 kg of solution was separated in the centrifuge and
361.7 kg of dried crystals got. Previous test on the centrifuge show
that it removed 60% of adhering solution. Calculate (a) the solubility
of the salt; (b) water removed in evaporator; (c) water removed in
drier; and (d) water leaving the evaporator.
Discussion: The solution leaving the evaporator and centrifuge
maintains the same concentration and is saturated. From evaporator,
the various streams leaving are water vapour, saturated solutions and
crystals with adhering solution.
116
PROCESS CALCULATIONS
Basis: One hour.
60% adhering solution (removed in centrifuge) = 198.7 kg
\
Total adhering solution entering centrifuge (along with crystal)
198.7
=
= 331.2 kg
0.6
Water vapour
10,000 kg/hr
Evaporator
Centrifuge
837.6 kg solution
198.7 kg
6% solution
Adhering
solution
+
crystal
Water
Drier
Dry crystal
361.7 kg
Let s be the solubility of salt, which is given by (s kg of salt/kg of solution)
Making salt balance, (10,000 ´ 0.06) = (837.6 + 198.7)s + 361.7
Solving, (a)
s = 0.23 kg of salt/kg of solution.
1 kg solution = 0.23 kg salt/0.77 kg H2O = 0.3 kg salt/kg H2O
(b) Weight of adhering solution entering drier = (331.2 – 198.7)
= 132.5 kg
(c) Water in it (removed in drier) = 132.5 ´ 0.77 = 102 kg (since,
1 kg of solution contains 0.77 kg water and 0.23 kg of salt)
(d) Weight of material entering drier = Final dry salt + Water evaporated
= (361.7 + 102) = 463.7 kg
Weight of material entering centrifuge = (463.7 + 198.7) = 662.4 kg
Weight of water leaving the evaporator = 10,000 – (837.6 + 662.4)
= 8,500 kg
Check:
Water balance = 8,500 + 102 + (837.6 + 198.7) ´ (0.77)
= 8,500 + 102 + 798 = 9,400 kg
6.9
In a solution of naphthalene in benzene, mole fraction of naphthalene
is 0.12. Calculate the weight of this solution necessary to dissolve 100 kg
of naphthalene at 40 °C (solubility of naphthalene is 57% by weight)
100 kg
X kg solution
0.817 wt fr. of benzene
0.183 wt fr. of naphthalene
Tank
(100 + X) kg
0.57 wt. fr. of naphthalene
CRYSTALLIZATION
117
Basis: 1 mole of feed solution
Component
mole
Molecular
weight
Weight, kg
Weight
fraction
Naphthalene (C10H8)
0.12
128
15.4
0.183
Benzene (C6H6)
0.88
78
68.6
0.817
Total
—
—
84.0
1.000
Let us assume that X kg of feed solution enters the tank.
Naphthalene balance in tank gives, 0.183X + 100 = (100 + X)0.57
X = 111 kg of original feed solution.
6.10 A crystallizer is charged with 7,500 kg of aqueous solution at 104 °C
containing 29.6% by weight of anhydrous Na2SO4. The solution is
then cooled to 20 °C. During this operation 5% of water is lost by
evaporation. Glauber salt crystallizes out. Find the yield of crystals.
Solubility at 20 °C = 194 g Na2SO4/100 g water
Molecular weight of Glauber’s salt (Na2SO4 ×10H2O) = 142 + 180 = 322
Basis: 7,500 kg of aqueous solution.
Na2SO4 present = 7500 ´ 0.296 = 2220 kg
H2O present = (7500 – 2220) = 5280 kg
H2O lost in evaporation = 5280 ´ 0.05 = 264 kg
H2O remaining = (5280 – 264) = 5016 kg
Let a kg of Glauber salt crystallize and b kg be the weight of the
solution.
a Ø È
b Ø
È
Na2SO4 balance: 2220 = É142 –
ÙÚ ÉÊ19.4 –
Ù
Ê
322
119.4 Ú
a Ø È
b Ø
È
H2O balance: 5016 = É180 –
Ù É100 –
Ù
Ê
322 Ú Ê
119.4 Ú
Solving, a = 3749.5 kg and b = 3486.5 kg
Weight of Glauber salt = 3749.5 kg
Weight of solution = 3486.5 kg
Check: (total balance) 3749.5 + 3486.5 + 264 = 7500 kg
6.11 An evaporator-crystallizer is to produce 13,000 kg dry salt/h from a
feed solution having 20% NaCl. The salt removed carries 20% of its
weight of brine containing 27% NaCl (All Composition in Weight %).
Calculate the feed rate kg/h.
118
PROCESS CALCULATIONS
Water
20% NaCl
Evaporatorcrystallizer
Feed ‘F’
13,000 kg dry salt/h
Basis: 1 h of operation
13,000 kg of salt with 20% brine
Weight of feed brine = 13,000 ´ 0.2 = 2,600
NaCl in this solution = 2,600 ´ 0.27 = 702
NaCl balance gives, 0.2F = (13,000 + 702)
\ Feed = 13,702/0.2 = 68,510 kg/h.
6.12 5000 kg of KCl are present in a saturated solution at 80 °C. The
solution is cooled to 20 °C in an open tank. The solubilities of KCl at
80 °C and 20 °C are 55 and 35 parts per 100 parts of water.
(a) Assuming water equal to 3% by weight of solution is lost by
evaporation, calculate the weight of crystals obtained.
(b) Calculate the yield of crystals neglecting loss of water by
evaporation; KCl crystallizes without any water of crystals.
3% Water
5000 kg KCl
80 °C
Saturated
solution
Crystallizer
20 °C
Saturated
solution
Crystal
Basis: 5,000 kg of KCl salt in a solution at 80 oC
(a) Solubility at 80 °C = 55 kg salt/100 kg water
or, solubility of KCl = 55 kg/155 kg solution = 0.355 kg/kg solution
5000 kg KCl is equivalent to 5000/0.355 = 140,90.91 kg of solution
Water present = 140,90.91 – 5000 = 9090.91 kg
Water evaporated = 3% weight if solution = 140,90.91 ´ 0.03 = 422.73 kg
Water remaining = 9,090.91 – 422.73 = 8668.18 kg
È 35 Ø
KCl in leaving solution at 20°C = 8668.18 ´ É
= 3033.86 kg
Ê 100 ÙÚ
\
KCl crystallized out = 5000 – 3033.86 = 1966.14 kg
CRYSTALLIZATION
119
(b) KCl in solution at 20 °C = 9090.91 ´ 0.35 = 3181.82 kg
\
KCl crystallized out = 1818.18 kg
100
= 36.37%
5000
6.13 A crystallizer is charged with 6,400 kg of an aqueous solution
containing 29.6% of anhydrous sodium sulphate. The solution is
cooled and 10% of the initial water is lost by evaporation.
Na2SO4.10H2O crystallizes out. If the mother liquor (after crystallization) is found to contain 18.3% Na2SO4, calculate the weight of the
mother liquor.
% Yield of crystals = 1818.18 ´
10% water
6,400 kg
Crystallizer
29.6%
18.3% Na2SO4
Mother liquor ‘M’
Na2SO4 × 10H2O
(142 + 180) = 322
Basis: 6,400 kg solution.
Weight of Na2SO4 = 6,400 ´ 0.296 = 1,894.4 kg
Water present in feed = 4,505.6 kg
Let M be the weight of mother liquor and C the weight of crystal
formed.
Water lost by evaporation = 10% of 4505.6 kg = 450.56 kg
142 Ø
È
Overall balance of Na2SO4 gives, 1894.4 = É C –
Ù + (0.183 ´ M)
Ê
322 Ú
C + M = (6400 – 450.56) = 5949.44 kg
Solving, M = mother liquor = 2826.7 kg
C = Crystal = 3122.7 kg
81.7 Ø È
180 Ø
È
Check: (water balance) 450.56 + É 2826.7 –
ÙÚ ÉÊ 3122.7 –
Ù
Ê
100
322 Ú
= 4505.6 kg
6.14 What will be the yield of Glauber salt if pure 32% solution is cooled
to 20 °C from hot condition? There is no loss of water. The solubility
of Na2SO4 in water at 20 °C is 19.4 g per 100 g of water.
120
PROCESS CALCULATIONS
Feed 32%
Crystallizer
Mother
Liquor, Y
Crystals, X
Basis: 1,000 g of the feed solution.
Na2SO4.10H2O (142 + 180 = 322)
Na2SO4 present in the feed = 320 g
Water present in the feed = 680 g
Overall balance is X + Y = 1000 g
142 X ˆ Ê 19.4Y ˆ
Na2SO4 balance = ÊÁ
= 320
+
Ë 322 ˜¯ ÁË 119.4 ˜¯
Ê 180 X ˆ Ê 100Y ˆ
Water balance = Á
= 680
+
Ë 322 ˜¯ ÁË 119.4 ˜¯
Solving, we get X = Weight of crystals = 566.761 g and
Y = Weight of mother liquor = 433.239 g
Yield of the crystals =
566.761
= 56.68%
1000
EXERCISES
6.1
Feed to a crystallizer contains 6,420 kg of aqueous solution containing
29.6% anhydrous sodium sulphate by weight at 104 °C. The solution
is cooled to 20 °C to crystallize out the desired Glauber’s salt. The
mother liquor is found to contain 16.1% anhydrous sodium sulphate.
Estimate the weight of mother liquor.
6.2
A vacuum crystallizer is to produce 10,000 kg of FeSO4 .5H2O
crystals per hour. The feed solution contains 38.9 parts FeSO4/100
parts water. The feed enters at 70 °C and is cooled to 27 °C. The
solubility at 27 °C is 30.3 parts FeSO4/100 parts water. Determine the
quantity of feed and water lost as vapour.
6.3
10,000 kg of a 30% aqueous solution of Na2CO3 is slowly cooled to
20 °C. During cooling, 3% water originally present evaporates. The
crystal is Na2CO3 .10H2O. If the solubility of anhydrous Na2CO3 at
20 °C is 21.5 kg/100 kg of water, what is the weight of
Na2CO3.10H2O formed?
CRYSTALLIZATION
121
6.4
1000 kg of a solution containing 25% of Na2CO3 by weight is slowly
cooled to 20 °C. During cooling 15% water originally present
evaporates. The crystal formed is Na2CO3.10H2O. If the solubility of
anhydrous Na2CO3 at 20 °C is 21.5 kg/100 kg of water, what is the
weight of Na2CO3.10H2O crystals formed?
6.5
1500 kg of a solution containing 20% of Na2SO4 and 80% water by
weight is subjected to evaporative cooling and 20% of original water
evaporates. Calculate the yield of Na2SO4.10H2O crystals formed from
solution, if the temperature of solution is reduced to 20 °C. The solubility
of anhydrous Na2SO4 in water at 20 °C is 19.4 gm per 100 gm of water.
6.6
An evaporator concentrates 10,000 kg/h of 20% KNO3 solution to
50% KNO3. The concentrated liquor is sent to a crystallizer where
crystals of KNO3 are formed and separated. The mother liquor from
the crystallizer is recycled and mixed with the evaporator feed. The
recycle stream is a saturated solution containing 0.6 kg KNO3/kg
water. The crystals carry 4% water. Compute water evaporated and
crystals obtained.
6.7
A crystallizer is charged with 9000 kg of an aqueous solution of 20%
Na2SO4 and it is subjected to evaporative cooling and 10% of original
water evaporates. Glauber’s salt crystallizes. The mother liquor leaves
with 16% salt. Calculate the weights of (a) water loss (b) the mother
liquor leaving and (c) crystals formed.
6.8
A solution of sodium sulphate in water is saturated at a temperature
of 40 °C. Calculate the weight of crystals formed and the percentage yield
obtained by cooling 100 kg of this solution to a temperature of 5 °C.
Solubility Data: At 40 °C : 32.6% Na2SO4
At 5 °C : 5.75% Na2SO4
6.9
A solution of K2Cr2O7 in water contains 16% K2Cr2O7 by weight.
1000 kg of this solution are evaporated to remove some water. The
resulting solution is cooled to 20 °C. If the yield of K2Cr2O7 crystals
is 80%, calculate the amount of water evaporated.
Solubility of K2Cr2O7 at 20 °C = 0.39 k mole/1000 kg H2O
6.10 A saturated solution of MgSO4 at 80 °C is cooled to 25 °C. During this
process 5% of the original water is lost. Find the weight of water lost,
mother liquor leaving and feed, if 1500 kg of MgSO4 .7H2O crystallizes.
Solubility data: At 80 °C : 64.28 g MgSO4 per 100 g water
At 25 °C : 40.88 g MgSO4 per 100 g water
6.11 Hypo crystals Na2S2O3 × 5H2O are to be produced at the rate of 2000
kg/h. A 60% Na2S2O3 solution is cooled to 293 K from 333 K. The
solubility at 293 K is 70 parts anhydrous salt per 100 parts of water.
Estimate the amount of feed needed.
7
Mass Balance
This chapter deals with the mass balance both with and without chemical
reactions. The principles given in Chapter 2 still hold good. The application
of mass balance in unit operations and processes is illustrated through
worked examples and exercise problems.
WORKED EXAMPLES
7.1
Soyabean seeds are extracted with hexane in batch extractors. The
seeds contain 18.6% oil, 69% solids and 12.4% moisture. At the end
of the extraction process, the residual cake is separated from hexane.
The analysis of cake reveals 0.8% oil, 87.7% solids and 11.5%
moisture. Find the % recovery of oil.
Basis: 100 kg of seeds
Solids free from oil is the tie element
Weight of solids in feed = 69 kg = 87.7% of cake
\
cake = 78.68 kg
Oil extracted = (18.6 – 78.68 ¥ 0.008) = 17.971 kg
Ê 17.971ˆ
¥ 100 = 96.6%
% oil extracted = Á
Ë 18.6 ˜¯
7.2
A multiple effect evaporator handles 100 tonnes/day of pure cane
sugar. The feed to the evaporator contains 30% solids. While the
concentrate is leaving with 75% solids concentration, calculate the
amount of water evaporated per day.
?
30% solids
Evaporator
Basis: 100 tonnes of feed = 105 kg
Solids in the feed = 30,000 kg
122
75% solids
MASS BALANCE
123
30,000
= 40,000 kg
0.75
Water evaporated/day = (1,00,000 – 40,000) = 60,000 kg
\
7.3
Concentrated liquid leaving =
A water soaked fabric is dried from 44% moisture to a final moisture
of 9%. Calculate the weight of water removed per 200 kg of dried
fabric.
?
44% moisture
Drier
9% moisture
200 kg
Basis: 200 kg of product.
Dry fabric = (200 ´ 0.91) = 182 kg
182 Ø
Wet fabric entering = ÈÉ
= 325 kg
Ê 0.56 ÙÚ
Water evaporated = (325 – 200) = 125 kg
7.4
The exit gas from a phosphate reduction furnace analyzes P4 : 8%,
CO : 89% and N2 : 3%. This gas is burnt with air under conditions
selectively to oxidize phosphorus. From the flue gas, the oxides of
phosphorus precipitate on cooling and is separated from the remaining
gas. Analysis of the latter shows CO2 : 0.9%, CO : 22.5%, N2 : 68.0%
and O2 : 8.6%. Assume oxidation of phosphorus is complete and that
it exists in the flue gas partly as P4O6 and partly as P4O10. Calculate:
(a) % of CO entering the furnace that is oxidized to CO2 and
(b) % of P4 that is oxidized to P4O10
P4 + 3O2 ® P4O6
P4 + 5O2 ® P4O10
CO + ½O2 ® CO2
Exit gas
from phosphate
reduction furnace
Air
Burner
CO2, O2,
CO, N2
Oxides of P
Basis: 100 g moles of exit gas from phosphate reduction furnace
P4 is 8 g moles.
Let x g mole of P4 get converted to P4O10
\
(8 – x) g mole of P4 gets converted to P4O6
‘Carbon’ is the tie element
124
PROCESS CALCULATIONS
89 g atoms of C º 23.4% of final gas. (Total carbon in CO and CO2)
89
= 380.342 g moles
0.234
(a) % CO oxidized to CO2 = (380.342 ´ 0.009/89) ´ 100
= 3.852 g moles
\
Total final gas moles =
(b) N2 in final gas = (380.342 ´ 0.68) = 258.632 g moles
N2 from air = (258.632 – 3) = 255.632 g moles
21
= 67.953 g moles
79
O2 for formation of P4O10 = 5x g mole
O2 from air = 255.632 ´
O2 for formation of P4O6 = 3(8 – x) g mole
O2 for formation of CO2 = 380.342 ´ 0.009 ´ 0.5 = 1.712 g moles
O2 leaving = 380.342 ´ 0.086 = 32.71 g moles
O2 reacted = (67.953 – 32.71) = 35.243 g moles
Making a balance for oxygen,
(5x + 24 – 3x + 1.712) = 35.243
7.5
\
x = 4.7654 g moles
\
P4 converted to P4O10 = (4.7654 ´ 100/8) = 60%
A tank of weak H2SO4 contains 12.43% acid. If 200 kg of 77.7%
H2SO4 are added to the tank and the final acid is 18.63%, how many
kg of weak acid is used?
Let the weight of weak acid be x kg and the weight of final acid be
y kg.
Overall balance is x + 200 = y
Acid balance is given as 0.1243x + (200 ´ 0.777) = 0.1863y
Solving, x = 1,905.5 kg and y = 2,105.5 kg
7.6
A cotton mill dries a water soaked fabric in a drier from 54% to 9%
moisture. How many kilogram of water are removed by drying
operation per 1200 kg of feed.
Basis: 1200 kg of feed.
Dry fabric = (1200 ´ 0.46) = 552 kg
Water
54% moisture
Drier
9% moisture
MASS BALANCE
125
552
= 606.6 kg
0.91
Water removed = 1200 – 606.6 = 593.4 kg
Weight of product =
Alternative method:
Water in feed = 1200 ´ 0.54 = 648 kg
Water in product = 606.6 ´ 0.09 = 54.6 kg
Water removed = 648 – 54.6 kg
= 593.4 kg
7.7
A gas containing 2% NH3, 25.4% N2 and the rest H2 is flowing in a
pipe. To measure the flow rate an ammonia rich gas containing 96%
NH3, 3% H2 and 1% N2 is sent at a rate of 100 cc/min. The
concentration of NH3 in the down stream is 6%. Find the flow rate of
gas at inlet.
Basis: Let x cc/min of feed gas enter the pipe
Ammonia rich gas entering = 100 cc/min
\
Exit gas = (100 + x) cc/min.
Making a balance for ammonia: 0.02x + 96 = (100 + x) 0.06
Solving, x = 2250 cc/min.
x cc/min. gas
Pipe
NH3 rich gas
(96%) 100 cc/min.
7.8
Exit (100 + x)
6% NH3
A lime stone analysis shows CaCO3 : 94.52%, MgCO3 : 4.16% and
rest insoluble. (a) How many kg of CaO could be obtained from
4 tonnes of limestone? (b) How many kg of CO2 are given off per kg
of limestone?
(a) Basis: 4 tonnes of limestone = 4,000 kg
CaCO3 = 0.9452 ´ 4000 = 3,780.8 kg
MgCO3 = 0.0416 ´ 4000 = 166.4 kg
CaCO3 ® CaO + CO2
100
56
44
MgCO3 ® MgO + CO2
84.3
CaO got =
40.3
44
3780.8 – 56
= 2,117.3 kg
100
126
PROCESS CALCULATIONS
(b) Basis: 1 kg of limestone
CaCO3 = 0.9452 kg, MgCO3 = 0.0416 kg
44 ˆ Ê
44 ˆ
Ê
CO2 produced = Á 0.9452 ¥
˜¯ + ÁË 0.0416 ¥
˜
Ë
100
84.3 ¯
= 0.4376 kg (by stoichiometry)
7.9
The gases from a sulphur burner have the following analysis:
SO2 : 9.86%, O2 : 8.54%, N2 : 81.6%. After passage of gases through a
catalytic converter, the analysis is 0.605% SO2; 4.5% O2; 94.895% N2.
What percentage of SO2 entering the converter has been oxidized to
SO3?
Basis: 100 g moles of gas entering the converter
N2 is the tie element.
81.6
= 86 g moles
0.94895
SO2 leaving = 86 ¥ 0.00605 = 0.5203 g moles
Exit stream =
SO2 converted to SO3
= SO2 entering – SO2 leaving the converter
= (9.86 – 0.52) = 9.34 g moles
100
% SO2 converted = 9.34 ¥
= 94.7%
9.86
7.10 Hydrochloric acid is commercially prepared in a Mannheim furnace
by the reaction between NaCl and H2SO4. The HCl gas is absorbed in
water. Calculate (a) the weight of HCl formed when 2 tonnes of 98%
salt reacts; (b) How much sodium sulphate is produced and how many
kg of 40% acid will be produced?
Basis: 2 tonnes of salt = 2000 kg
NaCl = 2000 ¥ 0.98 = 1960 kg
2NaCl + H 2 SO 4 → Na 2 SO 4 + 2HCl
(2 × 58.4)
98
142
(2 × 36.4)
È 2 ¥ 36.4 ˘
(a) HCl formed = 1960 ¥ Í
˙ = 1221.6 kg
Î 2 ¥ 58.4 ˚
È 142 ˘
(b) Na2SO4 formed = 1960 ¥ Í
˙ = 2382.9 kg
Î 2 ¥ 58.4 ˚
40% acid formed = 1221.6/0.4 = 3054 kg
7.11 Dolomite chiefly a mixture of calcium and magnesium carbonates is
to be leached with concentrated H2SO4 to recover Mg as MgSO4. The
MASS BALANCE
127
rock analysis indicates 20% Ca, 10% Mg and 5% SiO2. After reaction
the soluble material is filtered off and the solid are washed and
discarded. Analysis of the sludge cake shows 50% moisture, 2% SiO2;
0.5% Mg and 1% Al. What fraction of Mg was extracted?
Basis: 100 g of dolomite
SiO2 is the tie element
5 g SiO2 º 2% sludge cake
5
= 250 g
0.02
0.5
= 1.25 g
Mg in sludge cake = 250 ´
100
Mg extracted = (10 – 1.25) = 8.75 g
\
weight of sludge cake =
Fraction Mg extracted =
8.75
= 0.875
10
7.12 A certain soap contains 50% moisture on the wet basis when raw. The
moisture is reduced to 20% before the soap is pressed into cakes. How
many 300 g cakes can be pressed from 1 tonne of raw soap?
Basis: 1 tonne of raw soap = 1000 kg
Dry soap = 500 kg º 80% after drying
\
Weight of dried soap =
500
= 625 kg
0.8
625 Ø
No. of cakes pressed = ÈÉ
= 2083
Ê 0.3 ÙÚ
(The number of cakes has to be an integer)
7.13 Phosphate rock at ` 80/tonne is being added to a cheap fertilizer to
enrich it. If the cheap fertilizer costs ` 32/tonne and the final blended
product including 30% profit is to be marketed at ` 64/tonne, how
much phosphate rock should be added to each tonne of cheap
fertilizer?
Basis: 1 tonne of cheap fertilizer and x tonne of phosphate rock.
Blended fertilizer = (1 + x) tonne
Final cost of blended fertilizer = ` 64/tonne
64
= ` 49.23 = cost of production
1.3
Cost balance: 32 + 80x = (1 + x) 49.23
Actual cost (30% profit) =
Solving, x = 0.56 tonne (Phosphate rock)
128
PROCESS CALCULATIONS
7.14 Limestone is a mixture of Ca and Mg carbonates. Lime is made up
calcining the carbonates, i.e. driving away all CO2 gas. When pure
limestone is calcined, 44.8 g of CO2 is got per 100 g of limestone.
What is the composition of limestone?
Basis: 100 g of limestone (pure)
Let x g be the weight of CaCO3 and y g be the weight of
MgCO3
CaCO3 ® CaO + CO2
100
56
44
MgCO3 ® MgO + CO2
84.3
40.3
44
Overall balance: x + y = 100 (since pure)
CO2 balance = 0.44x + 0.52y = 44.8
Solving, x = 90 g (CaCO3) and y = 10 g (MgCO3)
7.15 The feed to a fractionating system is 30,000 kg/h of 50% benzene,
30% toluene and 20% xylene. The fractionating system consists of
two towers No. I and No. II. The feed enters tower I. The overhead
product from I is x kg/h of 95% benzene, 3% toluene and 2% xylene.
The bottom product from I is feed to II resulting in an overhead
product of y kg/h of 3% benzene, 95% toluene and 2% xylene while
the bottom from II tower is z kg/h of 1% benzene, 4% toluene and
95% xylene. Find x, y and z.
x
Feed
I
y
II
z
Overall balance yields
x + y + z = 30,000
Individual balance yields:
For benzene, 0.95x + 0.03y + 0.01z = 15,000
For toluene, 0.03x + 0.95y + 0.04z = 9,000
For xylene, 0.02x + 0.02y + 0.95z = 6,000
Solving, x = 15,452.2 kg/h
y =
8,741.3 kg/h
z =
5,806.5 kg/h
Total
30,000.0 kg/h
MASS BALANCE
129
7.16 A sample of fuel has the following composition by mass: C : 84%,
H2 : 15.2% and rest comprising noncombustibles. The fuel was
completely burnt with excess air in an internal combustion engine.
The dry exhaust gas has 8.6% CO2 by volume. Estimate the amount
of air supplied per kg of fuel and find the products of combustion (air
contains 23.1% oxygen by weight).
Basis: 1 kg of fuel
C : 0.84 kg = 0.07 kmole; H2 : 0.152 kg = 0.076 kmole;
Noncombustibles – 0.008 kg
C + O2
12
32
n CO ; H + ½O n H O
2
44
2
2
2
16
2
18
CO2 formed º 0.07 kmole
H2O formed = 0.076 kmole
Total kmole of exhaust stream =
0.07
= 0.814 (dry)
0.086
0.076 Ø
È
O2 needed for forming CO2 and H2O = É 0.07 Ù
Ê
2 Ú
= 0.108 kmole
79
21
= 0.4063 kmole
N2 entering (along with stoichiometrically needed O2) = 0.108 ´
0.108
= 0.5143 kmole
0.21
Exhaust gas contains CO2, N2 and O2
Air needed (stoichiometric) =
Excess air in exhaust = Total moles in exhaust stream (dry) –
(N2 entering as per stoichiometry + CO2 formed)
= 0.814 – (0.4063 + 0.07)
= 0.814 – 0.4763 kmole
= 0.3377 kmole
We know what air contains
O2 = 0.071 kmole (21%) N2 = 0.2667 kmole (79%)
N2 total = 0.4063 + 0.2667 = 0.673 kmole
Total air supplied = (0.5143 + 0.3377) = 0.8520 kmole
È 0.3377 Ø
´ 100 = 65.6%
% excess air = É
Ê 0.5143 ÙÚ
130
PROCESS CALCULATIONS
Weight of air supplied/kg of fuel = (0.852 ´ 28.84) = 24.57 kg
Components
Weight, kmole
Composition of dry exhaust gas
CO2
O2
N2
0.0700
0.0710
0.6730
8.6
8.7
82.7
Total
0.8140
100.0
7.17 A company buys its paper from the manufacturer at the contract price
of ` 3/kg on a specification that the paper should not have more than
5% moisture. It is also agreed that the price can be suitably adjusted if
the moisture content differs from the specified value.
A shipment of 5,000 kg of paper was found to contain 7.86%
moisture. (a) Find the price to be paid to manufacturer as per contract.
(b) If it were agreed that the freight was to be borne by the company
and adjusted according to the moisture content, what would the
company pay to the manufacturer if the freight rate is ` 40/1000 kg of
dry paper.
Basis: 100 kg of 5% moist paper.
Dry paper 95 kg; moisture 5 kg
3
= ` 3.16
95
In the case of 5,000 kg paper with 7.86% moisture
Cost of dry paper = 100 ´
Weight of dry paper = 4,607 kg
and weight of moisture = 393 kg
(a) Cost of paper = 4,607 ´ 3.16 = ` 14,558.12
40
= ` 185
1000
Total cost including freight = ` 14,743.12
(b) Freight charge = 4,607 ´
7.18 A mixture of 5% ethylene and 95% air is passed through a suitable
catalyst in a reactor. Some of the ethylene does not react, some form
oxide, some turn to CO2 and water. The entire gas mixture enters an
absorption tower where water is sprayed. The oxide is converted to
glycol. The gas leaving the absorber analyzes C2H4 : 1.085%,
CO2 : 4.345%, O2 : 13.055% and N2 : 81.515% on dry basis. The
partial pressure (pp) of H2O in this gas is 15.4 mm Hg while total
pressure is 745 mm Hg. If one mole of water is sprayed per 100 mole
of gas mixture, calculate the composition of ethylene glycol–water
product formed.
MASS BALANCE
5% C2H4
Reactor
Gases
131
Ethylene, CO2,
O 2, N 2
pp.H2O 15.4 mm Hg
Absorber
95% air
Ethylene glycol–water
Reactions:
1. 2C 2 H 4 + O 2 → 2C 2 H 4 O
Ethylene
Ethylene oxide
2. C2H4 + 3O2 ® 2CO2 + 2H2O
3. C 2 H 4 O + H 2 O → (CH 2 OH) 2
Ethylene glycol
Basis: 100 kmoles of feed gas.
C2H4 : 5 kmoles; O2 : (95 ´ 0.21) = 20 kmoles; N2 : 75 kmoles
N2 balance: Moles leaving the absorber =
75
= 92 kmoles
0.81515
C2H4 : 92 ´ 0.01085 = 1 kmole
CO2 : 92 ´ 0.04345 = 4 kmoles
O2 : 92 ´ 0.13055 = 12 kmoles
92 – 15.4
= 1.94 kmoles
745 15.4
C2H4 reacted : 5 – 1 = 4 kmoles
H 2O =
4
= 2 kmoles
2
C2H4 converted to C2H4O : 4 – 2 = 2 kmoles
C2H4 converted to CO2 :
Water formed by reaction 2 = 4 mole º CO2 kmole
[oxygen sent] – [used up oxygen] = 20 – [1 + 6] = 13 kmoles
Moles of gases entering absorber: C2H4 : 1; CO2 : 4; C2H4O : 2; H2O : 4;
O2 : 12; N2 : 75
Total number of moles = 98.
1
= 0.98
100
Total moles of water entering absorber = (4 + 0.98) = 4.98 kmoles
Water sprayed at the top of absorber = 98 ´
Water in exit gas = 1.94 kmoles
Water reacted in absorption column = 2.00 kmoles
132
PROCESS CALCULATIONS
Ethylene glycol formed = 2 kmoles
Water leaving along with ethylene glycol = (4.98 – 2 – 1.94) = 1.04
kmoles
Composition of liquid stream:
Component
moles
Molecular weight Weight
Weight %
(CH2OH)2
2.00
62
124.00
86.88
1.04
18
18.72
13.12
142.72
100.00
Ethylene glycol
(H2O)
Water
Total
3.04
7.19 The first step in the manufacture of H2SO4 from pyrites consists of
burning pyrites in air. The reactions taking place are:
FeS2 + 2.5O2 ® FeO + 2SO2
(1)
2FeS2 + 5.5O2 ® Fe2O3 + 4SO2
(2)
The flue gas analysis shows SO2 : 10.2%; O2 : 7.8%; N2 : 82% on dry
basis at 600 °C and 780 mm Hg.
(a) In what ratio does the two reactions take place?
(b) How much excess air was fed if the reaction (2) is desired?
Basis: 100 kmoles of flue gas
SO2 : 10.2%; O2 : 7.8%; N2 : 82%
Let x kmole of FeS2 by reaction (1) and
y kmole of FeS2 by reaction (2) be reacted.
21
= 21.8 kmoles (using nitrogen balance)
79
O2 leaving = 7.8 kmoles; O2 used = 14 kmoles
(a) O2 fed = 82 ´
SO2 balance = 2x + 2y = 10.2; \ x + y = 5.1
(a)
O2 used = 2.5x + 2.75y = 14.0
(b)
Solving (a) and (b), we get
x = 0.1 and y = 5; ratio of
x
y
0.1
= 0.02
5
(b) Total FeS2 reacted = 5.1 moles
O2 needed by Reaction (2) = 5.1 ´
5.5
= 14.025 moles
2
MASS BALANCE
133
Excess O2 supplied = 21.8 – 14.025 = 7.775 moles
È 7.775 Ø
´ 100 = 55.4%
% excess air = É
Ê 14.025 ÙÚ
7.20 A producer gas contains CO2 : 9.2%, C2H4 : 0.4%, CO : 20.9%,
H2 : 15.6%, CH4 : 1.9% and N2 : 52%, when it is burnt, the products
of combustion are found to contain CO2 : 10.8%, CO : 0.4, O2 : 9.2%,
N2 : 79.6%. Calculate the following:
(a) m3 of air used per m3 of producer gas
(b) % excess air
(c) % N2 that has come from producer gas.
Basis: 100 g mole of producer gas
Component
Weight, g mole
‘C’ atom
Oxygen needed for
combustion
CO2
C2 H 4
CO
H2
CH4
N2
9.2
0.4
20.9
15.6
1.9
52.0
9.2
0.8
20.9
—
1.9
—
—
0.4 ´ 3 = 1.2
20.9 ´ 0.5 = 10.45
15.6 ´ 0.5 = 7.8
1.9 ´ 2 = 3.8
—
Total
100.0
—
23.25
Reactions
C2H4 + 3O2 ® 2CO2 + 2H2O
CO + 0.5O2 ® CO2
H2 + 0.5O2 ® H2O
CH4 + 2O2 ® CO2 + 2H2O
Carbon is the tie element
C entering = 32.8 g atoms º 11.2% exit.
32.8
\ Total moles of exit gas =
= 293 g moles
0.112
N2 from air: N2 in exit = 293 ´ 0.796 = 233 g moles
N2 in feed = 52 g moles
N2 from air = 181 g moles
21
O2 from air = 181 ´
= 48; Total air = 229 g moles m3 of air/m3
79
229
feed = g mole of air/g mole of feed =
= 2.29
100
134
PROCESS CALCULATIONS
(b) O2 supplied = 48 g moles
needed = 23.25 g moles
Excess O2 supplied = (48 – 23.25) = 24.75 g moles
100
= 106.45%
23.25
(c) Total N2 leaving = 233 g moles
% excess air = 24.75 ´
100 Û
Ë
% N2 from air = Ì 52 –
= 22.3%
233 ÜÝ
Í
7.21 Formaldehyde is manufactured by the catalytic oxidation of methanol
using an excess of air. A secondary reaction also takes place:
CH3OH + 0.5O2 ® HCHO + H2O
(1)
HCHO + 0.5O2 ® HCOOH
(2)
The product gases have the following composition. CH3OH : 8.6%;
HCHO : 3.1%; HCOOH : 0.6%; H2O : 3.7%; O2 : 16%; N2 : 68%. Find
the following:
(a) the percentage conversion of CH3OH to HCHO
(b) % methanol lost in second reaction and
(c) molar ratio of feed to air and the % excess air used.
Basis: 100 g moles of product gases.
[From the product gas analysis, total quantity of HCHO formed is
equivalent to the sum of free HCHO available + HCOOH formed by
the secondary reaction (2)]
HCHO formed = HCHO formed by reaction (1) + HCOOH formed
from HCHO by reaction (2) = (3.1 + 0.6) = 3.7 g moles
CH3OH supplied = reacted by reaction (1) + unconverted at the end
= (3.7 + 8.6) = 12.3 g moles
(a) % conversion of CH3OH to HCHO = 3.7 ´
(b) % lost in second reaction = 0.6 ´
(c) Air fed (using nitrogen balance) =
Moles of feed/moles of air =
100
= 30%
12.3
100
= 4.9%
12.3
68
= 86 g moles; O2 = 18 g moles
0.79
12.3
= 0.143
86.0
O2 needed for converting CH3OH to HCHO =
12.3
= 6.15
2
MASS BALANCE
135
Excess O2 supplied = (18 – 6.15) = 11.85 g moles
100
= 192.7%
6.15
7.22 A rich copper ore analysis gives the following constituent percent:
CuS : 10%; FeS2 : 30% and inert : 60%. By crushing and floating,
2/3 inert is eliminated. The ore is roasted with carbon. Inert are
unchanged. In the reduction of the Cu2O to Cu, there is 5% loss. Find
the weight of copper obtainable from 1 tonne of ore.
% excess air = 11.85 ´
2CuS + 2.5O 2 → Cu 2 O + 2SO 2
2 × 95
2.5 × 32
(1)
2 × 64
142
2FeS2 + 5.5O2 ® Fe2O3 + 4SO2
(2)
Cu 2 O + C → 2Cu + CO
(3)
142
12
126
28
Basis: 1 tonne of ore = 1000 kg
Ore
Crushing and
Flotation
Inert (2/3)
P1
Roaster
P2
Cu, CO
CuS : 100 : 16.67%
FeS2 : 300 : 50.00%
Inert : 200 : 33.33%
600
2
= 400 kg
3
Inert remaining = 200 kg = 0.333 P1 \ P1 = 600 kg
Inert in ore = 600 kg. Removed = 600 ´
CuS in P1 = 600 ´ 0.166 = 100 kg; FeS2 = 600 ´ 0.5 = 300 kg
142
= 74.74 kg
190
(b) Cu2O reacted = 74.74 ´ 0.95 = 71.0 kg
(a) Cu2O got from CuS = 100 ´
126
= 63 kg
142
7.23 Pyrites is roasted in producing SO2. The gases leaving the roaster have
a temperature of 500 °C with composition SO2 : 9%, O2 : 8.6% and
N2 : 82.4%. Composition of pyrites by weight is Fe : 35%, S : 40%
and gangue 25%. Calculate for 100 kg of pyrites roasted.
(c) Cu got from Cu2O = 71 ´
(a) Excess air
(b) Volume of air supplied
(c) Volume of burner gas leaving.
136
PROCESS CALCULATIONS
Basis: 100 kg of pyrites
weight of FeS2 = Total weight – weight of inert = 100 – 25 = 75 kg
4FeS2 + 11O 2 → 2Fe 2 O3 + 8SO 2
480
FeS2 = 75 kg =
352
512
320
75
= 0.625 kmole
120
8
= 1.25 kmoles
4
SO2 = 1.25 kmoles º 9% of exit gas.
SO2 produced = 0.625 ´
1.25
= 13.89 kmoles
0.09
O2 in exit gas = 13.89 ´ 0.086 = 1.19 kmoles
\ Total exit gas in mole =
N2 in exit gas = 13.89 ´ 0.824 = 11.44 kmoles
O2 consumed = 0.625 ´
11
= 1.72 kmoles
4
(a) % excess air = 1.19 ´
100
= 69.2%
1.72
1.19 1.72
= 13.86 kmoles
0.21
Volume of air supplied = 13.86 ´ 22.414 = 310.6 m3 at NTP.
(b) Air supplied =
773
= 881.5 m3
273
7.24 A fuel oil with the following composition C : 84%, H2 : 13%, O2 : 1%,
S : 1% and H2O : 1%; is burnt and the flue gas obtained gives the
following analysis: CO2 : 9.9%, CO : 1.6%, H2O : 10.75%, SO2 : 0.05%,
O2 : 3.7% and N2 : 74%. Calculate the % excess air used.
(c) Volume of exit gas = 13.89 ´ 22.414 ´
Basis: 100 g of oil
g
g atoms
O2 needed
C
84
7.00
7.00
C + O2 ® CO2
H2
13
6.50
3.25
H2 + ½O2 ® H2O
S
1
0.03
0.03
S + O2 ® SO2
10.28
O2
1
0.03
H2O
1
0.06
Net oxygen needed
– 0.03
10.25
Reaction
MASS BALANCE
137
Carbon in exit stream = (9.9 + 1.6) = 11.5%
7
= 60.87 g moles
0.115
N2 in exit = (60.87 ´ 0.74) = 45.04 g moles
\ g moles of exit gas =
45.04 Ø
Air supplied = ÈÉ
= 57.02 g moles
Ê 0.79 ÙÚ
\ O2 supplied = 11.98 g moles and
excess O2 = 11.98 – 10.25 =1.73 g moles
100
= 16.9%
10.25
7.25 Pure sulphur is burnt in air. Even when 20% excess dry air is passed
only 30% of the S burns to SO3 and the remaining goes to SO2. S to
SO3 is the desired reaction
Hence % excess air = 1.73 ´
(a) What is the analysis of resulting gases?
(b) The gases from the burner are passed through a converter where
SO2 is converted to SO3 (without addition of any more air) if the
gases leaving the converter has 4.3% SO2. Calculate the molar
ratio of SO3 to SO2 in the exit gas.
Basis: 1 g atom of S = 32 g
S + O 2 → SO 2
S + 1.5O 2 → SO 3
32
32
32
O2 needed
64
48
80
= 1.5 g mole
O2 supplied = 1.5 ´ 1.2 = 1.8 g moles
N2 supplied = 1.8 ´ 79/21 = 6.8 g moles
O2 reacted
= 0.7 + (0.3 ´ 1.5) = 1.15 g moles
Exit gas stream:
Component
SO2
SO3
Weight, g mole
0.7
0.3 (1.8 – 1.15) = 0.65
Mole %
8.28
3.55
Let x moles of SO2 react to form SO3
SO2 + 0.5 O2 ® SO3
Exit gas stream consists of the following:
SO2 = 0.7 – x
SO3 = 0.3 + x
O2
N2
7.70
Total
6.8
8.45
80.47
100.0
138
PROCESS CALCULATIONS
O2 = 0.65 – 0.5x
N2 = 6.8
Total = 8.45 – 0.5x
SO2 =
0.7 x
º 0.043
8.45 0.5 x
Solving, we get x = 0.34
È SO3 Ø
ÊÉ SO ÚÙ
2 exit
È 0.64 Ø
= 1.78
ÊÉ 0.36 ÚÙ
7.26 500 kg/h of pure sulphur is burnt with 20% excess air (based on S to
SO2) 5% S is oxidized to SO3 and rest to SO2. Find the exit gas
analysis.
500
Basis: 500 kg sulphur =
= 15.625 katoms
32
S + O 2 → SO 2
32
32
64
S + 1.5 O 2 → SO3
32
48
80
1 katom of sulphur requires 1 kmole of oxygen to form SO2
Therefore oxygen supplied = 15.625 ´ 1.2 = 18.75 kmoles
79
= 70.54 kmoles
21
S to SO2 = 15.625 ´ 0.95 = 14.845 kmoles
N2 entering = 18.75 ´
S to SO3 = 15.625 ´ 0.05 = 0.780 kmole
Total O2 consumed = [14.845 + (1.5 ´ 0.78)] = 16.015 kmoles
O2 remaining = (18.75 – 16.015) = 2.735 kmoles
SO2
SO3
O2
N2
Total
Weight, kmole
14.845
0.78
2.735
70.54
88.9
mole %
16.70
0.88
3.08
79.34
100.00
Exit gas
7.27 The composition of the gas entering a converter is SO2 : 7.2%,
O2 : 13.2% and N2 : 79.6% and that of the gas leaving is SO2 : 2.8%,
O2 : 11.7% and N2 : 85.5%. Determine the % of SO2 oxidized to SO3
Basis: 100 g moles entering
79.6
\ Total moles in exit =
0.855
= 93.1 g moles (making nitrogen balance)
MASS BALANCE
139
\ SO2 in exit stream = 93.1 ´ 0.028 = 2.61 g moles
SO2 converted to SO3
= moles of SO2 entering – moles of SO2 in exit stream
= 7.2 – 2.61 = 4.59 g moles
100
= 63.75%
7.2
7.28 Limestone containing (on dry basis) 42.5% CO2 is burnt with coal
having an analysis of 81% C; 4.7% H2; 1.8% N2; 4.6% O2 and 7.9%
ash. The stack gas analyzes 24.4% CO2, 4.1% O2 and 71.5% N2.
Compute
% SO2 oxidized = 4.59 ´
(a) Ratio of lime produced to coal burnt
(b) Stack gas produced/tonne of lime produced.
Reactions:
CaCO3 + C + O 2 → CaO + 2CO 2
100
12
56
32
88
MgCO 3 + C + O 2 → MgO + 2CO 2
84.3
12
40.3
32
2 × 44
Stack gas
Limestone
z coal
Burner
CaO Lime
y Air
Basis: 100 kmoles of stack gas
Let z kg of coal and y kg of air enter the burner
(air contains 23% O2; 77% N2 by weight %)
C + O2 ® CO2
H2 + 0.5O2 ® H2O
On weight basis,
N2 balance gives, 0.018z + 0.77y = 71.5 ´ 28 = 2002
O2 balance gives,
32
16 Ø
È
0.047z – Ù = (4.1 ´ 32) = 131.2
(0.046z + 0.23y) – É 0.81z –
Ê
12
2Ú
Solving the above equations, we get
z = 187 kg
y = 2,595.6 kg
140
PROCESS CALCULATIONS
CO2 balance gives: 24.4 kmoles CO2 = 1,073.6 kg
12 kg C gives 44 kg CO2
Carbon in coal = 187 ´ 0.81 kg C gives
44 – 187 – 0.81
12
= 555.4 kg of CO2
\ CO2 obtained from limestone = total CO2 – CO2 from coal
= (1,073.6 – 555.4) = 518.2 kg
518.2
= 1,219.3 kg
0.425
56
Lime obtained from limestone = 518.2 ´
= 659.5 kg
44
\ Limestone needed =
(a) Lime produced/coal burnt =
659.5
= 3.5
187
N2
O2
CO2
(b) Weight of stack gas 2,002 + 131.2 + 1,073.6 = 3,206.8 kg
\
659.5 kg lime produces 3,206.8 kg stack gas
Hence, for 1 tonne or 1000 kg lime, the stack gas produced is 4,862.5 kg
7.29 A gas containing 80% ethane and 20% oxygen is burnt with 200%
excess air. 80% ethane goes to CO2; 10% to CO; and 10% remains
unburnt. Calculate the stack gas composition.
Basis: 100 g moles of gas
we have,
C2H6 : 80 g moles
O2 : 20 g moles
C2H6 + 3.5O2 ® 2CO2 + 3H2O
C2H6 + 2.5O2 ® 2CO + 3H2O
O2 needed = 80 ´ 3.5 = 280 g moles
O2 available in gas = 20 g moles
O2 theoretically needed = 260 g moles
200% excess oxygen is supplied
\ O2 supplied = (260 ´ 3) = 780 g moles
Total O2 available = 20 + 780 = 800 g moles
N2 entering = (780 ´ 79/21) = 2,934 g moles
CO2 formed = 80 ´ 0.8 ´ 2 = 128 g moles
CO formed = 80 ´ 0.1 ´ 2 = 16 g moles
MASS BALANCE
141
O2 used = (80 ´ 0.8 ´ 3.5) + (80 ´ 0.1 ´ 2.5) = 244 g moles
O2 remaining = Oxygen available – oxygen used = (800 – 244)
= 556 g moles
H2O formed = (80 ´ 0.8 ´ 3) + (80 ´ 0.1 ´ 3) = 216 g moles
C2H6 remaining = 80 ´ 0.1 = 8 g moles
Composition of stack gases:
Gases
g moles
mole %
CO2
CO
O2
N2
H2O
C2H6
128
16
556
2,934
216
8
3.318
0.415
14.412
76.050
5.598
0.207
Total
3,858
100.000
7.30 Pure CO2 may be prepared by treating limestone with sulphuric acid. The
limestone used in the process contains CaCO3, MgCO3 and inert
compounds. The acid used contains 12% H2SO4 by weight. The residue
from the process had the following composition: CaSO4 : 8.56%,
MgSO4 : 5.23%, H2SO4 : 1.05%, Inert : 0.53%, CO2 : 0.12%, H2O : 84.51%.
During the process, the mass was warmed where CO2 and H2O got
removed. Calculate the following:
(a) The analysis of limestone
(b) The % excess acid used
(c) The weight and analysis of the material distilled from the reaction
mass per 1000 kg of limestone treated.
Basis: 100 kg of residue
Inert : 0.53 kg, CaSO4 : 8.56 kg, MgSO4 : 5.23 kg
CaCO3 + H2SO4 ® CaSO4 + CO2 + H2O
100
98
136
44
18
MgCO3 + H2SO4 ® MgSO4 + CO2 + H2O
84.3
98
120.3
44
8.56 kg of CaSO4 obtained from 8.56 ´
5.23 kg of MgSO4 obtained from 5.23 ´
18
100
= 6.3 kg of CaCO3
136
84.3
= 3.67 kg of MgCO3
120.3
142
PROCESS CALCULATIONS
(a) Limestone analysis
Weight, kg
Weight %
CaCO3
MgCO3
Inert
Total
6.3
60
3.67
34.95
0.53
5.05
10.50
100.0
98 Ø
È
(b) 6.3 kg of CaCO3 requires É 6.3 –
Ù kg of H2SO4 = 6.174 kg
Ê
100 Ú
98 Ø
È
3.67 kg of MgCO3 requires É 3.67 –
Ù kg of H2SO4 = 4.27 kg
Ê
84.3 Ú
Total acid needed = 6.174 + 4.27 = 10.444 kg
Excess acid remaining = 1.05 kg
100 Ø
È
% Excess acid used = É1.05 –
Ù = 10.05%
Ê
10.444 Ú
Total acid used = (10.444 + 1.05) = 11.494 kg
È 11.494 Ø
(i.e.) 12% acid supplied = É
= 95.78 kg
Ê 0.12 ÙÚ
44 Ø È
44 Ø
È
CO2 formed = É 6.3 –
ÙÚ ÉÊ 3.67 –
Ù = 4.689 kg
Ê
100
84.3 Ú
18 Ø È
18 Ø
È
H2O formed = É 6.3 –
Ù É 3.67 –
Ù = 1.918 kg
Ê
100 Ú Ê
84.3 Ú
H2O in acid = (95.78 – 11.494) = 84.286 kg
Total water = water from acid + water formed from reaction
= (84.286 + 1.918) = 86.204 kg
H2O in residue = 84.51 kg
CO2 in residue = 0.12 kg
\ Amount of H2O vapours = (86.204 – 84.51) = 1.694 kg
and CO2 in gas phase = (4.689 – 0.12) = 4.569 kg
vapours:
H2O
CO2
1.694 kg
4.569 kg
27 % (weight %)
73 % (weight %)
47.56 mole %
52.44 mole %
Total
6.263 kg
100
100
Check Limestone + acid = (10.5 + 95.78) = 106.28 kg
Residue + vapours = (100 + 6.263) = 106.263 kg
For 10.5 kg limestone, vapours formed = 6.263 kg
(c) \ 1000 kg limestone, vapours formed = 6.263 ´
100
= 596.5 kg
10.5
MASS BALANCE
143
7.31 A coal containing 87% C and 7% unoxidized hydrogen is burnt in air.
If 40% excess air is used;
(a) Calculate kg of air per kg of coal burnt
(b) Assuming complete combustion calculate the composition of gases
by weight %
Basis: 1 kg of coal
Carbon = 0.87 kg = 0.0725 katom
H2 = 0.07 kg = 0.035 kmole
32
16
(a) O2 needed = ÈÉ 0.87 – ØÙ ÈÉ 0.07 – ØÙ = 2.88 kg [by stoichiometry]
Ê
12 Ú Ê
2Ú
O2 supplied = (2.88 ´ 1.4) = 4.032 kg
4.032
= 17.38 kg
0.232
kg of air/kg of coal = 17.38
Air supplied =
(b) Gases leaving are: CO2, H2O, O2 and N2
CO2 produced = 0.87 ´
44
= 3.19kg = 17.41 wt %
12
18
= 0.63kg = 3.44 wt %
2
O2 remaining = (4.032 – 2.88) = 1.152 kg = 6.29 wt %
H2O produced = 0.07 ´
N2 leaving = (17.38 – 4.032) = 13.348 kg = 72.86 wt %
Total
18.32 kg
100.00
7.32 A furnace using hydrocarbon fuel oil has a dry stack gas analysis as
follows:
CO2 : 10.2%, O2 : 8.3%, N2 : 81.5%; Find the following:
(a) % excess air used
(b) the composition of fuel oil in weight %
(c) m3 of air supplied/kg of fuel.
Basis: 100 kmoles dry stock gas
N2 in stack gas = 81.5 kmoles
Air supplied =
81.5
= 103.16 kmoles
0.79
O2 entering = 81.5 ´
21
= 21.66 kmoles
79
144
PROCESS CALCULATIONS
O2 leaving = 8.3 kmoles
O2 consumed = (21.66 – 8.3) = 13.36 kmoles
8.3 Ø
´ 100 = 62.13%
(a) % excess air = ÈÉ
Ê 13.36 ÙÚ
(b) Let the fuel be Cx Hy
CxHy + (x + y/4)O2 ® xCO2 + y/2H2O
x = 10.2 kmoles (CO2);
x + y/4 = 13.36 kmoles (O2 reacted)
\
y = 12.64 katoms
and x = 10.2 katoms
Composition of fuel:
Element
katom
Weight, kg
Weight %
C
H
10.20
12.64
122.40
12.64
90.64
9.36
135.04
100.00
Total
(c) Air supplied = 103.16 kmoles
22.414 Ø
È
m3 of air/kg of fuel = É103.16 –
Ù = 17.12 at NTP
Ê
135.04 Ú
7.33 The waste acid from a nitrating process contains 23% HNO3; 57%
H2SO4; 20% water. This acid is to be concentrated to 27% HNO3,
60% H2SO4 by addition of 93% H2SO4 and 90% HNO3. Calculate the
weight of acids needed to obtain 1000 kg of desired acid.
Basis: 1000 kg of desired acid
Let x kg be the weight of waste acid;
y kg be the weight of H2SO4 and
z kg be the weight of HNO3
Overall balance: x + y + z = 1000
Balance for H2SO4 gives, 0.57x + 0.93y = 1000 ´ 0.60 = 600
Balance for HNO3 gives, 0.23x + 0.9z = 1000 ´ 0.27 = 270
Solving, x = waste acid = 418 kg
y = con. H2SO4 = 390 kg
z = con. HNO3 = 192 kg
Total = 1,000 kg
MASS BALANCE
145
Check (using water balance):
(0.2 ´ 418) + (0.07 ´ 390) + (0.1 ´ 192) = (1000 ´ 0.13) = 130
7.34 In order to obtain barium in a form that may be put into solution, the
natural sulphate (barites) containing only pure barium sulphate and
infusible matter is fused with an excess of pure anhydrous soda ash.
Upon analysis of the fusion mass it is found to contain 11.3% barium
sulphate, 27.7% sodium sulphate and 20.35% sodium carbonate. The
remainder is barium carbonate and infusible matter. Calculate the
following:
(a) % conversion of barium sulphate to the carbonate.
(b) composition of the original barites
(c) % excess of the sodium carbonate used in excess of the
theoretically needed amount for all the barium sulphate.
Basis: 100 kg of fusion mass.
Fused
BaSO4 + IM + Na2CO3
(IM = infusible matter)
BaCO3 + Na2SO4 + IM
Analysis of product: 11.3% BaSO4; 27.7% Na2SO4; 20.35% Na2CO3
Therefore, (BaCO3 + IM) = 100 – (11.3 + 27.7 + 20.35) = 40.65%
BaSO 4 + Na 2 CO 3 → BaCO 3 + Na 2 SO 4
233
106
197
142
Na2SO4 = 27.7 kg
Weight of
BaSO4 reacted = 27.7 ´
233
= 45.45 kg
142
Na2CO3 reacted = 27.7 ´
106
= 20.68 kg
142
197
= 38.43 kg
142
BaSO4 initially present = (45.45 + 11.3) = 56.75 kg
Na2CO3 supplied = (20.68 + 20.35) = 41.03 kg
BaCO3 formed = 27.7 ´
È 45.45 Ø
´ 100 = 80%
(a) % conversion of BaSO4 to BaCO3 = É
Ê 56.75 ÙÚ
(b) Composition of original barites
BaSO4 = 56.75, IM = (40.65 – 38.43) = 2.22 kg
Also, total mass of products = total mass of reactants = 100 kg
Therefore, IM = 100 – [56.75 + 41.03] = 2.22 kg
146
PROCESS CALCULATIONS
Therefore original barites = 56.75 + 2.22 = 58.97 kg
È 2.22 Ø
Hence %IM = É
´ 100 = 3.76%
Ê 58.97 ÙÚ
È 56.75 Ø
%BaSO4 = É
´ 100 = 96.24%
Ê 58.97 ÙÚ
(c) Excess Na2CO3:
106
= 25.82 kg
233
Excess Na2CO3 = (41.03 – 25.82) = 15.21
Na2CO3 needed for all BaSO4 = 56.75 ´
È 15.21 Ø
% excess = É
´ 100 = 58.9%
Ê 125.82 ÙÚ
7.35 A fuel oil containing 88.2% C and 11.8% H2 is burnt with 20%
excess air. 95% of carbon is burnt to CO2 and the rest to CO. All the
Hydrogen is converted to water. Determine the orsat analysis of the
flue gas (dry flue gas).
Basis: 100 kg of fuel oil
Component
Weight, kg
kmole or katom
C
H2
88.2
11.8
7.35 katoms
5.90 kmoles
Total
100.0
—
C + O2 ® CO2
C + 0.5O2 ® CO
H2 + 0.5O2 ® H2O
95% C is converted to CO2 = 7.35 ´ 0.95 = 6.9825 katoms
Oxygen used for this reaction = 6.9825 kmole
5% C is converted to CO = (7.35 – 6.9825) = 0.3675 katoms
0.3675
= 0.18375 kmoles
2
Conversion of H2 to H2O = 5.9 kmoles
Oxygen used for this reaction is =
5.9
= 2.95
2
O2 needed (theoretically) = (7.35 + 2.95) = 10.3 kmoles
Oxygen used for this reaction is =
O2 supplied = 10.3 ´ 1.2 = 12.36 kmoles
MASS BALANCE
147
O2 remaining = [12.36 – (6.9825 + 0.18375 + 2.95)]
= 2.24375 kmoles
79
= 46.5 kmoles
N2 entering = 12.36 ´
21
Component
CO2
CO
O2
N2
Total
mole
6.9825
0.3675
2.24375
46.50
56.0937
mole %
12.45
0.6600
4.00000
82.89
100.0000
7.36 A furnace uses a natural gas which consists entirely of hydrocarbons.
The flue gas analysis is: CO2 : 9.5%, O2 : 1.4%, CO : 1.9% and rest is
N2. Calculate the following:
(a) the atomic ratio of hydrogen to carbon in the fuel
(b) % excess air
(c) the composition of the fuel gas in the form Cx Hy
Basis: 100 mole of the flue gas
Let us assume the hydrocarbon (HC) to be Cx Hy
Let a g mole of HC get oxidized to CO2
and b g mole of HC get oxidized to CO
The reactions to take place are
a(Cx Hy) + a(x + y/4)O2 ® axCO2 + ay/2H2O
b(Cx Hy) + b(x/2 + y/4)O2 ® bxCO + by/2H2O
The composition in percentage of the gases are:
CO2 : 9.5%, O2 : 1.4%, CO : 1.9%, N2 : 87.2%, Total = 100.0%
Carbon balance = a + b = (9.5 + 1.9) = 11.4; ax = 9.5; bx = 1.9
ax 9.5
=5
bx 1.9
a = 5b
By nitrogen balances oxygen supplied =
87.2
´ 0.21 = 23.18 kmoles
0.79
Oxygen left = 1.40 kmoles
Oxygen reacted = 21.78 kmoles
È ay Ø È bx Ø È by Ø
Oxygen reacted = ax + É Ù É Ù É Ù = 21.78
Ê 4Ú Ê 2Ú Ê 4Ú
È ay Ø È 1.9 Ø È by Ø
Oxygen reacted, 9.5 + É Ù É Ù É Ù = 21.78
Ê 4Ú Ê 2 Ú Ê 4Ú
148
PROCESS CALCULATIONS
È yØ
(i.e.) É Ù (a + b) = 11.33
Ê 4Ú
\ y(11.4) = 45.32; \ y = 3.975
a + b = 11.4 = 6b; \ b = 1.9; a = 9.5
since ax = 9.5; x = 1
H 3.975
= 3.975
C
1
(b) For % of excess air, let us consider the following:
(a) Atomic ratio of
(c) CO + 0.5O2 ® CO2
One kmole of CO requires 0.5 kmole of oxygen
Therefore, 0.95 mole of O2 will be used for converting CO to CO2
Excess O2 remaining = 1.4 – 0.95 = 0.45 kmole
Theoretical O2 required = 11.4 (from CO and CO2 values) + 11.33
(for H2O formation) = 22.78 kmoles
È 0.45 Ø
´ 100 = 1.98%
\ Excess air = É
Ê 22.78 ÙÚ
Hence, Hydrocarbon = C1H3.975 = CH4 (methane)
7.37 The analysis of gas entering the converter in a contact H2SO4 plant is
SO2 : 4%, O2 : 13% and N2 : 83%. The gas leaving the converter
contains 0.45% SO2 on SO3 free basis. Calculate the % of SO2
entering the converter getting converted to SO3.
Basis: 100 g mole of gas entering the converter
Let x g mole SO2 get converted to SO3
SO2 : 4
O2 : 13
N2 : 83
SO2 : 0.45%
O2
N2
Converter
SO 2 + 0.5O 2 → SO 3
x
x /2
gases leaving
SO2 ® (4 – x)
(SO3 free basis)
O2 ® (13 – x/2)
x
N2 ® 83
Total
% SO2 in exit =
(100 – 1.5x)
4x
100 1.5 x
È 0.45 Ø
ÉÊ
Ù of exit gas.
100 Ú
MASS BALANCE
149
Solving the above equations, we get x = 3.57
100
= 89.25%
4
7.38 A producer gas made from coke has the composition CO : 28%,
CO2 : 3.5%, O2 : 0.5% and N2 : 68%. This gas is burnt with 20%
excess of the net O2 needed for combustion. If the combustion is 98%
complete, find the weight and volumetric composition of gases
produced per 100 kg of gas burnt.
% SO2 converted to SO3 = 3.57 ´
Basis: 100 kmoles of fuel gas
Component
mole
Molecular weight
CO
28.0
28
28 ´ 28 =
784
CO2
3.5
44
3.5 ´ 44 =
154
O2
0.5
32
0.5 ´ 32 =
16
N2
68.0
28
68 ´ 28 = 1,904
Total
100.0
—
2,858
we have
CO + 0.5O2 ® CO2
Oxygen balance
O2 needed = 28 ´ 0.5 = 14 kmoles
O2 available in feed = 0.5 kmoles
Net O2 needed = 13.5 kmoles
O2 supplied = 13.5 ´ 1.2 = 16.2 kmoles
O2 consumed = 14 ´ 0.98 = 13.7 kmoles
O2 remaining = (16.2 + 0.5 – 13.7) = 3 kmoles
Carbon balance
CO2 formed = (28 ´ 0.98) = 27.44 kmoles
CO2 total = (27.44 + 3.5) = 30.94 kmoles
CO unreacted = (28 ´ 0.02) = 0.56 kmole
Nitrogen balance
79
= 60.9 kmoles
21
N2 in exit = (68 + 60.9) = 128.9 kmoles
N2 from air = 16.2 ´
Weight, kg
150
PROCESS CALCULATIONS
Exit gas
kmole
Molecular
weight
mole % =
volume %
Weight,
kg
Weight %
CO2
30.94
44
18.93
1,359.0
26.60
CO
0.56
28
0.34
15.7
0.31
O2
3.00
32
1.84
96.0
1.88
N2
128.90
28
78.89
3,637.0
71.21
Total
163.40
100.00
5,107.7
100.00
100 kmoles of fuel gas = 2,858 kg
\ Weight of product/100 kg of feed = 5,107.7 ´
100
= 178.7 kg
2,858
7.39 In the manufacture of soda ash by Le Blanc process, sodium sulphate
is heated with charcoal and calcium carbonate. The resulting black ash
has the following composition: Na2CO3 : 42%; other water soluble
material : 6% and insoluble 52%. The black ash is treated with water
to extract the sodium carbonate. The solid residue left behind has
the composition Na2CO3 : 4%; other water soluble material : 0.5%;
insoluble : 85% and moisture : 10.5%
(a) Calculate the weight of residue remaining from the treatment of
1 tonne black ash.
(b) Calculate the weight of Na2CO3 extracted per tonne of black ash.
Feed
I Unit
Black
ash
Water
Residue x
II Unit
Na2CO3: 4%
Water solubles: 0.5%
Insoluble: 85%
Moisture: 10.5%
Na CO
2
3
Solution y
Basis: 1 tonne (1000 kg) of black ash
Let x be the weight of residue and y be the weight of solution.
Insoluble: (0.52 ´ 1000) = 0.85x \ x = 612 kg
Na2CO3 extracted = (0.42 ´ 1000) – (612 ´ 0.04) = 395.5 kg
7.40 A laundry can purchase soap containing 30% water at the rate of
` 7 per 100 kg f.o.b. the factory. The same manufacturer offers
another soap having 5% water. If the freight rate is ` 0.6 per 100 kg
of soap what is the maximum price that the laundry should pay the
manufacturer for the soap with 5% water? (f.o.b.: freight on board)
MASS BALANCE
151
Basis: 100 kg of dry soap
30
= 42.85 » 43 kg
70
Case I: (30% water) water present = 100 ´
Total weight = (100 + 43) = 143 kg
7.6
100
= ` 10.88
Cost of 100 kg dry soap including freight at laundry = 143 ´
5
= 5.25 kg
95
Case II: (5% water) water present = 100 ´
Total weight = 105.25 kg
0.6
= ` 0.63
100
Cost of 100 kg dry soap = (10.88 – 0.63) = ` 10.25 = cost of 105.25 kg
of wet soap
Freight charge alone = 105.25 ´
100
= ` 9.72
105.25
7.41 Phosphorus is prepared by heating in an electric furnace a thoroughly
mixed mass of calcium phosphate, sand and charcoal. It may be
assumed that in a charge the following conditions exist. The amount
of silica used is 10% in excess. Charcoal is 40% in excess of that
required to combine with P2O5 liberated, to form CO.
\ Cost of 100 kg wet soap = 10.25 ´
(a) Calculate the weight % of feed
(b) Calculate the weight of phosphorus obtained per 100 kg of charge
when the reaction I is 90% complete and reaction II is 70% complete.
I
Ca 3 (PO 4 ) 2 + 3SiO 2 → 3CaSiO 3 + P2 O 5
(3 × 60)
310
II
5C
(5 × 12)
(3 × 116)
142
+ P2 O 5 → 2P + 5CO
142
(2 × 31)
(5 × 28)
(a) Basis: 1 kmole of calcium phosphate = 310 kg
Silica charged = 3 ´ 1.1 = 3.3 kmoles
Carbon charged = 5 ´ 1.4 = 7.0 kmoles
Feed
Weight,
kmole or katom
Mol.wt
or At.wt
Weight,
kg
Composition
in weight %
Ca3(PO4)2
SiO2
C
1.0
3.3
7.0
310
60
12
310
198
84
52.3
33.5
14.2
Total
11.3
592
100.0
152
PROCESS CALCULATIONS
(b) Basis: 100 kg of charge: calcium phosphate present = 52.3 kg
Calcium phosphate consumed = (52.3 ´ 0.9) = 47.07 kg
P2O5 formed = 47.07 ´
142
= 21.6 kg
310
62
= 6.6 kg
142
7.42 In the Deacon process for the manufacture of chlorine, a dry mixture
of HCl gas and air is passed over a hot catalyst, which promotes
oxidation of the acid. 30% excess air is used. Calculate the following.
Phosphorus produced = (21.6 ´ 0.7) ´
(a) The weight of air supplied per kg of acid.
(b) The composition (weight %) of gases entering
(c) The composition (weight %) of gases leaving, assuming that 60%
of the acid is oxidized in the process
4HCl
+
(4 × 36.5) =146
O 2 → 2Cl 2 + 2H 2 O
(2 ×16) = 32
(2 × 71) =142
(2 ×18) = 36
Basis: 4 kmoles of HCl º 146 kg
HCl
O2
Converter
N2
Cl2
HCl
O2
N2
H 2O
(a) Oxygen supplied = 1 ´ 1.3 = 1.3 kmoles
È 100 Ø
´ 28.84 = 178.5 kg
Weight of air supplied = 1.3 ´ É
Ê 21 ÙÚ
178.5
= 1.22 kg of air
146
(b) O2 entering = 1.22 ´ 0.23 = 0.281 kg
N2 entering = (1.22 – 0.281) = 0.939 kg
Weight of air/weight of acid =
Gases
Weight, kg
Weight %
HCl
O2
N2
1.000
0.281
0.939
45.0
12.7
42.3
2.220
100.0
(c) HCl consumed = 0.6 kg HCl remaining = 0.4 kg
Ë
32 Ø Û
O2 remaining = Ì 0.281 ÈÉ 0.6 –
ÙÚ Ü = 0.1495 kg
Ê
146
Í
Ý
MASS BALANCE
153
142
= 0.585 kg
146
36
= 0.148 kg
H2O formed = 0.6 ´
146
Cl2 formed = 0.6 ´
Gases
Weight, kg
Weight %
HCl
Cl2
O2
N2
H2O
0.4000
0.5850
0.1495
0.939
0.1480
18.0
26.33
6.73
42.27
6.66
2.2215
100.00
7.43 In the manufacture of H2SO4 by contact process, iron pyrites are burnt
in dry air. Iron is oxidized to Fe2O3. SO2 formed is further oxidized to
SO3 by passing the gases mixed with air over hot catalyst. Air
supplied is 40% in excess of the sulphur actually burnt to SO3. Of the
pyrites charged 15% is lost in grate and not burnt. Calculate the
following:
(a) Weight of air to be used/100 kg of pyrite charged
(b) In the burner 40% of S burnt is converted to SO3. Calculate the
composition (by weight %) of gases leaving I unit.
(c) The catalyst converts 96% of SO2 to SO3. Calculate the weight of
SO3 formed.
(d) Calculate the composition (by weight %) of gases leaving the II
unit.
(e) Calculate the overall degree of completion of sulphur to SO3
FeS2
Unit I
(Burner)
Air
SO3, SO2
O 2, N 2
Fe2O 3
FeS2
Basis: 100 kg of iron pyrites
Pyrites burnt = 85 kg
lost = 15 kg
(1) 2FeS 2 + 5.5 O 2 → Fe 2 O 3 + 4SO 2
(2 × 120)
(5.5 × 32)
160
(2) 4SO 2 + 2O 2 → 4SO 3
(4 × 64)
(2 × 32)
(4 × 80)
(4 × 64)
Unit II
(Converter)
Gases
154
PROCESS CALCULATIONS
(3) 2FeS2 + 7.5O 2 → Fe 2 O 3 + 4SO 3 [Reaction (1) + Reaction (2)]
(2 × 120)
(7.5 × 32)
(4 × 80)
160
In Reaction (3) equal weight of oxygen is needed for FeS2.
85 – 1.4
= 517.4 kg; (40% excess oxygen)
0.23
O2 = 119 kg; N2 = 398.4 kg
(a) Air supplied =
(b) SO3 produced (Unit I) = (85 ´ 0.4) ´
320
= 45.3 kg
240
SO2 produced (Unit I) = (85 ´ 0.6) ´
256
= 54.4 kg
240
32 Ø
È
O2 used for forming SO3 = É 85 – 0.4 – 7.5 –
Ù = 34.0 kg
Ê
240 Ú
32 Ø
È
O2 used for forming SO2 = É 85 – 0.6 – 5.5 –
Ù = 37.4 kg
Ê
240 Ú
Total oxygen used
= 71.4 kg
O2 remaining (entering Unit II) = (119 – 71.4) = 47.6 kg
Gases
Weight, kg
Weight %
SO2
54.4
9.97
SO3
45.3
8.30
O2
47.6
8.72
N2
398.4
73.01
Total
545.7
100.00
320 Ø
È
(c) SO3 produced (II Unit) = É 54.4 – 0.96 –
Ù = 65.28 kg
Ê
256 Ú
\ Total SO3 produced = (45.3 + 65.28) = 110.58 kg
Ë
2 – 32 Û
= 13.05 kg
(d) O2 consumed = Ì 54.4 – 0.96 –
4 – 64 ÜÝ
Í
O2 remaining = (47.6 – 13.05) = 34.55 kg
SO2 remaining = (54.4 ´ 0.04) = 2.176 kg
MASS BALANCE
Weight, kg
Weight %
SO2
2.176
0.400
SO3
110.580
20.264
O2
34.550
6.332
N2
398.400
73.004
Total
545.706
100.000
Gases
(e) Sulphur in FeS2 = 100 ´
155
64
= 53.3 kg
120
Sulphur in SO3 = 110.58 ´
32
= 44.32 kg
80
100
= 83.4%
53.3
7.44 In the common process for the production of nitric acid sodium nitrate
is treated with 95% H2SO4. In order that the resulting ‘niter cake’ may
be fluid, it is desirable to use excess acid, so that final cake contains
34% sulphuric acid. It may be assumed that the cake will contain
1.5% water and that the reaction will go to completion. 2% of HNO3
formed will remain in the cake.
% conversion of S to SO3 = 44.32 ´
(a) Calculate the composition of niter cake by weight %, formed per
100 kg of sodium nitrate charged.
(b) Calculate the weight of sulphuric acid to be used.
(c) Calculate the weight of HNO3 and water vapour distilled from the
niter cake.
2NaNO 3 + H 2 SO 4 → Na 2 SO 4 + 2HNO 3
(2 × 85)
(98)
(142)
(2 × 63)
Basis: 100 kg of NaNO3 charged.
142
= 83.5
170
kg
126
= 74.0
170
kg
Na2SO4 formed = 100 ´
HNO3 formed = 100 ´
98
= 57.65 kg
170
HNO3 remaining in cake = 74 ´ 0.02 = 1.48 kg
(H2SO4 + H2O)% in cake = (34 + 1.5) = 35.5%
H2SO4 required = 100 ´
156
PROCESS CALCULATIONS
Niter cake has Na2SO4, H2SO4, HNO3 and H2O
\ (Na2SO4 + HNO3) % cake = 100 – 35.5 = 64.5%
(83.5 + 1.48) = 84.98 kg º 64.5% cake.
\ Weight of niter cake = 84.98/0.645 = 131.75 kg
(a) Composition of niter cake:
Components
Weight, kg
Weight %
HNO3
Na2SO4
H2SO4
H2O
1.48
83.50
44.80
1.97
1.12
63.38
34.00
1.50
Total
131.75
100.00
(b) H2SO4 in the cake = 131.75 ´ 0.34 = 44.8 kg
H2SO4 for the reaction = 57.65 kg
Total acid to be supplied = 102.45 kg
102.45
= 107.84 kg
0.95
(c) Water in the acid = (107.84 – 102.45) = 5.39 kg
95% acid needed =
Water remaining in cake = 1.97 kg
Water distilled off = (5.39 – 1.97) = 3.42 kg
57.65
= 74.12 kg
98
HNO3 in cake = 1.48 kg
HNO3 formed = 126 ´
HNO3 distilled off = 72.64 kg
Composition of vapours removed:
Components
Weight, kg
Weight %
HNO3
H2O
72.64
3.42
95.50
4.50
Total
76.06
100.00
7.45 Barium carbonate is a commercially important chemical. In its
manufacture, BaS is first prepared by heating the barites the natural
sulphate with carbon. The BaS is extracted from this mass with water
and the solution treated with Na2CO3 to precipitate BaCO3.
In such a process, it is found that the solution of BaS formed also
contains some CaS originating from impurities in barites. The solution
MASS BALANCE
157
is treated with Na2CO3 and the precipitated mass of CaCO3, BaCO3 is
filtered off. It is found that 16.45 kg of dry precipitate is removed
from each 100 kg of filtrate collected. The analysis of the precipitate
is BaCO3 90.1% and CaCO3 9.9%. The analysis of filtrate is reported
to be Na2S : 6.85%, Na2CO3 : 2.25% and H2O : 90.9%. The Na2CO3
for the precipitation used contained CaCO3 as impurity.
(a) Determine the percentage excess Na2CO3 used above than that
required for BaS and CaS.
(b) Calculate the composition of the original solution of BaS and CaS
(c) Calculate the composition of dry soda ash used.
Na2CO3
H 2O
Barites
(X)
Carbon
(CaS
impurity)
Reactions in (X):
Cake
Filter
Filtrate
BaS + Na2CO3 ® BaCO3 + Na2S
CaS + Na2CO3 ® CaCO3 + Na2S
Molecular weights Na2CO3 : 106, BaCO3 : 197.4, BaS : 169.4,
CaCO3 : 100, CaS : 72, Na2S : 78.
Basis: 100 kg of filtrate
Precipitate 16.45 kg
Amount of BaCO3 in precipitate = 16.45 ´ 0.901 = 14.82 kg
Amount of CaCO3 in precipitate = (16.45 – 14.82) = 1.63 kg
(a) Na2CO3 required for forming 14.82 kg BaCO3
= 14.82 ´
106
= 7.96 kg
197.4
78
= 5.85 kg
197.4
\ Na2S formed along with CaCO3 = (6.85 – 5.85) = 1.00 kg
Na2S formed along with BaCO3 = 14.82 ´
100
= 1.28 kg
78
CaCO3 impurity in soda ash = (1.63 – 1.28) = 0.35 kg
CaCO3 formed along with Na2S = 1 ´
106
= 1.357 kg
100
Na2CO3 present in filtrate = 2.25 kg (2.25% in filtrate)
Na2CO3 required for 1.28 kg CaCO3 = 1.28 ´
158
PROCESS CALCULATIONS
Total Na2CO3 needed = (7.96 + 1.357) = 9.317 kg
\ Excess Na2CO3 = 2.25 ´
(b) BaS formed = 14.82 ´
100
= 24.1%
9.317
169.4
= 12.72 kg
197.4
72
= 0.9216 kg
100
Components of original solution
CaS formed = 1.28 ´
Components
Weight, kg
Weight %
BaS
CaS
Water
12.7200
0.9216
90.9000
12.17
0.82
87.01
Total
104.5416
100.00
Total Na2CO3 = Na2CO3 used in reaction + Na2CO3 in filtrate
= 9.317 + 2.25 = 11.567 kg
We have CaCO3 as impurity in Na2CO3 = 0.35 kg
Composition of dry Na2CO3 is shown as follows:
Components
Weight, kg
Weight %
Na2CO3
CaCO3
11.567
0.350
97.06
2.94
Total
11.917
100.00
7.46 In the manufacture of straw pulp for the production of cheap straw
board paper, a certain amount of lime is carried into the beater. It is
proposed to neutralize this lime with acid of 67% H2SO4. In a beater
containing 5000 gallons of pulp it is found that there is lime
equivalent to 0.5 g of CaO per litre.
(a) Calculate the kmole of lime in the beater.
(b) Calculate kmole and kg of H2SO4 that must be added to beater in
order to provide an excess of 1% above needed to neutralize the
lime.
(c) Find the weight of acid.
(d) Find CaSO4 formed in kg (1 gallon = 4.4 litres).
MASS BALANCE
159
Basis: 5000 gallons = (5000 ´ 4.4) = 22,000 litres
CaO + H 2 SO 4 → CaSO 4 + H 2 O
56
98
136
18
(a) Lime in this solution = 22,000 ´ 0.5 = 11,000 g
11,000 g = 196.43 g moles
(b) Acid needed to neutralize = 196.43 g moles
Acid used is, 1% excess = (196.43 ´ 1.01) = 198.4 g moles
= 198.4 ´ 98 = 19.4432 kg
(c) 67% acid needed =
19.4432
= 29.02 kg
0.67
136
= 26,714 g = 26.714 kg
56
7.47 The available nitrogen content in a urea sample is 45%. Find the
actual urea content in the sample.
(d) CaSO4 formed = 11,000 ´
CO(NH2)2 = Molecular weight 60
Therefore, N2 in urea is =
28
= 0.4666
60
100
= 96.43%
0.4666
7.48 Carbon tetrachloride is made as follows:
Purity of sample = 0.45 ´
CS2 + 3Cl 2 → CCl 4 + S2 Cl 2
212.76
153.84
The product gases are found to contain CCl4 33.3%; S2Cl2 33.3%;
CS2 1.4% and Cl2 32%. Calculate the following:
(a) the percentage of the excess reactants used.
(b) the percentage conversion.
(c) kg of CCl4 produced per 100 kg Cl2 converted.
Basis: 100 kmoles of product gas
(a) CS2 reacted = 33.3 kmoles
CS2 remaining = 1.4 kmoles
CS2 taken = 34.7 kmoles
100
= 4.2%
33.3
CS2 is the limiting reactant and Cl2 is the excess reactant.
% excess reactant, CS2 = 1.4 ´
(based on the kmole left in the product)
160
PROCESS CALCULATIONS
Cl2 required (theoretical) = 34.7 ´ 3 = 104.1 kmoles (100% conversion)
Cl2 reacted = 99.9 kmoles
But Cl2 unreacted = 32.0 kmoles
\ Cl2 taken = 99.9 + 32.0 =131.9 kmoles
% excess reactant = 32.0 ´
100
= 30.73%
104.1
(b) % Conversion:
CS2 = 33.3 ´
100
= 95.97%
34.7
100
= 75.74%
131.9
(c) Cl2 reacted = 99.9 kmoles = 7084.91 kg
\ Cl2 = 99.9 ´
212.76 kg Cl2 gives 153.84 kg CCl4
\7084.91 kg Cl2 gives 153.84 ´
7087.91
= 5122.87 kg CCl4
212.76
100
= 72.31 kg
7084.91
7.49 Limestone is a mixture of calcium and magnesium carbonates and
inert. Lime made by calcining the limestone by heating until the CO2
is driven off. When 100 kg of limestone is calcined 44 kg of CO2 is
obtained. If the limestone contains 10% inert, calculate the following:
CCl4 produced/100 kg Cl2 reacted = 5122.87 ´
(a) Compute the analysis of limestone. 10 kg of above limestone is
mixed with 2 kg of coke and is burnt with 100% excess air. The
calcination is complete.
(b) Calculate the composition of gases leaving the kiln. Analysis of
coke C : 76%, ash : 21% and moisture : 3%
Basis: 100 kg of limestone containing 10 kg inert.
(a) Let the limestone contain x kg of CaCO3 and y kg of MgCO3
\ x + y = 90
(1)
CaCO 3 → CaO + CO 2
100
56
44
MgCO3 → MgO + CO 2
84.3
40.3
CO2 balance gives,
44
0.44x + 0.52y = 44.0
(2)
Eq. (1) ´ 0.44 gives, 0.44x + 0.44y = 39.6
(3)
Eq. (2) – Eq. (3)
0.08y =
\ y = 55 kg MgCO3; x = 35 kg CaCO3
4.4
MASS BALANCE
161
(b) CaCO 3 + C + O 2 → CaO + 2CO 2
100
12
32
56
88
MgCO 3 + C + O 2 → MgO + 2CO 2
84.3
12
32
40.3
88
10 kg of limestone = 5.5 kg MgCO3; 3.5 kg CaCO3; and 1 kg inert
2 kg Coke = 1.52 kg Carbon; 0.06 kg moisture.
Gases leaving kiln CO2, O2, N2 and H2O
CO2
From CaCO3 = 3.5 ´
88
= 3.08 kg
100
From MgCO3 = 5.5 ´
88
= 5.74 kg
84.3
8.82
= 0.2 kmole
44
O2 Carbon present = 1.52 kg = 0.1266 k atom
Total
CO2 = 8.82 kg =
O2 supplied (100% excess) = (0.1266 ´ 2) = 0.2532 kmole
N2 entering = 0.2532 ´
79
= 0.9525 kmole
21
0.06
= 0.00333 kmole
18
Total CO2 in gases leaving
H2O (3%) = 0.06 kg =
= CO2 from limestone + CO2 from Carbon (Coke)
= 0.2 kmole + 0.1266 kmole
= 0.3266 kmole
Composition of gases leaving:
Gas
kmole
mole %
CO2
O2 (Excess)
N2
H2O
0.3266
0.1266
0.9525
0.0033
23.18
8.99
67.60
0.23
Total
1.4090
100.00
7.50 A chemical manufacturer produces ethylene oxide (EO) by burning
ethylene gas with air in the presence of catalyst. If the conditions are
carefully controlled, a substantial fraction of the ethylene converted to
ethylene oxide, some unconverted, some completely oxidized to form
162
PROCESS CALCULATIONS
CO2 and H2O. Formation of CO2 is negligible. After the gases leave,
they are passed through an absorber in which the ethylene oxide is
removed. A typical orsat analysis of the gases leaving the absorber
is CO2 : 9.6%, O2 : 3%, C2H4 : 6.4% and N2 : 81%. Of the ethylene
entering the reactor, what percent is converted to oxide?
C 2H 4
Air
Reactor
CO2, O2,
N2
EO,
H 2O
Reactor
Exit Gases
CO2, O2, C2H4, N2
Reaction (1): 2C2H4 + O2 ® 2C2H4O
Reaction (2): C2H4 + 3O2 ® 2CO2 + 2H2O
Basis: 100 g moles of exit gas
6.4 g moles of C2H4 unreacted.
9.6 g moles of CO2 º 4.8 mole of C2H4 converted to CO2
81 g moles of N2 º (81 ´ 21/79) = 21.53 g moles of O2 supplied.
O2 consumed = oxygen supplied – oxygen remaining
= (21.53 – 3) = 18.53 g moles
O2 consumed by Reaction (2) = (4.8 ´ 3) = 14.4 g moles
Therefore, O2 consumed by Reaction (1) = 18.53 – 14.4 = 4.13 g moles
\
C2H4 converted to C2H4O = 8.26 g moles
C2H4 supplied = C2H4 remaining + C2H4 consumed by Reaction (1)
+ C2H4 consumed by Reaction (2)
= 6.4 + 8.26 + 4.8 = 19.46 g moles
100
= 42.4%
19.46
7.51 A spent dye sample obtained from a soap-making unit contains 9.6%
glycerol and 10.3% salt. It is concentrated at a rate of 5,000 kg/h in a
double effect evaporator until the solution contains 80% glycerol and
6% salt. Assume that 4.5% glycerol is lost by entrainment. Find:
C2H4 converted to C2H4O = 8.26 ´
(a) the amount of salt crystallized out in the salt box of the
evaporator and
(b) the evaporation taken place in the system.
Basis: One hour
9.6
= 480 kg
100
Salt = 5,000 ´ 0.103 = 515 kg
Glycerol = 5,000 ´
Water = 4,005 kg
MASS BALANCE
163
Entrained Vapour
5,000 kg/hr
Evaporator
Solution
Salt
Final solution: Loss of glycerol = 480 ´ 0.045 = 21.6 kg
Glycerol remaining = (480 – 21.6) = 458.4 kg
458.4
= 573 kg
0.8
Salt in it = 573 ´ 0.06 = 34.38 kg
\ Solution leaving =
Water in this leaving solution = (573 – 458.4 – 34.38) = 80.22 kg
Salt crystallized = (515 – 34.38) = 480.62 kg
Vapours leaving = Water in vapour + Glycerol in vapour
= (4,005 – 80.22) + 21.6 = 3,946.38 kg
Check: Vapour + solution + salt = (3,946.38 + 573 + 480.62)
= 5,000 kg = Feed
7.52 Coal with 90% purity and rest ash is burnt with 25% excess air. Find
the analysis of the flue gases.
Basis: 100 g of coal:
carbon present is 90 g = 7.5 g atoms
C + O2 ® CO2
O2 needed for the above reaction is 7.5 g moles
O2 supplied: 7.5 ´ 1.25 = 9.375 g moles (25% excess)
9.375 – 79
= 35.27 g moles
21
O2 remaining 9.375 – 7.5 = 1.875 g moles
N2 from air:
Exit gases
CO2
O2
N2
Total
g mole
mole %
7.5
16.8
1.875
4.2
35.27
79.0
44.645
100.0
7.53 Determine the weight of water removed while drying 1,000 kg of wet
substance from 35% to 5%.
Basis: 1,000 kg of wet material.
Dry material is the ‘Tie element’.
Moisture present is 1,000 ´ 0.35 = 350 kg
Dry material is 1,000 – 350 = 650 kg
164
PROCESS CALCULATIONS
Dry material appears as 95% in the exit.
650
= 684.21 kg
0.95
So water removed during the drying is 1,000 – 684.21 = 315.79 kg
Therefore the total weight of material leaving is
7.54 A mixture containing 47.5% of acetic acid is being separated by
extraction in a counter current multistage unit. The operating
temperature is 24 °C and the solvent used is iso-propyl ether. Using
the solvent in the ratio of 1.3 kg/kg of feed, the final extract
composition on a solvent free basis is found to be 82% of acid. The
raffinate is found to contain 14% of acid on solvent free basis. Find
the percentage of acid unextracted?
Acetic acid
Isopropyl ether
Solvent
Extraction unit
Extract ‘E’
Raffinate ‘R’
Basis: 1 kg of feed contains 0.475 kg of acid and 0.525 kg of water
Solvent used = 1.3 kg
Let E kg and R kg be the weight of extract and raffinate
Acid balance: 0.475 = 0.82E + 0.14R
Water balance: 0.525 = 0.18E + 0.86R
Solving the above, we have, E = 0.493 kg and R = 0.507 kg
0.507 Ø
È
\ Acid unextracted: É 0.14 –
Ù = 14.94%.
Ê
0.475 Ú
7.55 A plant makes liquid carbon-dioxide by treating Dolomite with
commercial sulphuric acid. The ore analyzes CaCO3 : 68%,
MgCO3 : 30% and rest silica. Acid used is 94% pure. Find:
(a) CO2 produced
(b) acid required per tonne of the ore and
(c) the composition of the solid left behind.
Basis: 1 tonne of the ore
Weight of CaCO3 : 680 kg, MgCO3 : 300 kg and SiO2 : 20 kg
The reactions taking place are:
Reaction (i): CaCO 3 + H 2 SO 4 → CaSO 4 + CO 2 + H 2 O
100
98
136
44
18
Reaction (ii): MgCO 3 + H 2 SO 4 → MgSO 4 + CO 2 + H 2 O
84.3
98
120.3
44
18
MASS BALANCE
165
(a) CO2 produced by
Reaction (i):
680 – 44
= 299.2 kg
100
300 – 44
= 156.6 kg
84.3
Total weight of CO2 produced = 299.2 + 156.6 = 455.8 kg
Reaction (ii):
(b) H2SO4 required during
Reaction (i):
680 – 98
= 666.4 kg
100
300 – 98
= 348.8 kg
84.3
Total weight of H2SO4 required = 1,015.2 kg
Reaction (ii):
È 1,015.2 Ø
= 1,080 kg
Commercial acid required = É
Ê 0.94 ÙÚ
(c) The solid residue contains CaSO4, MgSO4 and SiO2
136 Ø
È
CaSO4 formed: É 680 –
Ù = 924.8 kg
Ê
100 Ú
120.3 Ø
È
MgSO4 formed: É 300 –
Ù = 428.1 kg
Ê
84.3 Ú
Silica remaining from the ore = 20 kg
Composition of the solid left behind
CaSO4 : 67.36%, MgSO4 : 31.18% and SiO2 : 1.46.%
7.56 A fuel gas contains 70% methane, 20% ethane and 10% oxygen. The
fuel-air mixture contains 200% excess O2 before combustion. 10% of
the hydrocarbon remains unburnt. Of the total carbon burnt 90% forms
CO2 and the rest forms CO. Calculate the composition of the flue gas
on dry and wet basis.
Basis: 100 g moles of the fuel gas
The reactions taking place are:
CH4 + 2O2 ® CO2 + 2H2O
CH4 + 1.5O2 ® CO + 2H2O
C2H6 + 3.5O2 ® 2CO2 + 3H2O
C2H6 + 2.5O2 ® 2CO + 3H2O
Oxygen required for complete combustion = (70 ´ 2) + (20 ´ 3.5) – 10
= 200 g moles
166
PROCESS CALCULATIONS
Oxygen supplied: 200 ´ 3 = 600 mole (200% excess)
Nitrogen entering from air = (600 ´ 79/21) = 2,257 g moles
Methane burnt: (70 ´ 0.9) = 63 mole, unburnt = 7 g moles
Ethane burnt: (20 ´ 0.9) = 18 mole, unburnt = 2 g moles
CO2 formed: (63 ´ 0.9) + (18 ´ 0.9 ´ 2) = 89.1 g moles
CO formed: (63 ´ 0.1) + (18 ´ 0.1 ´ 2) = 9.9 g moles
O2 used: (63 ´ 0.9 ´ 2) + (63 ´ 0.1 ´ 1.5) + (18 ´ 0.9 ´ 3.5)
+ (18 ´ 0.1 ´ 2.5) = 184.1 moles
H2O formed: (63 ´ 0.9 ´ 2) + (63 ´ 0.1 ´ 2) + (18 ´ 0.9 ´ 3)
+ (18 ´ 0.1 ´ 3) = 180 g moles
Component
gases
Weight, g mole
CH4
C2 H 6
CO2
CO
O2
N2
Total (dry)
H2O
7.0
2.0
89.1
9.9
415.9
2,257.0
2,780.9
180.0
0.236
0.067
3.010
0.334
14.049
76.224
—
6.080
0.252
0.072
3.204
0.356
14.959
81.157
100.00
—
Total (wet)
2,960.9
100.000
—
Composition %
Wet basis
Dry basis
7.57 In a catalytic incinerator a liquid having a composition of 88% carbon
and 12% hydrogen is vaporized and burnt with dry air to a flue gas of
the following composition on a dry basis. CO2 : 13.4%, O2 : 3.6% and
N2 : 83%. Find:
(a) How many kmole of dry flue gas are produced per 100 kg of the
liquid feed and
(b) What was the % excess air?
Basis: 100 kg of feed. C : 88 kg = 7.334 katoms
H2 : 12 kg = 6 kmoles
The reactions are:
C + O2 ® CO2
H2 + 0.5O2 ® H2O
CO2 produced will be 7.334 kmoles and it appears as 13.4% of exit
gases.
MASS BALANCE
167
È 7.334 Ø
Thus the exit gas moles is É
= 54.73 kmoles
Ê 0.134 ÙÚ
12
= 6 kmoles
2
Dry flue gas leaving is = 54.73 kmoles
Water present in the exit gases =
Oxygen reacted: (7.334 + 3) = 10.334 kmoles
È 3.6 Ø
Percentage excess air: É
´ 100 = 34.84%
Ê 10.334 ÙÚ
7.58 Aviation gasoline is isooctane C8H18. It is burned with 20% excess air
and 30% of the carbon forms CO and rest goes to carbon dioxide.
What is the analysis of the exit gases (on dry basis)?
Basis: 1 kmole of the isooctane. The reactions that are taking place
are:
C8H18 + 8.5O2 ® 8CO + 9H2O;
C8H18 + 12.5O2 ® 8CO2 + 9H2O.
Oxygen required for complete combustion is 12.5 kmoles
Oxygen supplied is (12.5 ´ 1.2) = 15 kmoles
79 Ø
È
Nitrogen coming from air É15 – Ù = 56.43 kmoles
Ê
21 Ú
CO2 formed by the reaction is (1 ´ 0.7 ´ 8) = 5.6 kmoles
CO formed by the reaction is (1 ´ 0.3 ´ 8) = 2.4 kmoles
Oxygen reacted: (1 ´ 0.7 ´ 12.5) + (1 ´ 0.3 ´ 8.5) = 11.3 kmoles
Oxygen remaining is (15 – 11.3) = 3.7 kmoles
Gases
CO2
CO
O2
N2
Total
mole
5.6
2.4
3.7
56.43
68.13
mole %
8.22
3.52
5.43
82.83
100.0
7.59 Isothermal and isobaric absorption of SO2 is carried out in a packed
tower containing Raschig rings. The gases enter the bottom of
the tower with 14.8% of SO2. Water is distributed at the top of the
column at the rate of 1,000 litres per minute. The total volume of
gas handled at 1 atm and 30 °C is 1,425 m3/h. The gases leaving the
tower are found to contain 1% of SO2. Calculate the percentage
of SO2 in the outlet water and express it in weight %.
168
PROCESS CALCULATIONS
Basis: One hour.
Volume of the gases at standard conditions:
273 Ø
È
3
1425 –
Ù = 1283.9 m
ÊÉ
Ú
303
È 1283.9 Ø
Amount of gases = É
= 57.281 kmoles
Ê 22.414 ÙÚ
SO2 entering absorber: (57.281 ´ 0.148) = 8.4776 kmoles
Inert gases are the tie element
Inert gases: (57.281 – 8.4776) = 48.8034 kmoles
È 48.8034 Ø
= 0.493 kmole
If 99% is equal to 48.8034 kmoles then 1% is É
Ê 99 ÙÚ
SO2 absorbed in the tower: 8.4776 – 0.493 = 7.9846 kmoles
= 511 kg
Amount of water flowing = 1,000 ´ 60 = 60,000 kg/h
Weight of the total liquid leaving the tower is 60,511 kg
100 Ø
È
= 0.84%.
Weight % of SO2 = É 511 –
Ê
60,511ÙÚ
7.60 A pure gaseous hydrocarbon is burnt with excess air. The Orsat
analysis of the flue gas: CO2 : 10.2%, CO : 1%, O2 : 8.4% and rest
nitrogen. What is the atomic ratio of H to C in the fuel? Find the %
excess oxygen supplied.
Basis: 100 kmoles of the flue gases.
N2 in flue gas = 100 – (10.2 + 1 + 8.4) = 80.4 kmoles
Let the hydrocarbon be CxHy.
Let a kmole of it get oxidized to CO2 and b kmole of it get oxidized
to CO.
The reactions are:
CxHy + (x + y/4)O2 ® xCO2 + y/2H2O;
CxHy + (x/2 + y/4)O2 ® xCO + y/2H2O.
We also know that,
2C + O2 ® 2CO
i.e. 2 kmoles of CO requires 1 kmole of O2.
\ 1 kmole of CO requires 0.5 kmole of O2 for conversion to CO2
\
21 Ø
È
O2 supplied = É 80.4 – Ù = 21.37 kmoles
Ê
79 Ú
MASS BALANCE
169
O2 reacted = (21.37 – 8.4) = 12.97 kmoles for forming CO and CO2
For converting CO to CO2 additional O2 required is 0.5 kmoles
\ O2 excess = 8.4 – 0.5 kmoles
= 7.9 kmoles
i.e. O2 required = 21.37 – 7.9
= 13.47 kmoles
7.9 t 100
58.65
13.47
Carbon balance = a + b = 11.2
% excess air =
CO2 formed = ax = 10.2; CO formed: bx = 1
ax/bx = 10.2
ay
bx
by
Oxygen consumed; ax + ÉÈ ÙØ ÉÈ ÙØ ÉÈ ÙØ = 12.97
Ê 4Ú Ê 2Ú Ê 4Ú
Solving, we get x = 1 and y = 0.8
Hence, y is approximated to a full number “1”
Molar ratio, H/C = y/x = 1;
The hydrocarbon will be C2H2, acetylene.
7.61 A feed of 100 kmoles/h containing 40 mole % A is to be distilled
to yield a product containing 95 mole % A and a residue containing
90 mole % B. Estimate the flow rate of distillate and residue.
Making a total material balance,
F=D+W
(1)
Making a component balance,
F.Xf = D.XD + W.Xw
(2)
Substituting,
F = 100 kmoles/h, Xf = 0.4, XD = 0.95 and Xw = 0.1
100 = D + W
(3)
1000(0.4) = D(0.95) + W(0.1)
(4)
Solving (3) and (4), we get
D = 353 kmoles/h
W = 647 kmoles/h
7.62 One hundred kilograms of liquid mixture containing 30% A and 70%
B is extracted with a solvent mixture containing C and D. After
thorough mixing and allowing the system to reach equilibrium, two
separate layers are observed. The composition of both the layers have
been analyzed and given below:
170
PROCESS CALCULATIONS
Layer
Component
A
B
C
D
Top
10
05
60
25
Bottom
20
60
05
15
Estimate (i) the weight of each layer, (ii) weight of solvent, and
(iii) composition of C and D in solvent.
Basis: 100 kg of feed
Solvent
Feed
x
Extractor
y
Let the weight of top layer be x, and the weight of bottom layer be y
from the extractor.
Let the weight of solvent used be s
Total material balance gives
100 + s = x + y
(or)
s = [x + y] – 100
Making component balance for A, we get
100(0.3) = x(0.1) + y(0.2)
30 = 0.1x + 0.2y
Similarly, making component balance for B, we get
100(0.7) = x(0.05) + y(0.6)
70 = (0.05)x + 0.6y
Solving, we get
x = 80 kg, and
y = 110 kg
Solvent used is
s = [x + y] – 100
= 110 + 80 – 100 = 90 kg
Making a component balance for C and D
Weight of C = 80(0.6) + 110(0.05) = 48 + 5.5 = 53.5 kg
MASS BALANCE
171
Weight of D = 80(0.25) + 110(0.15) = 20 + 16.5 = 36.5 kg
Since, the total weight of solvent is 90 kg
53.5
= 0.5945
90
36.5
= 0.4055
xD =
90
7.63 One hundred kg of a mixture containing 80% alcohol and 20% water
is mixed with another mixture containing 40% alcohol and 60% water.
If it is desired to produce a mixture containing 50% alcohol and 50%
water, estimate the quantity of 40% alcohol and 60% water mixture
needed.
xC =
Basis: 100 kg of feed mixture
Let the weight of 40% alcohol + 60% water mixture to be mixed be x kg
Making a material balance for alcohol:
100(0.8) + x(0.4) = (100 + x)(0.5)
i.e. 80 + 0.4x = 50 + 0.5x
Solving, x = 300 kg
Check: by making water balance
100(0.2) + x(0.6) = (100 + x)(0.5)
i.e. 0.1x = 30
Solving, x = 300 kg
7.64 A gas mixture has CO : 10%, CO2 : 50%, O2 : 5% and rest nitrogen by
volume. It is desired to have the nitrogen composition as 40% in the
final mixture by mixing it with fresh air. Estimate the gas/air ratio to
be maintained to achieve the composition as 40% in the final air. Also,
estimate the composition of leaving air.
Basis: 1 kmole of incoming gas mixture
Since the composition of gas is given in volume %, it is to be taken as
mole %.
Air is assumed to contain 79% N2 and 21% O2 by mole %.
Let 1 kmole of gas be mixed with y kmole of air to give final product
containing 40 mole % N2 in (1 + y) kmole of leaving air
Making a material balance for nitrogen
(1) (0.35) + y(0.79) = (1 + y) (0.4)
0.05 = 0.39y
Solving, y = 0.128 kmole
Gas to air ratio is 1/0.128 = 7.813
172
PROCESS CALCULATIONS
y kmole
O2 : 21%
N2 : 79%
1 kmole
CO = 10%
CO2 = 50%
O2 = 5%
N2 = 35%
Mixer
(1 + y) kmole
N2 = 40%
Leaving stream:
Component
Weight, kmoles
mole %
CO
CO2
O2
N2
0.1
0.5
0.05 + 0.128 × 0.21 = 0.07688
0.35 + 0.128 × 0.79 = 0.45112
8.86
44.32
6.82
40.00
Total
1.128
100.00
Average molecular weight:
(0.0886) × (28) + (0.4432) × (40) + (0.0682) × (32) + (0.4) × (28) = 33.5912
7.65 A gas containing 5% SO2, 10% O2, and rest 85% N2 enters a catalytic
chamber where the leaving gas contains only 0.5% SO2. Estimate the
fractional conversion of SO2 to SO3 and also the composition of gases
leaving catalytic chamber.
Basis: 100 kmoles of gas entering the chamber
The reaction is
SO2 + ½ O2 ® SO3
Let x be the kmoles of SO2 reacted
So, unreacted SO2 = 5 – x
Then, SO3 formed is x and the corresponding O2 reacted is x/2 moles
(by stoichiometry)
xÛ
Ë
Balance O2 = Ì10 Ü
2Ý
Í
xÛ
Ë
Total amount of gas leaving = 85 + Ì10 Ü + x + [5 – x]
2
Í
Ý
x
2
Mole fraction of SO2 in the leaving stream = 0.005
= 100 –
MASS BALANCE
i.e.
173
5- x
= 0.005
x
2
Solving, we get
x = 4.511
100 -
Ê 4.511ˆ
× 100 = 90.22%
Fractional conversion of SO2 to SO3 = Á
Ë 5 ˜¯
Gas analysis:
Component
Weight, kmoles
mole %
SO2
SO3
O2
N2
5 – 4.511 = 0.489
4.511
10 – 4.511/2 = 7.744
85.000
0.500
4.615
7.923
86.962
Total
97.744
100.000
EXERCISES
7.1
The waste acid from a nitrating process contains 25% HNO3, 50%
H2SO4 and 25% water. This acid is to be concentrated to 30% HNO3,
60% H2SO4 by addition of 95.3% H2SO4 and 90% HNO3. Calculate
the weight of acids needed to obtain 10,000 kg of desired acid.
7.2
The gas obtained from a furnace fired with a hydrocarbon fuel oil
analyses
CO2: 10.2%, O2: 7.9%, N2: 81.9%.
Calculate: (a) percentage excess air, (b) C:H ratio in the fuel and
(c) kg of air supplied per kg of fuel burnt.
7.3
The flue gas from an industrial furnace has the following composition
by volume. CO2: 11.73%, CO: 0.2%, H2: 0.09% ,O2: 6.81% N2:
81.17%. Calculate the percentage excess air used, if the loss of carbon
in the clinker and the ash is 1% of the fuel used. The fuel gas has the
following composition by weight:
C: 74%, H2: 5%, O2: 5%, N2: 1%, H2O: 9% S: 1% and ash: 5%.
7.4
A fuel gas contains CO2: 2%, CO: 34%, H2: 41%, O2: 1%, C2H4: 7%,
CH4: 11% and rest N2. It is burnt with 25% excess air. Assuming
complete combustion, estimate the composition of leaving gases.
7.5
Limestone is burnt with coke having 85% carbon, producing a gas of
28% CO2, 5% O2, and rest N2. Calculate the amount of lime produced
per 100 kg of coke burnt and the amount of excess air.
174
PROCESS CALCULATIONS
7.6
Pure S is burnt in a furnace with 65% excess air. During combustion
90% of S is burnt to SO2 and rest to SO3. Estimate the composition of
gases leaving.
7.7
In the manufacture of nitric acid, ammonia is reacted with air at
650 °C and 7 bar. The composition of the mixed vapour is nitrogen:
70%, oxygen: 18.8%, ammonia: 10% and rest water. Find the average
molecular weight, composition of leaving gases in weight % and the
density of gases.
7.8
Thirty kilograms of coal analyzing 80% carbon and 20% hydrogen are
burnt with 600 kg of air yielding a gas having an orsat analysis in
which the ratio of CO2 to CO is 3 : 2. What is the percentage of
excess air?
7.9
Pure sulphur is burnt in a burner at a rate of 1,000 kg/h. Fresh air is
supplied at 30 °C and 755 mm Hg. Gases from the burner contain
16.5% SO2 and 3% O2 and rest nitrogen on SO3 free basis. Gases
leave the burner at 800 °C and 760 mm Hg pressure. Calculate
(a) fraction of sulphur burnt to SO3
(b) % excess air over the amount required to oxidize sulphur to SO2.
7.10 Butane is burnt with 80% of the theoretical air. Calculate the analysis
of the gases leaving assuming that all H2 present is converted to water.
7.11 Determine the combustion gas analysis when a medium fuel oil with
84.9% carbon, 11.4% hydrogen, 3.2% sulphur, 0.4% oxygen and 0.1%
ash by weight is burnt with 25% excess air. Assume complete
combustion.
7.12 A synthetic fuel oil is known to contain only H and C, gives on
combustion an Orsat analysis of CO2: 2%, O2: 2.8% and N2: 80.6%.
Calculate the C:H ratio in the fuel.
7.13 A low grade pyrites containing 32% sulphur is mixed with 10 kg of
pure sulphur per 100 kg of pyrites so that the mixture will burn
readily forming a burner gas that analyzes 13.4% SO2, 2.7% O2 and
83.9% N2. No sulphur is left in the cinder. Calculate the % of sulphur
fired that burnt to SO3.
7.14 A mixture containing 20 mole % butane, 35 mole % pentane and rest
hexane, is to be separated by fractional distillation into a distillate
containing 95 mole % butane, 4 mole % pentane and rest hexane and
a bottom product. The distillate is expected to contain 90% of the
butane in the feed. Calculate the composition of the bottom product.
7.15 Gypsum (plaster of Paris: CaSO4 × 2H2O) is produced by the reaction
of calcium carbonate and sulphuric acid. A certain limestone analyzes:
CaCO3 : 96.89%, MgCO3 : 1.41% and inerts : 1.70%. For 5 metric
tonne of limestone reacted completely, determine:
MASS BALANCE
175
(a) kg of anhydrous gypsum (CaSO4) produced.
(b) kg of sulphuric acid solution (98 weight %) required.
(c) kg of carbon dioxide produced.
7.16 The synthesis of ammonia proceeds according to the following reaction
N2 + 3H2 ® 2NH3
In a given plant, 4,202 kg of nitrogen and 1,046 kg of hydrogen are
fed to the synthesis reactor per hour. Production of pure ammonia
from this reactor is 3,060 kg per hour.
(a) What is the limiting reactant?
(b) What is the percent excess reactant?
(c) What is the percent conversion obtained (based on the limiting
reactant)?
7.17 A triple effect evaporator is designed to reduce water from an
incoming brine stream from 25 weight % to 3 weight %. If the
evaporator unit is to produce 14,670 kg/h of NaCl (along with 3
weight % H2O), determine:
(a) the feed rate of brine in lb/h.
(b) the water removed from the brine in each evaporator.
7.18 A natural gas analyzes CH4 : 80.0% and N2 : 20.0%. It is burnt under
a boiler and most of the CO2 is scrubbed out of the flue gas for the
production of dry ice. The exit gas from the scrubber analyzes
CO2 : 1.2%, O2 : 4.9% and N2 : 93.9%.
Calculate:
(a) percentage of the CO2 absorbed.
(b) percent excess air used.
7.19 A synthetic gas generated from coal has the following composition:
CO2 : 7.2%, CO : 24.3%, H2 : 14.1%, CH4 : 3.5% and N2 : 50.9%.
(a) Calculate the cubic metre of air necessary for complete combustion per cubic metre of synthetic gas at the same conditions.
(b) If 38% excess air were used for combustion, what volume of flue
gas at 400 °C and 738 mm Hg would be produced per cubic foot
of synthetic gas at standard conditions?
(c) Calculate the flue gas analysis for (a) and (b).
7.20 The gas obtained by burning pure carbon in excess oxygen analyzes
75% CO2, 14% CO, and rest O2 in mole %.
(a) What is the percentage of excess oxygen used?
(b) What is the yield of CO2 in kg per kg of carbon burnt?
176
PROCESS CALCULATIONS
7.21 A producer gas has the following composition by volume:
CO: 23%, CO2: 4.3%, oxygen: 2.7%, and nitrogen: 70%
(a) Calculate the average molecular weight.
(b) Calculate the gas (in m3) at 30 °C and 760 mm Hg pressure
formed per kg of carbon burnt.
(c) Calculate the volume of air at 30 °C and 760 mm Hg pressure per
100 m3 of the gas at the same conditions if the total oxygen
present be 20% in excess of that theoretically required, and
(d) Calculate the composition of gases leaving for part (c) assuming
complete combustion.
7.22 A sample of coke contains 80% C, 5.8% hydrogen, 8% oxygen, and
1.4% nitrogen. Rest is ash. It is gasified and the gas produced has 5%
CO2, 32% CO, 12% H2, and 51% N2.
(a) Estimate the volume at 25 °C and 750 mm Hg of the gas formed
per 100 kg of coke gasified, and
(b) Calculate the volume of air used per unit volume of gas produced,
both measured under same conditions.
7.23 A coal containing 87.5% total carbon and 7% unoxidised hydrogen is
burnt in air
(a) If 40% excess air is used than that of theoretically needed,
calculate the kg of air used per kg of coal burned.
(b) Calculate the composition by weight of gases leaving the furnace
assuming complete combustion.
7.24 In a lime manufacturing process, pure limestone is burnt with coke having
87% carbon, producing a gas of 27% CO2, 3% O2, and rest N2. Calculate
(a) the amount of lime produced to coke burnt.
(b) the percentage of excess air, and
(c) the amount of stack gas obtained per tonne of lime produced.
7.25 A liquid hydrocarbon feed is passed into a flash vaporizer where it is
heated and separated into vapour and liquid streams. The analysis of
various streams in weight % is as follows:
Stream component
Feed
Vapour
Liquid
C4H10
20
71.2
8.6
C5H12
30
23.8
—
C6H14
50
4.8
—
Estimate the flow rates of liquid and vapour stream for a feed rate of
100 kg/h. Also, evaluate the composition of other two components in
liquid phase.
MASS BALANCE
177
7.26 The petrol used for petrol engine contains 84% carbon and 16%
hydrogen. The air supplied is 80% of that required theoretically for
complete combustion. Assuming that all the hydrogen is burnt and that
carbon is partly burnt to CO and to CO2 without any free carbon
remaining, find the volumetric analysis of the dry exhaust gas.
7.27 Producer gases are produced by burning coke with a restricted supply
of air so that more CO is produced than CO2. The producer is
producing gas having CO:CO2 mole ratio as 5:1 from a coke
containing 80% carbon and 20% ash. The solid residue after
combustion carries with it 2% unburnt carbon. Calculate:
(a) moles of gas produced per 100 kg of coke burnt,
(b) moles of air supplied per 100 kg of coke burnt, and
(c) percentage of carbon lost in the ash
7.28 A furnace uses coke containing 80% carbon and 0.5% hydrogen and
the rest ash. The furnace operates with 50% excess air. The ash
contains 2% unburnt carbon. Of the carbon burnt 5% goes to form
CO. Calculate,
(a) the composition of the flue gas,
(b) the ash produced, and
(c) the carbon lost per 100 kg of coke burnt.
7.29 A petroleum refinery burns a gaseous mixture containing
C5H12 : 7%
C4H10 : 10%
C3H8 : 16%
C2H6 : 9%
CH4 : 55%
N2 : 3%
at rate of 200 m3/h measured at 4.5 bars and 30 °C. Air flow rate is so
adjusted that 15% excess air is used and under these conditions the
ratio of moles of CO2: moles of CO in the flue gas is 20:1. Calculate
(a) m3/h of air being introduced at 1 atm and 30 °C, and
(b) the composition of the flue gas on dry basis.
7.30 The off gas from a phosphate reduction furnace analyses
P4 : 8%
CO : 89%
N2 : 3%
178
PROCESS CALCULATIONS
and is burnt with air under the conditions such that phosphorus is
selectively oxidized. From the flue gas analysis, the oxides of
phosphorus precipitate on cooling and are separated from the
remaining gas. Analysis of the latter shows that:
CO2 : 0.9%
CO : 22.5%
N2 : 68%
O2 : 8.6%
It may be assumed that oxidation of phosphorus is complete and
phosphorus exists in the flue gas partly as P4O6 and partly as P4O10.
Calculate what % of CO entering the burner is oxidized to CO2, and
what % of P4 is oxidized to P4O10?
7.31 Determine the flue gas analysis, air–fuel ratio by weight, and the
volume of the combustion products at 250 °C, when the coal refuse of
the following composition burns with 50% excess air:
Proximate analysis
Moisture
Ash
Volatile matter
Fixed carbon
Air dried %
8
20
28.5
43.5
Ultimate analysis
Carbon
Hydrogen
Nitrogen
Sulphur
Air dried %
81.0
4.6
1.8
0.6
Balance is oxygen
If the rate of burning of coal is 3 tonnes/h, what is the capacity of the
air blower used? Assume complete combustion.
7.32 Determine the flue gas analysis and the air–fuel ratio by weight when
a medium viscosity of fuel–oil with 84.9% C, 11.4% H2, 3.2% S,
0.4% O2 and 0.1% ash is burnt with 20% excess air. Assume complete
combustion.
7.33 A furnace burns producer gas with 10% excess air at a rate of 7200
Nm3/h and discharges flue gases at 400 °C and 760 mm Hg. Calculate
the flue gas analysis, air requirement, and the volume of flue gases
per hour. The gas is supplied from the gas holder and its orsat analysis
is as follows:
MASS BALANCE
179
CO2 : 4%
CO : 29%
N2 : 52.4%
H2 : 12%
CH4 : 2.6%
Normal temperature = 30 °C. Assume complete combustion.
7.34 The following is the ultimate analysis of a sample of petrol by weight:
C
H2
85%
15%
Calculate the ratio of air to petrol consumption by weight if the
volumetric analysis of the dry exhaust gas is:
Composition
CO2
O2
CO
N2
Volume %
11.5
0.9
1.2
86
Also find the % excess air.
Recycle and Bypass
8.1
8
RECYCLE
In industries, sometimes a part of the main product stream or the
intermediate product stream comprising both reactants and products
or the intermediate product is sent back along with feed to the system or
somewhere in the middle of the system. Such a stream is called Recycle
stream. This is done to improve the conversion whenever the conversion is
low and to have energy economy in operations. This also improves the
performance of an equipment as in the case of absorption of sulphur trioxide
using sulphuric acid rather than water, as the solubility is low in pure water.
8.2 BYPASS
Bypassing of a fluid stream is dividing it into two streams, and is often used
in industries to have a closer control in operation. This is done if there is a
sudden change in the property of a fluid stream like excessive heating
(or cooling) as it passes through a preheater (cooler) before entering another
unit. In such cases this conditioned stream is mixed with a portion stream at
its original condition and then used in the process. This is called bypassing
operation.
8.3 PURGE
One of the major problems encountered during recycling is the gradual
increase in the concentration of inert or impurities in the system. A stage
may reach when the concentration of these components may cross
permissible levels. By bleeding off a fraction of the recycle stream, this
problem can be overcome. This operation is known as purging. This is quite
common in the synthesis of ammonia and electrolytic refining of copper.
The above (8.1, 8.2, and 8.3) definitions have been shown in Figure 8.1.
180
RECYCLE AND BYPASS
Bypass
Fresh
feed
Feed
Mixing
unit
181
Gross product
Separator
Process
Net product
Recycle
Purge
Figure 8.1 A scheme indicating recycle, bypass and purge.
WORKED EXAMPLES
8.1
A distillation column separates 10,000 kg/h of a 50% benzene and
50% toluene. The product recovered from the top contains 95%
benzene while the bottom product contains 96% toluene. The stream
entering the condenser from the top of the column is 8,000 kg/h. A
portion of the product is returned to the column as reflux and the
remaining is withdrawn as top product. Find the ratio of the amount
refluxed to the product taken out.
8,000 kg/h
Condenser
V
10,000 kg/h ‘F’
R
Distillation
column
Benzene 95% B
50% B, 50% T
W
96% Toluene
Figure 8.2
Overall balance
F =D+W
or, 10,000 = D + W
Benzene balance gives,
5000 = 0.95D + 0.04W
Solving, D = 5,050 kg/h
W = 4,950 kg/h
D
182
PROCESS CALCULATIONS
Balance around condenser gives,
V
=D+R
or, 8,000
= 5,050 + R
\ R
= 2,950 kg/h
Refluxed quantity R
= 0.584
Actual product
D
Reflux ratio =
8.2
What is the flow rate in recycle stream in Figure 8.3 shown below?
Water W
300°F
Evaporator
50% KNO3
M
Crystallizer
F Feed
10,000 kg/h
20% KNO3
0.6 kg KNO3/kg water
(i.e. 0.6/1.6 = 0.375 KNO3 solution)
R
100°F
C
Crystal with 4% H2O
Figure 8.3
Basis: One hour. KNO3 entering = 2,000 kg/h
2000 µ
¶
¨ 0.96 ·
\ Crystal leaving crystallizer, C = ¦§
Overall balance
since
8.3
= 2,080 kg/h
F =C+W
F = 10,000 kg/h
C = 2,080 kg/h
\ W = 7,920 kg/h
Crystallizer balance gives, M = C + R = (2,080 + R)
KNO3 Balance gives 0.5M = 0.96C + 0.375R
Thus R = 7,680 kg/h
Metallic silver may be obtained from sulphide ores by roasting to
sulphates and leaching with water and subsequently precipitating
silver with copper. In the Figure 8.4 shown below, the material leaving
the second separator was found to contain 90% silver and 10%
copper. What percentage excess copper was used? If the reaction goes
to 75% completion based on the limiting agent Ag2SO4, what is the
recycle stream in kg/tonne of product?
RECYCLE AND BYPASS
183
CuSO4
Cu
S1
Reactor
F
S2
90% Ag;
10% Cu
Ag2SO4
Recycle “R” Ag2SO4
Figure 8.4
Ag 2 SO 4 + Cu →
312
63.5
2Ag + CuSO 4
(2 × 107.9)
159.5
Basis: 1 tonne of product = 1000 kg product
(i.e.) 900 kg silver; 100 kg of copper
Ag2SO4 needed =
900 t 312
= 1300 kg
2 t 107.9
CuSO4 formed =
1,300 t 159.5
= 665 kg
312
1,300 t 63.5
= 265 kg (for forming CuSO4)
312
Total copper supplied = 265 + 100 = 365 kg
Cu needed =
100
= 37.7%
265
We know the reaction is 75% complete. 25% of the limiting reactant
(Ag2SO4) was unconverted which goes in the recycle steam.
% Excess copper used = 100 ´
Ag2SO4 balance: F = R + 1,300 or, F = 1,733.3 kg
0.25F = R
or, 0.25 (R + 1,300) = R \ R = 433 kg
8.4
In the diagram shown in Figure 8.5, what fraction of dry air leaving is
recycled?
‘x’ g dry air
A
H1 = 0.0152 g
of wv/g
C
B
0.099 g H2O/g
dry solid
52.5 g dry air (e)
Drier
Figure 8.5
Basis: 1 g of dry solid.
Let x g of dry air be recycled.
D H = 0.0525 g
2
of wv/g DA
1.562 g H2O/g
dry solid
184
PROCESS CALCULATIONS
Water removed by drying = (1.562 – 0.099) = 1.463 g
Water removed/g of dry air = 0.0525 – 0.0152 = 0.0373 g
1.463
= 39.22 g
0.0373
Dry air passing through drier (e) = 52.5 g (between B and C)
\ Dry air needed =
\ Dry air recycled (x) = 52.5 – 39.22 = 13.28 g
x 13.28
= 0.253
e 52.5
What is recycle, feed and waste for the system shown in Figure 8.6?
\ Fraction recycled =
8.5
Basis: 100 units of feed.
Material balance for A at  = 20 + x = (100 + x)0.4
Solving, x = 33.3 units.
‘Aw’ waste
20% A; Feed 
80% B
40% A
Product P
A 5% and
B 95%
‚
Recycle A only
x
Figure 8.6
Recycle 33.3
= 0.333
Feed
100
Material balance for A at ‚ = (100 + 33.3)0.4 = Aw + 33.3 + 0.05P
Overall balance: 100 = Aw + P
Solving: Aw = 15.81 units Product, P = 84.19 units
8.6
Methanol is produced by the reaction of CO with H2 according to the
equation CO + 2H2 ® CH3OH. Only 15% of the CO entering the
reactor is converted to methanol. The methanol product is condensed
and separated from the unreacted gases, which are recycled. The feed
to the reactor contains 2 kmoles of H2 for every kmoles of CO. The
fresh feed enters at 35 °C and 300 atm. To produce 6,600 kg/h of
methanol calculate:
(a) Volume of fresh feed gas, and
(b) The recycle ratio.
RECYCLE AND BYPASS
35 °C 300 atm
Reactor
CO, 2H2
185
CH3OH
6,600 kg/h
Unreacted CO, H2
Figure 8.7
Basis: One hour of operation
Methanol produced =
6,600
= 206.25 kmoles
32
CO + 2H2 ® CH3OH
15% conversion of CO entering gives 206.25 kmoles of methanol
206.25
0.15
and H2 entering the reactor = 1,375 ´ 2
\ CO entering the reactor =
= 1,375 kmoles
= 2,750 kmoles
= 4,125 kmoles
Total amount of CO and H2
CO unconverted (goes into recycle)
= CO entering – CO converted
= (1,375 – 206.25) = 1,168.75 kmoles
H2 unconverted (goes into recycle) = (2,750 ´ 0.85) = 2,337.50 kmoles
\ Total moles of feed unconverted = 3,506.25 kmoles
Fresh CO needed = 206.25 kmoles
Fresh H2 needed = 412.50 kmoles
\ Total moles of fresh feed = 618.75 kmoles
¦ 308 µ ¦ 1 µ
¶ t§
¶
¨ 273 · ¨ 300 ·
(a) Volume of feed gas = 618.75 ´ 22.414 ´ §
= 52.16 m3
(b) Amount of gas leaving reactor = (3,506.25 + 206.25)
= 3,712.50 kmoles
Amount of gas recycled = 3,506.25 kmoles
3,506.25 µ
¶
¨ 3,712.5 ·
\ Recycle ratio = ¦§
= 0.944 (mole ratio)
By weight 1,168.75 kmoles of CO = 32,725 kg;
2,337.5 kmoles of H2 = 4,675 kg
Recycle stream of CO and H2 = 32,725 + 4675 = 37,400 kg
186
PROCESS CALCULATIONS
Products leaving reactor = (37,400 + 6,600) = 44,000 kg
37, 400 µ
¶
¨ 44,000 ·
\ Recycle ratio = ¦§
8.7
= 0.85 (weight ratio)
Limestone containing 95% of CaCO3 and 5% SiO2 is being calcined.
Heat for the reaction is supplied from a furnace burning coke. The hot
flue gases analyze 5% CO2. The kiln gas contains 8.65% CO2. In
order to conserve some of the sensible heat a portion of the kiln gas is
continuously recycled and mixed with fresh hot flue gas. After mixing
the gas entering the kiln analyzes 7% CO2
(a) Find the kg of CaO produced/kg of coke burnt
(b) Find the recycle ratio, (i.e.) moles of gas recycled per mole of gas
leaving the kiln.
Limestone
K
Kiln
P
X
Air
F
Burner
Coke
R
Figure 8.8
Basis: 12 kg of coke
represent F kmole of flue gas (5% CO2)
K, kmole of kiln gas (8.65% CO2)
R, kmole of gas recycled (8.65% CO2)
P, kmole of product gas (8.65% CO2)
X, kmole of gas entering the kiln (7% CO2)
X
F
R
Figure 8.9
X = F + R (overall balance)
(a)
0.07X = 0.05F + 0.0865 R (CO2 balance)
(b)
CaCO 3 → CaO + CO 2
100
56
44
RECYCLE AND BYPASS
187
Assuming complete combustion
0.05F = 1 kmole CO2 \ F = 20 kmoles
(Q 1 kmole of CO2 comes from 12 kg of Carbon)
C + O2 Æ CO2
Solving (a) and (b), we have R = 24.2 kmoles; X = 44.2 kmoles
Let m kmole of CO2 be added in kiln from the calcinations of limestone
\ K = 44.2 + m
Making CO2 balance, (0.07 X) + m = K ¥ 0.0865
or, 44.2 ¥ 0.07 + m = (44.2 + m)(0.0865)
\ m = 0.8 kmole
\ K = 44.2 + 0.8 = 45.0 kmoles
CaO formed = (0.8 ¥ 56) = 44.8 kg
44.8
12
(a) CaO formed/kg of coke burnt =
= 3.73
R 24.2
= 0.538
=
K
45
Sea water is desalinated by reverse osmosis using the scheme shown
in Figure 8.10D stream has 500 ppm salt = 0.05%
(b) recycle ratio (mole) =
8.8
Find
(a) rate of B
(b) rate of D
(c) recycle R
“R”, Recycle
1,000 kg/h
A
3.1 salt %
sea water
4%
“B”, Waste 5.25% salt
Reverse osmosis cell
Desalinated water “D”, 0.05% salt
Figure 8.10
Basis: One hour
Overall balance, 1,000 = B + D
Salt balance = (1,000 ¥ 3.1) = 5.25B + 0.050
Solving, B = 586.5 kg; D = 413.5 kg
R = 5.25%
188
PROCESS CALCULATIONS
Overall balance (At entry to cell), 1,000 + R = A
Salt balance, 1,000 ´ 0.031 + R ´ 0.0525 = A ´ 0.04
Solving, A = 1,720 kg and R = 720 kg
8.9
In the feed preparation section of an ammonia plant, hydrogen is
produced by a combination of steam-reforming/partial oxidation
process. Enough air is used in partial oxidation to give a 3:1 H2-N2
molar ratio in the feed to the ammonia unit. The H2-N2 mixture is
heated to reaction temperature and fed to a fixed bed reactor where
20% conversion of reactants to NH3 is obtained per pass. The products
from the reactor are cooled and NH3 is removed by condensation. The
unreacted H2-N2 mixture is recycled and mixed with fresh feed. On
the basis of 100 kmoles per hour of fresh feed determine the NH3
produced and recycle rate.
Fresh
Reactor
feed
NH3
Figure 8.11
Basis: 100 kmoles of feed
Given that H2 : N2 = 3 : 1
hydrogen: 75 kmoles and nitrogen: 25 kmoles
N2 + 3H2 ® 2NH3
Overall balance gives:
Feed
NH3
NH3 formed = 50 kmoles
One mole of N2 gives 2 moles of NH3
Reactor balance gives:
3x, H2
x, N2
Reactor
NH3, N2 and H2
Let 3x kmole of H2 and x kmole of N2 enter the reactor
Since 20% conversion takes place,
NH3 formed = x ´ 0.2 ´ 2 = 0.4x = 50 kmoles
\ x = 125 kmoles of N2; and 3x = 375 kmoles of H2
Unreacted N2 = 100 kmoles, and H2 = 300 kmoles
RECYCLE AND BYPASS
189
Ans.: NH3 produced = 50 kmoles
Recycle = 400 kmoles
8.10 Find S, A, R and B from Figure 8.12 shown below, if 1 kg of grease/
100 m2 area is present and the degreased surface per day is 105 m2.
Greased
metal
Solvent S
0% grease
Cleaned
metal
G
Degreasing
B
15% Grease solvent Separator
A
40%
grease
with
solvent
1% grease R
Solvent Recycle
Figure 8.12
Grease removed/day = 1,00,000/100 = 1,000 kg
(i) Overall balance gives:
G
S
100%
100%
Process
A
40% Grease
60% Solvent
S + G = A (overall)
(1)
G = 0.4A = 1000 (grease balance) (2)
Solving Eqs. (1) and (2) A = 2,500 kg; S = 1,500 kg; G = 1,000 kg.
‘S’ (100%
free from
grease)
Degreasing
B (15%)
R-1% (recycle)
(ii) B = S + G + R
B = 1,500 + 1,000 + R (overall)
0.15B = 0 + 1,000 + 0.01R (grease balance)
Solving, (4) and (5), B = 6,964.29 kg; R = 4,464.29 kg
Check B = A + R; 6,964.29 = 2500 + 4,464.29
(0.15 ´ 6,964.29) = 1,000 + (0.01 ´ 4,464.29)
1,044.6435 = 1,044.6429
(3)
(4)
(5)
190
PROCESS CALCULATIONS
8.11 A solution containing 10% NaCl, 3% KCl and water is fed to the process
shown in Figure 8.13 at the rate of 18,400 kg/h. The compositions of the
streams are as follows: Evaporator product P—NaCl : 16.8%,
KCl : 21.6% and water. Recycle product R—NaCl : 18.9% and water.
Calculate the flow rates in kg/h and compute the composition of feed to
the evaporator (F)
R
Fresh
feed
W
F
Evaporator
P
Crystallizer
NaCl only
KCl only
Figure 8.13
Basis: One hour
Water in feed = 18,400 ¥ 0.87 = 16,008 kg/h
KCl in feed = 18,400 ¥ 0.03 = 552 kg/h
NaCl in feed = 18,400 ¥ 0.1 = 1,840 kg/h
Overall balance: water vapour
Feed, F
Process
KCl
NaCl
Balance around crystallizer:
Overall balance: P = R + 552
NaCl balance: 0.168P = 0.189R
P
Crystallizer
KCl
R
RECYCLE AND BYPASS
191
Solving, R = 4,416 kg
and
P = 4,968 kg
Balance around evaporator:
W
Evaporator
F
P
NaCl
F = W + NaCl + P
F = 16,008 + 1,840 + 4,968
= 22,816 kg/h
Feed to evaporator = recycle + fresh feed
= 18,400 + 4,416 = 22,816 kg/h
Recycle, R
Fresh feed
Feed to evaporator, F, m (Conc. of NaCl)
n (Conc. of KCl)
Making a balance for NaCl, we have
18,400 ´ 0.1 + 0.189 ´ 4,416 = m ´ 22,816
Solving, m = 11.72%
Similarly, for KCl, we have 18,400 ´ 0.03 = n ´ 22,816
or, n = 2.42%
Check:
Making water balance, 18,400 ´ 0.87 + 4,416 ´ 0.811
= 22,816 ´ (100 – 11.72 – 2.42)
19,589.6 kg = 19,589.6 kg
8.12 Ethylene oxide is produced by catalytic oxidation of ethylene and
oxygen. The total feed to the catalytic bed of the reactor contains 10:1
volume ratio of oxygen to ethylene and the conversion per pass is
23%. Ethylene oxide is removed from the products completely and the
unreacted ethylene is recycled. The oxygen for the reaction is supplied
from air. Calculate: (a) inlet and outlet composition of the streams and
(b) moles of fresh oxygen required for recycle gases.
192
PROCESS CALCULATIONS
Reaction: C2H4 + 0.5O2 ® C2H4O.
Separator
Ethylene
Air
Reactor
Gases
Figure 8.14
Basis: 1 kmole of ethylene.
Oxygen supplied: 10 kmoles
79
= 37.62 kmoles
21
C2H4 reacted: 1 ´ 0.23 = 0.23 kmole
Nitrogen entering: 10 ´
C2H4O formed: 0.23 kmole
0.23
= 0.115 kmole
2
Oxygen remaining = 10 – 0.115 = 9.885 kmoles
Oxygen reacted =
(a)
Inlet gases
mole
mole %
C2H4
O2
N2
1.0
10.0
37.62
2.056
20.560
77.384
Total
48.62
100.000
Ethylene recycled = (1 – 0.23) = 0.77 mole
(b) Moles recycled/mole of feed =
0.77
= 0.0159
48.62
oxygen required = 0.115 mole
(c)
Outlet gases
mole
mole %
C2H4O
O2
N2
0.230
9.885
37.620
0.48
20.71
78.81
Total
47.735
100.000
RECYCLE AND BYPASS
193
8.13 In the diagram shown in Figure 8.15 find E, P, A and B. Also, find the
composition of A.
The compositions are: F = 20% C2, 40% C3, 40% C4,
E = 95% C2, 4% C3, 1% C4,
P = 99% C3, 1% C4,
B = 8.4% C3, 91.6% C4.
Basis: 100 kg of feed
P
E
Feed, F
A
I Unit
II Unit
B
Figure 8.15
Let us assume that the compositions given are in weight %.
Overall balance: feed = 100 = E + P + B.
C2 balance: 20 = 0.95E. So, the value of E = 20/0.95 = 21.053 kg
C3 balance: 40 = (0.04 ´ 21.053) + 0.99P + 0.0084B
C4 balance: 40 = (0.01 ´ 21.053) + 0.01P + 0.916B
Solving the above, we find P = 35.9006 kg and B = 43.0464 kg
Since we know that F = E + A, we substitute the values of F and E,
and observe, 100 = 21.053 + A. Solving A is found to be 78.947 kg
Composition of A:
C3 balance: (0.99 ´ 35.9006) + (0.084 ´ 43.0464) = 39.1575
Weight % = 49.6.
C4 balance: (0.01 ´ 35.9006) + (0.916 ´ 43.0464) = 39.7895
Weight % = 50.4.
8.14 A contact sulphuric acid plant produces 98% acid. A gas containing
8% SO3 (rest inert) enters a SO3 absorption tower at the rate of
28 kmoles/h 98.5% of SO3 is absorbed in this tower by 97.3% acid
introduced at the top and 95.9% acid is used as the make up acid.
Compute tonne/day of
(a) make up acid required
(b) acid fed at the top of the tower and
(c) acid produced.
194
PROCESS CALCULATIONS
Basis: In one day, gas entering = 28 ´ 24 = 672 kmoles
SO3 entering = 672 ´ 0.08 = 53.76 kmoles
= 53.76 ´ 80 = 4,301 kg
SO 3 + H 2 O → H 2 SO 4
80
18
98
SO3 absorbed = 4,301 ´ 0.985 = 4,236.3 kg
Acid formed = 4,236.3 ´
98
= 5,189.5 kg
80
Water reacted = 4,236.3 ´
18
= 953.17 kg
80
97.3%
SO3 Exit
r
y 95.9%
Absorption
tower
z
(Acid)
SO3 in 28 kmoles/h
x
98%
Figure 8.16
Let us label x, y, z and r as shown.
Overall balance gives: z = (r + 4,236.3)
Acid balance gives, 0.98z = (0.973r + 5,189.5)
Solving the above, we get r = 1,48,271.5 kg;
z = 1,52,507.7 kg (overall)
Another balance of stream gives, z + y = x + r
Acid balance in this stream gives: 0.98z + 0.959y = 0.98x + 0.973r
or, (0.98 ´ 1,52,507.7) + 0.959y = 0.98x + (0.973 ´ 1,48,271.5)
Solving the above, we get x = 53,661.5 kg; y = 49,425.05 kg
Check: H2O balance, 0.021r – 953.17 = (0.02 ´ z)
or, (0.027 ´ 1,48,271.5) – 953.17 = (0.02 ´ 1,52,507.7)
4,003.33 – 953.17 = 3,050.16.
RECYCLE AND BYPASS
195
EXERCISES
8.1
NO is produced by burning gaseous NH3 with 20% excess O2:
4NH3 + 5O2 Æ 4NO + 6H2O
The reaction is 70 percent complete. The NO is separated from the
unreacted NH3, and the latter recycled. Compute (a) moles of NO
formed per 100 moles of NH3 fed, and (b) moles of NH3 recycled per
mole of NO formed.
8.2
In a particular drier, 100 kg of a wet polymer containing 1.4 kg water/kg
of dry polymer is dried to 0.25 kg of water per kg of dry polymer per
hour. 5,000 kg of dry air is passed into the drier. The air leaving the
drier is having a humidity of 0.0045 kg of water vapour per kg of dry
air and fresh air supplied at a humidity of 0.011 kg of water vapour
per kg of dry air. Calculate the mass rate of fresh air supplied and
fraction of air recycled per hour.
Energy Balance
9
9.1 DEFINITIONS
The following definitions are frequently used since the study of energy
balance concerns conversion of our resources into energy effectively and
utilize the same properly. In order to understand the basic principles
pertaining to the generation, transformation and uses of energy, the
following terms need to be discussed first.
9.1.1
Standard State
A substance at any temperature is said to be in its standard state when its
activity is equal to one. The activity may be looked upon as a thermodynamically corrected pressure or concentration. For pure solids, liquids and
gases the standard state corresponds to the substances at one atmosphere
pressure. For real gases, the pressure in the standard state is not 1 atmosphere but the difference from unity is not large. In the case of dissolved
substances the standard state is the concentration in each instance at which
the activity is unity. The enthalpies of substances in standard states are
designated by the symbol H°, while the DH of a reaction where all reactants
and products are at unit activity is represented by DH°.
9.1.2 Heat of Formation
The thermal change involved in the formation of 1 mole of a substance from
the elements is called the heat of formation of a substance. The standard
heat of formation is the heat of formation when all the substances involved
in the reaction are each at unit activity. The enthalpies of all elements in
their standard states at 298 K are zero.
196
ENERGY BALANCE
197
9.1.3 Heat of Combustion
It is the heat liberated per mole of substance burned. The standard heat of
combustion is that resulting from the combustion of a substance, in the state
that is normal at 298 K and atmospheric pressure, with the combustion
beginning and ending at 298 K.
9.1.4 The Heat of Reaction
Heat of reaction from enthalpy data
It is defined as the enthalpy of products minus the enthalpy of reactants. The
standard heat of reaction for the reaction,
aA + bB Æ cC + dD
is given by
DH°r = [c DH°r, C + d DH°r, D] – [a DH°r, A + b DH°r, B]
where DH°r, i is the standard heat of formation of i th component.
By convention,
negative sign indicates that heat is given out and denotes exothermic
reaction
positive sign indicates that heat is absorbed, i.e. endothermic reaction
Heat of reaction from heats of combustion data
For reactions involving organic compounds, it is more convenient to
calculate the standard heat of reaction directly from the standard heats of
combustion instead of standard heats of reaction. The standard heat of
reaction under such circumstances is the standard heat of combustion of the
reactants minus the standard heat of combustion of products.
i.e.
[DHreaction = SDHc, (reactants) – SDHc, products]25°C
9.1.5 Heat of Mixing
When two solutions are mixed, the heat evolved or absorbed during the
mixing process is known as heat of mixing.
9.2
HESS’S LAW
If a reaction proceeds in several steps, the heat of the overall reaction will
be the algebraic sum of the heats of the various stages, and this sum in turn
will be identical with the heat, the reaction would evolve or absorb if it were
to proceed in a single step.
198
9.3
PROCESS CALCULATIONS
KOPP’S RULE
The heat capacity of a solid compound is approximately equal to the sum of
the heat capacities of the constituent elements. As per Kopp’s rule, the
following atomic heat capacities are assigned to the elements at 20 °C:
Carbon: 1.8; Hydrogen: 2.3; Boron: 2.7; Silicon: 3.8; Oxygen: 4.0;
Fluorine: 5.; Phosphorus: 5.4 and all others: 6.2.
Since the heat capacities of solids increase with temperature, it is clear
that these values do not apply over a wide range of temperature.
9.4
ADIABATIC REACTION TEMPERATURE
Adiabatic reaction temperature is the temperature attained by reaction
products, if the reaction proceeds without loss or gain of heat and if all the
products of the reaction remain together in a single mass or stream of
materials.
9.5
THEORETICAL FLAME TEMPERATURE
The temperature attained when a fuel is burnt in air or oxygen without loss
or gain of heat is called the Theoretical flame temperature.
WORKED EXAMPLES
9.1
Calculate the enthalpy of sublimation of Iodine from the following
reactions and data
(a) H2 (g) + I2 (s) Æ 2HI(g) DH = 57.9 kJ
(b) H2 (g) + I2 (g) Æ 2HI(g) DH = –9.2 kJ
The desired reaction is I2(s) Æ I2 (g)
Solution: (a) – (b) = DH = 67.1 kJ
9.2
Find the enthalpy of formation of liquid ethanol from the following
data:
–DH, Heats of reaction, kJ
(1) C2H5OH (l) + 3O2(g) Æ 2CO2 (g) + 3H2O(l) – 1367.8
– 393.5
(2) C (graphite) + O2(g) Æ CO2(g)
– 285.8
(3) H2(g) + ½O2(g) Æ H2O(l)
Solution: [2 ¥ (2) – (1) + (3 ¥ (3))] = 2C + 3H2 + ½O2 Æ C2H5OH.
The enthalpy of formation of ethanol = –276.6 kJ
ENERGY BALANCE
9.3
199
200 kg of Cadmium at 27 °C is to be melted. (The melting point is
320.9 °C). The heat supply is from a system, which supplies 210 kcal/
kg, at steady state. Find the quantity of heat to be supplied by the
system.
Atomic weight of Cadmium = 112.4, Cp = (6 + 0.005T) kcal/kmole
°C and T in °C.
Latent heat of fusion = 2050 kcal/kmole
Basis: 200 kg of Cadmium º 1.78 katoms
©
Sensible heat = 1.78 ª(6 t 320.9)
ª
«
¦
320.92 µ ¸
t
0.005
¹
§
2 ¶· ¹º
¨
= 3885.5 kcal
Latent heat of fusion = 1.78 ´ 2050 = 3649.0 kcal
Total heat to be supplied = 7534.5 kcal
7534.5
= 35.88 kg
210
An evaporator is to be fed with 10,000 kg/h of a solution having 1%
solids. The feed is at 38 °C. It is to be concentrated to 2% solids.
Steam at 108 °C is used. Find the weight of vapour formed and the
weight of steam used. Enthalpies of feed are 38.1 kcal/kg, product
solution is 100.8 kcal/kg, steam is 540 kcal/kg and that of the vapour
is 644 kcal/kg.
Quantity of steam to be supplied =
9.4
Basis: One hour.
Feed = 10,000 kg/h
Vapour formed is 5000 kg/h
Thick liquor is 5000 kg/h
Enthalpy of feed = 10,000 ´ 38.1 = 38.1 ´ 104 kcal
Enthalpy of the thick liquor = 100.8 ´ 5,000 = 5,04,000 kcal.
Enthalpy of the vapour = 644 ´ 5,000 = 32,20,000 kcal.
Heat supplied by steam = Msls = Ms ´ 540 kcal.
Heat balance:
Heat input by steam + heat in, by feed
= Heat out, in vapour + Heat out, thick liquor
or, [Ms.(540) + 38.1 ´ 104] = (32,20,000 + 5,04,000)
\ Ms (540) = 33,43,000.
Thus the weight of steam required, Ms = 6,190.75 kg/hr.
9.5
Calculate the standard heat of reaction:
CaC2 + 2H2O ® Ca(OH)2 + C2H2
DHf cal/mole –15,000 –68,317.4 –2,35,800 54,194
200
PROCESS CALCULATIONS
Solution:
DHrxn = (–2,35,800) + (54,194) – (–15,000) – (2 ´ 68,317.4)
The standard heat of reaction is –29,971.2 cal/mole
9.6
How much heat must be added to raise the temperature of 1 kg of a
20% caustic solution from 7 °C to 87 °C? Take datum temperature as
0 °C.
Data:
Specific heat at 7 °C = 3.56 and at 87 °C = 3.76 kJ/kg K
Solution: Q = (m.Cp.t)1 – (mCpt)2 = 1 [(3.76 ´ 87) – (3.56 ´ 7)]
= 302.2 kJ
9.7
How many Joules are needed to heat 60 kg of sulphur trioxide from
273.16 K to 373.16 K?
CpSO3 = 34.33 + 42.86 ´ 10–3T – 13.21 ´ 10–6T2 J/mole K
Solution:
Number of moles of the trioxide =
60
= 0.75 kmole
80
At 373.16 K,
Q = nòCpSO3 dt
At 273.16 K
Q = [{34.33 ´ (373.16 – 273.16)} + {(42.86 ´ 10–3/2)(373.162 – 273.162)}
+ {(–13.21 ´ 10–6/3)(373.163 – 273.163)}]
Q = 3,509.25 kJ/kmole.
9.8
Using the following data of heats of combustion in cal/g mole,
calculate the following:
(a) Heats of combustion of benzene to water
(b) Heat of vaporization of benzene – cal/g mole
(i) C6H6 (l) to CO2 (g) and H2O (l) = 7,80,980
(ii) C6H6 (g) to CO2 (g) and H2O (g) = 7,59,520
(iii) H2 (g) to H2O (l)
=
68,317
(iv) H2 (g) to H2O (g)
=
59,798
(v) Graphite to CO2 (g)
=
94,052
Desired reactions:
(a) C6H6 (l) + 7.5O2 ® 6CO2 (g) + 3H2O (l)
(b) C6H6 (l) ® C6H6 (g)
ENERGY BALANCE
201
(a) Equation (i) itself gives value, DHc = – 7,80,980 cal/g mole.
(ii) C6H6 (g) + 7.5O2 ® 6CO2 (g) + 3H2O (g)
(iii) H2 (g) + ½O2 ® H2O (l)
(iv) H2 (g) + ½O2 ® H2O (g)
(v) C + O2 ® CO2 (g)
9.9
We can obtain the reaction (b) from the reaction (i) to (v) using
suitable multiplication factor for each step and adding or subtracting
the equations as shown below:
i.e. Steps for equation. (b) = (i) + 3(iv) – (ii) – 3(iii) l = 8,097 cal/g mole
Find the heat of formation of ZnSO4 from its elements and from these
data:
kcal/mole
(i) ZnS ® Zn + S
44
–221.88
(ii) 2ZnS + 3O2 ® 2ZnO + 2SO2
(iii) 2SO2 + O2 ® 2SO3
–46.88
55.1
(iv) ZnSO4 ® ZnO + SO3
Desired equation: Zn + S + 2O2 ® ZnSO4 kcal/mole
Steps: ½ [(ii) + (iii) – 2(i) – 2(iv)] = –233.48 kcal/mole.
9.10 Steam that is used to heat a batch reaction vessel enters the steam
chest, which is segregated from the reactants, at 250 °C, is saturated
and completely condensed. The reaction absorbs 1000 Btu/lb of
charge in the reactor. Heat loss from the steam chest to the
surroundings is 5000 Btu/h. The reactants are placed in the vessel at
70 °F. At the end of the reaction, the materials are at 212 °F. If the
charge contains 325 lb of material and the products and reactants have
an average Cp of 0.78 Btu/1b °F, how many lb of steam are needed
per lb of charge. The charge remains for an hour in the vessel.
Basis: One hour: Datum 70 °F
Btu
Reaction absorbs heat = 1000 ´ 325
= 3,25,000
Heat loss to surroundings
=
5,000
Heat in products: 325 ´ 0.78 (212 – 70) = 36,000
\ Q = total heat
= 3,66,000
From steam tables at 482 °F(250 °C) ls = 734.9 Btu/lb
we have, Q = msls = (ms) (734.9)
= 3,66,000 Btu
\ ms = 498.2 lb/h
498.2
lb of steam/lb of charge =
= 1.533
325
202
PROCESS CALCULATIONS
9.11 Pure ethylene is heated from 30 °C to 250 °C at a constant pressure.
Calculate the heat added per kmole
Cp = 2.83 + 28.601 ´ 10–3T – 87.26 ´ 10 –7T2 where
Cp is in kcal/kmole K and T in K
T2
DH = n
∫ C dT
p
T1 = 303 K, T2 = 523 K
T1
n = 1 kmole
Or, DH = [2.83 {T2 – T1} + (28.601 ´ 10–3/2) {T22 – T12}
– {87.26 ´ 10–7/3}{T23 – T13}]
Heat added = 2,886.11 kcal
9.12 Calculate the amount of heat given off when 1 m3 of air at standard
conditions cools from 500 °C to –100 °C at constant pressure.
Cp air = 6.386 + 1.762 ´ 10–3 T – 0.2656 ´ 10–6 T2, where
Cp is in kcal/kmole K and T in K.
1 m3 =
1
= 0.0446 kmole
22.414
173
Q = 0.0446
∫ C dT = 0.0446 [6.386 ´ 600 + (1.762 ´ 10 /2)(173 – 773 )
–3
2
2
p
773
– (0.2656 ´ 10–6/3) (1733 – 7733)]
Q = –191.345 kcal,
Hence, heat is given off
9.13 Air being compressed from 2 atm and 460 °C (enthalpy 210.5 Btu/lb)
to 10 atm and 500 °R (enthalpy 219 Btu/lb). The exit velocity of air is
200 ft/s. What is the horse power required for the compressor if the
load is 200 lb of air/hour?
Basis: One hour.
(Ws = shaft work, Btu/lb v: velocity, ft/s)
Ws = (219 – 210.5) = 8.5 Btu/lb.
¦ 'v 2 µ
(200)2
§
¶
(2 t 32.2 t 778)
¨ 2 gc ·
= 0.8 Btu/lb
200
= 0.73 HP
2545
[1 Btu = 778 ft. lbf; 1 HP = 2545 Btu/h]
Horse power needed = (8.5 + 0.8) ´
ENERGY BALANCE
203
9.14 Find the heat of reaction at 1200 K.
C 2H 6 ® C2H 4 + H 2
' H f, C 2 H 6 84,720 kJ/kmole
' H f, C 2 H 4 52,280 kJ/kmole
DH°rxn, 298 K: 52,280 – (–84,720) = 1,37,000 kJ
DHrxn = DH°rxn – nCpReactants(1200 – 298) + nCpProducts (1200 – 298)
= (1,37,000) – [1 ´ 100 ´ (1200 – 298)] + [{(1 ´ 78.7)
+ (1 ´ 29.7)}(1200 – 298)]
= 1,37,000 – 90,200 + 97,776.8
= 1,44,576.8 kJ/kmole (Heat to be supplied)
9.15 Calculate the heat input to raise the temperature of 132 kg of CO2
from 100°C to 1000°C. Perform the calculation in the following ways.
(a) by integrating the expression for Cp and (b) by using mean heat
capacity value Cp in kcal/kmole K.
Cp = 6.85 + 8.533 ´ 10–3 T – 2.475 ´ 10–6 T2, kcal/kmole K
Basis: 132 kg of CO2 º 132/44 = 3 kmoles
T2
(a) DH = n
∫ C dT = 3 [6.85 ´ 900 + (8.533 ´ 10 /2)(1273 – 373 )
–3
2
2
p
T1
– (2.475 ´ 10–6 /3)(12733 – 3733)]
DH = 3[10,826.29] = 32,478.87 kcal
Cp values at 1273 K and 373 K are:
Cp at 1273 K = 13.702 kcal/kmole K
and Cp at 373 K = 9.6884 kcal/kmole K
Cpav = 11.6952 kcal/kmole K
DH = òm Cpav dT
= 3 ´ 11.6952 ´ (1273 – 373)
= 31,577.04 kcal
9.16 SO2 gas is oxidized in 100% excess air with 70% conversion to SO3.
The gases enter the converter at 400 °C and leave at 450 °C. How
many kcals are absorbed in the heat exchanger of the converter per
kmole of SO2 sent?
Basis: 1 kmole SO2.
SO2 + ½O2 ® SO3
204
PROCESS CALCULATIONS
O2 sent = 0.5 ´ 2 = 1.
N2 in air = 1 ´ 79/21 = 3.76 kmole.
gases leaving kmoles
Cp mean, cal/g mole °C
SO3 formed = 0.7 kmole
SO2 remaining = 0.3 kmole
SO3
SO2
O2
N2
Total
0.7
15.5
0.3
11.0
0.65
7.5
3.76
7.1
5.41
—
DHrxn = –23, 490 cal/g mole (given)
DHrxn = –23,490 ´ 0.7 = –16,443 kcal
Datum: 0°C
Heat in
Heat out
SO2 = 1
´ 11 ´ 400 = 4,400 kcal SO2 = 0.3 ´ 11 ´ 450 = 1,485.00 kcal
´ 7.5 ´ 400= 3,000 kcal O2 = 0.65 ´ 7.5 ´ 450 = 2,193.75 kcal
O2 = 1
N2 = 3.76 ´ 7.1 ´ 400 = 10,676 kcal N2 = 3.76 ´ 7.1 ´ 450 = 12,013.20 kcal
SO3 = 0.7 ´ 15.5 ´ 450 = 4,882.50 kcal
Total
= –18,076 kcal Total
DHrxn
= + 20,574.45 kcal
–16,443 kcal \ Heat in
–34,519.00 kcal
Hence, heat absorbed in heat exchanger = –13,944.5 kcal
9.17 From the following data compute the enthalpy change of formation for
NH3 at 480 °C.
DHf at 25°C for NH3 = –10.96 kcal/kmole
Cp N2 = 6.76 + (6.06 ´ 10–4T) + (13 ´ 10–8T2)
Cp H2 = 6.85 + (2.8 ´ 10–5T) + (22 ´ 10–8T2)
Cp NH3 = 6.703 + (0.0063T) where T is in K,
Reaction:
DHf /kmole
N2
+
3H2 ®
2NH3
­
­
­
0
0
–10.96
Basis: 1 mole of N2 (Feed at 273 K)
DHrxn 298 K: (2 ´ –10.96) = –21.92 kcal.
Find DCp = 2NH3 – (N2 + 3H2)
Da = (2 ´ 6.703) – [7.76 + (3 ´ 6.85)] = –13.9
Db = (2 ´ 0.0063) – [6.06 ´ 10–4 + (3 ´ 2.8 ´ 10–5)] = 0.0119
Dg = (2 ´ 0) – (13 ´ 10–8 + 3 ´ 22 ´ 10–8) = –7.9 ´ 10–7
¦ 'C µ 2 ¦ ' H µ 3
DHo= DHrxn – Da T – §
¶ T –§
¶T
¨ 2 ·
¨ 3 ·
DHo= –21.92 – (–13.9 ´ 298) –
7.9 t 10 7 t 2983
0.0119 t 2982
–
3
2
ENERGY BALANCE
205
DHo= 3,598.87 kcal
480 °C = 753 K
¦ 'C µ 2
¦ 'H µ 3
¶ T + §
¶T
¨ 2 ·
¨ 3 ·
DHrxn, 480°C = DHo + Da T + §
¦ 0.0119 µ
¶
¨
2 ·
= 3,598.87 + (–13.9 ´ 753) + §
´ (753)2
¦ 7.9 t 10 7 µ
¶
3
¨
·
+ §
´ (753)3
\ DHrxn 480°C = –3,606.56 kcal/kmole
9.18 Calculate the calorific value of a blast furnace gas analyzing 25% CO,
12.5% CO2 and 62.57% N2.
(a) C + O2 ® CO2; DHrxn: –94 kcal
(b) C + ½O2 ® CO; DHrxn: –26 kcal
Also, calculate the theoretical flame temperature for the combustion of
this gas assuming theoretical amount of air is used, the combustion
reaction is complete and reactants enter at 25 °C.
Cp = a + bT + cT2, cal/kmole K
where a, b and c are all dimensional constants and available in literature
Gas
a
b ´ 103
c ´ 105
CO2
N2
10.55
6.66
2.16
1.02
–2.04
—
C
I Combustion
Air
CO
CO2
II Combustion
O2
N2
CO2
N2
Basis: 100 g moles of inlet gas
CO entering 25 g moles, CO2 exit = 25 + 12.5 = 37.5 g moles
O2 needed 12.5 g moles, N2 = §¦ 62.5 12.5 t
¨
79 µ
¶ = 109.52 g moles
21 ·
CO + ½O2 ® CO2
Reaction (a) – (b) gives
DHrxn = –94 + 26 = –68 kcal/kmole
Calorific value: Heat given out = 68 ´ 25 = 1,700 kcal
Exit gases carry this heat away.
206
PROCESS CALCULATIONS
This gas temperature is called Theoretical flame temperature which is
calculated as follows:
–17 ´ 105 cal = 37.5 [{10.55 (T – 298)} + 2.16 ´ 103(T2 – 2982)/2
– 2.04 ´ 10–5(T3 – 2983)/3]
+ 109.52[6.66 (T – 298) + 1.02 ´ 10–3(T2 – 2982)/2]
Solving the above equation we have T = 2721.085 K 2721 K = 2448 °C.
9.19 An inventor thinks he has developed a new catalyst which can make
the gas phase reaction CO2 + 4H2 ® CH4 + 2H2O proceed to 100%
conversion. Estimate the heat that must be provided or removed if the
reactants enter and products leave at 500 °C (in effect, we have to
calculate the heat of reaction at 500 °C).
DHf
CO2
CH4
H2O
kcal/kmole
–94,052
–17,889
–57,798 at 298 K
\ DHrxn = [–17,889 – (2 ´ 57,798)] – [–94,052]
= –39,433 kcal/kmole of CO2
Cp for CO2 = 6.339 + 10.14 ´ 10–3T – 3.415 ´ 10–6T2
H2
= 6.424 + 1.039 ´ 10–3T – 0.078 ´ 10–6T2
H2O = 6.97 + 3.464 ´ 10–3T – 0.483 ´ 10–6T2
CH4 = 3.204 + 18.41 ´ 10–3T – 4.48 ´ 10–6T2
Da = [3.204 + (2 ´ 6.97) – 6.339 – (4 ´ 6.424)] = –14.891
Db = [18.41 + (2 ´ 3.464) – 10.14 – (4 ´ 1.039) = 11.047 ´ 10–3
Dg = [–4.48 – (2 ´ 0.483) – {–3.415 – (4 ´ 0.078)}] = –1.719 ´ 10–6
DCp = –14.891 + 11.047 ´ 10–3T – 1.719 ´ 10–6 T2
We find DHo using data at 298 K
¦ 'C µ 2
¦ 'H µ 3
¶T – §
¶T
¨ 2 ·
¨ 3 ·
DHo = DHrxn – DaT – §
¦
= –39,433 – (–14.891 ´ 298) – § 11.047 t 10 3 t
¨
¦
– § 1.716 t 10 6 t
¨
2982 µ
2 ¶·
2983 µ
3 ¶·
DHo = –39,433 + 4,430 – 491 + 15.16 = –35,479 kcal
Next we find DHrxn at 500 °C (or) 773 K:
¦ 'H µ 3
¦ 'C µ 2
¶ T
¶ T + §¨
¨ 2 ·
3 ·
DH773 = DHo + Da T + §
ENERGY BALANCE
= –35,479 + (–14.891 ´ 773) +
+
207
11.047 t 10 3 t 7732
2
1.719 t 10 6 t 7733
3
= – 43,943 kcal/kmole
\ 43,943 kcal of heat must be removed.
9.20 CO at 50 °F is completely burnt at 2 atm pressures with 50% excess
air, which is at 1000 °F. The products of combustion leave the
combustion chamber at 800 °F. Calculate the heat evolved from
the combustion chamber in terms of Btu/lb of CO entering.
Basis: 1 lb mole of CO = 28 lb, O2 needed = 0.5 lb mole
CO + ½O2 ® CO2
O2 supplied = 0.5 ´ 1.5 = 0.75 lb mole (50% excess)
Air supplied = 3.57 lb mole and N2 = 2.82 lb moles
DHrxn = –1,21,745 Btu/lb mole
Q = DHrxn + DHproducts – DHreactants
O2 remaining = 0.25 lb mole
CO2 : 1 lb mole, N2 : 2.82 lb moles
Datum: 32°F
Data: DH (Btu/lb mole)
Temperature, °F
CO
Air
O2
N2
CO2
50
125.2
—
—
—
—
77
313.3
312.7
315.1
312.2
392.2
800
—
—
5,690
5,443
8,026
1000
—
6,984
—
—
—
DHproducts = DH800°F – DH77°F
= 1(8,026 – 392.2) + 2.82(5,443 – 312.2) + 0.25(5,690 – 315.1)
= 23,446 Btu/lb mole
DH reactants: (DH1000°F – DH77°F)air + (DH50°F – DH77°F)CO
= 3.57 (6,984 – 312.7) + 1(125.2 – 313.3)
= 23,612 Btu/lb mole
Q = –1,21,745 + 23,446 – 23,628 = –1,21,927 Btu/lb mole
Heat evolved by combustion = 1,21,927/28
= 4,354.5 Btu/lb of CO
208
PROCESS CALCULATIONS
9.21 Pure CO is mixed with 100% excess air and completely burnt at
constant pressure. The reactants are originally at 200 °F. Determine
the heat added or removed, if the product temperatures are 200 °F,
500 °F, 1000 °F, 1500 °F, 2000 °F and 3000 °F.
Basis: 1 lb mole of CO
CO + ½O2 ® CO2
O2 supplied = 1 lb mole, N2 = 3.76 lb moles
Exit: CO2 : 1 lb mole, O2 : 0.5 lb mole, N2 : 3.76 lb moles
Assuming a base temperature of 25 oC, (77 oF) and using mean heat
capacities,
DH = Hp – HR; Q = DH
DH = SnCppr (77 – 200) + DHrxn77 °F + SnCpR (t – 77)
Reactants: DHrxn = –1,21,745 Btu/lb mole
Gas
n
Cp
nC p
CO
O2
N2
1
1
3.76
6.95
7.10
6.95
6.95
7.10
26.13
Total
40.18
\ SnCppr (77 – 200) = –(40.18 ´ 123) = – 4,942 Btu
DH = – 4,942 –1,21,745 + SnCpR (t – 77°)
200 °F
500 °F
1000 °F
1500 °F
—
Cp
—
nCp
—
Cp
2000 °F
3000 °F
—
nCp
—
Cp
—
Cp
—
nCp
12.75
n
—
Cp
—
nCp
—
Cp
—
nCp
CO2 1.0
9.15
9.15
9.9
9.9
10.85 10.85
11.5
11.5
12.05 12.05
12.75
O2
0.5
7.10
3.55
7.25
3.63
7.15
3.88
7.8
3.9
8.0
4.0
8.3
4.15
N2
3.76
6.95 26.13
7.0
26.32
7.15 26.88
7.35
27.64
7.55 28.39
7.88
29.42
—
—
39.85
—
—
43.04
—
—
46.32
SnCp —
R
38.83
41.51
—
nCp
44.44
SnCp
(t–77)
4,776
16,857
38,314
61,246
83,758
1,35,810
Q = DH
–21,911
–1,09,830
–88,373
–65,441
–41,229
+9,123
9.22 Coal is burnt to a gas of the following composition: CO2 : 9.2,
CO : 1.5, O2 : 7.3, N2 : 82%. What is the enthalpy difference for this
gas between the bottom and the top of the stack if the temperature at
the bottom is 550 °F and at the top is 200 °F?
Cp of N2
= 6.895 + 0.7624 ´ 10–3 T – 0.7 ´ 10–7 T2
Cp of O2
= 7.104 + 0.7851 ´ 10–3 T – 0.5528 ´ 10–7 T2
ENERGY BALANCE
209
Cp of CO2 = 8.448 + 5.757 ´ 10–3 T – 21.59 ´ 10–7 T2 + 3 ´ 10–10 T3
Cp of CO = 6.865 + 0.8024 ´ 10–3 T – 0.736 ´ 10–7 T2
Basis: 1 lb mole of CO2:
Multiplying these equations by the respective mole fractions of each
component and adding them together, we have
for N2
= 0.82 ´ CpN2
for O2
= 0.073 ´ CpO2
for CO2 = 0.092 ´ CpCO2
for CO = 0.015 ´ CpCO
= 7.049 + 1.2243 ´ 10–3 T – 2.6164 ´ 10–7 T2 + 0.2815 ´ 10–10 T3
Cpnet
200
\ DH =
∫C
p net dT
550
¦
= 7.049 (200 – 550) + § 1.2243 t
¨
¦
– § 2.6164 t
¨
10 3 µ
(2002 – 5502)
2 ¶·
¦
10 7 µ
10 10 µ
3
3
0.2815
(200
–
550
)
+
t
(2004 – 5504)
§
3 ¶·
4 ¶·
¨
or, DH = –2,465 – 160.6 + 13.8 – 0.633 = –2,612 Btu
9.23 Calculate the theoretical flame temperature for CO burnt at constant
pressure with 100% excess air? The reactants enter at 200 °F.
CO + ½O2 ® CO2
Basis: 1 g mole CO
Temperature of reactants: 200 °F = 93.3 °C
Gases entering: CO–1, O2–1, N2–3.76 (all in moles)
Gases leaving: CO2–1, O2–0.5, N2–3.76 (all in moles)
\ DHrxn 25 °C = – 67,636 cal.
Gas
mole
DT
Cpm
DH = nCpmDT
CO
1.0
(93.3 – 25)
6.981
476
Air
4.76
(93.3 – 25)
6.993
2,270
Total
2,746 cal
Next we have DHproduct = – (DHreactants – DHrxn)
= – (2,746 + 67,636) = –70,382 cal
210
PROCESS CALCULATIONS
Let us assume exit temperature as 1800 °C, then DT = (1800 – 25)
= 1775 °C
Gas
mole
DT
Cpm
DH
CO2
O2
N2
1.0
0.5
3.76
1775
1775
1775
12.94
8.35
7.92
23,000
7,400
52,900
This total of –83,300 cal is not matching with –70,382 cal, the value
calculated.
Let the Theoretical flame temperature be 1500 °C, then DT = 1475 °C
DH = (1 ´ 12.7 ´ 1475) + (0.5 ´ 8.31 ´ 1475) + (3.76 ´ 7.88 ´ 1475)
= 68,460 cal
Making linear interpolation for the theoretical flame temperature, we
have,
Theoretical flame temperature
Ë 70,382 68,460 Û
= 1500 + Ì
Ü ´ (1800 – 1500)
Í 83,300 68,460 Ý
= 1500 + 39 = 1539 °C º 2798 °F
9.24 Calculate the theoretical flame temperature of a gas having 20% CO
and 80% N2 when burnt with 150% excess air. Both air and gas being
at 25 °C.
Data: Heat of formation of CO2 = – 94,052 cal/g mole, CO = –26,412
cal/g mole at 25 °C.
Cpm: CO2 : 12.1, O2 : 7.9, N2 : 7.55 cal/g mole K (from literature)
Basis: 1 g mole CO, CO + 0.5O2 ® CO2
O2 supplied = (0.5 ´ 2.5) = 1.25 g moles (150% excess)
N2 in feed
È 80 Ø
= 1 É Ù = 4 g moles
Ê 20 Ú
N2 in air
79 Ø
È
= É1.25 – Ù = 4.7 g moles
Ê
21 Ú
Exit gas: CO2 : 1 g mole, O2 : 0.75 g mole, N2 : 8.7 g moles
Q = SHproducts + SHrxn – SHreactants. (Datum 298 K)
SHreactants is zero, since air and gas are at 25 °C.
DHrxn = DHCO2 – DHCO
= –94,052 – (–26,412) = –67,640 cal/g mole.
Let the “Theoretical Flame Temperature” be T, K
67,640 = [1 ´ 12.1 ´ (T – 298)] + [8.7 ´ 7.55 ´ (T – 298)]
+ [0.75 ´ 7.9 ´ (T – 298)],
ENERGY BALANCE
211
67,640 = 83.71 T – 24,945.6
\ T = 1106.03 K ∫ 833.03 °C
1106 K ∫ 833 °C.
9.25 Find the theoretical flame temperature of a gas containing 30% CO
and 70% N2 when burnt with 100% excess air. The reactants enter at
298 K.
DHf CO2 = –3,93,700 kJ/kmole; DHf CO = –1,10,600 kJ/kmole
Mean molar specific heat, kJ/kmole K at different temperatures is
given below:
Temperature K
CO2
O2
N2
800
1000
1200
1400
1600
1800
45.4
47.6
49.4
50.8
52.0
53.2
31.6
32.3
33.0
33.6
34.0
34.4
30.3
30.6
31.2
31.8
32.3
32.7
Basis: 1 kmole of CO, Datum: 298 K
N2 in feed = 70/30 = 2.34 kmoles
O2 supplied = 0.5 ¥ 2 = 1 kmole
N2 from air = 3.76 kmoles
Exit gas consists of:
CO2 : 1, O2 : 0.5, N2 : (3.76 + 2.34) = 6.1 kmoles
Let us consider the equation
Q = SHproducts + DHrxn – SHreactants
where SHreactants = Zero at 298 K (Q Datum is 298 K)
\ DHrxn = (–3,93,700) – (–1,10,600) = –2,83,100 kJ/mole
By iteration method:
Let the theoretical flame temperature be 1400 K:
DT = (1400 – 298) = 1102 K
DHpr = (1 ¥ 50.8 ¥ 1102) + (0.5 ¥ 33.6 ¥ 1102) + (6.1 ¥ 31.8 ¥ 1102)
= 2,88,261 kJ/kmole π 2,83,100 kJ/kmole
Let the theoretical flame temperature be 1200 K
\ DT (1200 – 298) = 902 K
DHpr = (1 ¥ 49.4 ¥ 902) + (0.5 ¥ 33 ¥ 902) + (6.1 ¥ 31.2 ¥ 902)
= 2,31,110 kJ/kmole π 2,83,100 kJ/kmole
212
PROCESS CALCULATIONS
So theoretical flame temperature lies in between these two values
(by interpolation )
\ Theoretical flame temperature
Ë 2,83,100 2,31,110 Û
= 1,200 + Ì
Ü ´ (1400 – 1200)
Í 2,88,261 2,31,110 Ý
So, the temperature of the exit gases is 1382 K = 1109 °C.
9.26 The analysis of 15,000 lit of a gas mixture at standard condition is as
follows: SO2 : 10%, O2 : 12% and N2 : 78%. How much heat must be
added to this gas to change its temperature from 30 °C to 425 °C?
The Cpm values are in cal/g mole °C
Gas
SO2
O2
N2
Cpm 30 °C
10
6.96
6.80
Cpm 425 °C
11
7.32
7.12
The amount of gas mixture = 15,000 litres º
15,000
22.414
= 669.2 g moles
we can then write the amount of each component
SO2 : 669.2 ´ 0.1 = 66.92 g moles
O2 : 669.2 ´ 0.12 = 80.30 g moles
N2 : 669.2 ´ 0.78 = 521.98 g moles
669.20 g moles
Reference temperature: 0 °C
\ Q = 66.92 [(11 ´ 425) – (10 ´ 30)] + 80.3 [(7.32 ´ 425) – (6.96 ´ 30)]
+ 521.98 [(7.12 ´ 425) – (6.8 ´ 30)] = 19,98,849.22 cal
9.27 Estimate the theoretical flame temperature of a gas containing 20%
CO and 80% N2 when burnt with 100% excess air. Both air and gas
are initially at 25 °C.
Cp CO2 = 6.339 + 10.14 ´ 10–3 T – 3.415 ´ 10–6 T2
Cp O2
= 6.117 + 3.167 ´ 10–3 T – 1.005 ´ 10–6 T2
Cp N2
= 6.457 + 1.389 ´ 10–3 T – 0.069 ´ 10–6 T2
The values of Cp are in kcal/kmole K and temperature is in K
DHrxn 25 °C = –67,636 kcal
ENERGY BALANCE
213
Basis: 1 kmole of CO; N2 = 4 kmoles
Air supplied: O2 : 1 kmole, N2 from air : 3.76 kmoles
Exit gas: CO2 : 1, O2 : 0.5, N2 : 7.76 kmoles
Datum 25 °C = 298 K
DHrxn = –DH products + 67,636 = [1 ´ òCpCO2 ´ (T – 298)]
+ [0.5 ´ òCpO2 ´ (T – 298)]
+ [7.76 ´ òCpN2 ´ (T – 298)]
67,636 = 6.339(T – 298) + 10.14 ´ 10–3
T 2 2982
2
– 3.415 ´ 10–6
T 298
T 3 2983
+ 6.117 ´
2
3
+ 3.167 ´ 10–3
T 3 2983
T 2 2982
– 1.005 ´ 10–6
6
4
+ 7.76 ´ 6.457 ´ (T – 298) + 7.76 ´ 1.389 ´ 10–3
T 2 2982
2
T 3 2983
3
Solving for theoretical flame temperature = T = 1216 K = 943 °C
– 7.76 ´ 0.069 ´ 10–6
9.28 Dry methane and dry air at 298 K and 1 bar pressure are burnt with
100% excess air. The standard heat of reaction is –802 kJ/g mole of
methane. Determine the final temperature attained by the gaseous
products if combustion is adiabatic and 20% of heat produced is lost
to the surroundings.
Data: Cpm values (J/g mole K) for the components are:
O2 : 31.9, N2 : 32.15, H2O : 40.19, CO2 : 51.79.
Basis: 1 g mole of methane.
Datum: 298 K
CH4 + 2O2 ® CO2 + 2H2O
\ Oxygen supplied = 2 ´ 2 = 4 g moles
79
= 15.05 g moles
21
Gases leaving are: CO2 : 1, H2O : 2, O2 : 2, and N2 : 15.05 g moles.
N2 entering = 4 ´
Heat given out = 802 kJ
Heat loss = (802 ´ 0.2) = 160.4 kJ
\ Q = Heat in exit gases = (802 – 160.4) = 641.6 kJ
214
PROCESS CALCULATIONS
Q = [1 ´ 51.79 ´ (T – 298)] + [2 ´ 40.19 ´ (T – 298)]
+ [2 ´ 31.9 ´ (T – 298)] + [15.05 ´ 32.15 ´ (T – 298)]
= [679.8275 ´ (T – 298)] = 641.6 ´ 103 J.
\ T = 1242 K = 969 °C.
9.29 An iron pyrite ore contains 85% FeS2 and 15% gangue. It is roasted
with 200% excess air to get SO2. The reaction is given. All the gangue
plus Fe2O3 end up in the solid waste produced which analyzes 4%
FeS2. Determine the standard heat of reaction in kJ/kg of ore roasted
and the analysis of the solid waste. Heat of formation data is in kJ/g
mole.
4FeS2
+
®
11O2
2Fe2O3 + 8SO2
Weights
479.4
352
319.4
512
Heats of formation
(kJ/g mole)
–177.9
0
–88.2
–296.9
Basis: 1 kg of ore containing FeS2 = 0.85 kg and gangue = 0.15 kg
Let x kg of FeS2 be in the solid waste,
319.4 Û
Ë
then, FeS2 reacted: (0.85 – x) kg; Fe2O3 formed Ì(0.85 x ) –
479.4 ÜÝ
Í
= 0.666(0.85 – x)
x = 0.04{(0.15) + 0.666(0.85 – x) + x};
Solving the above, we find x = 0.029 kg
Solid waste = (0.15) + (0.029) + [0.666 ´ (0.85 – 0.029)] = 0.725786 kg
A summary of the composition of the solid waste is given:
Solid waste
kg
Weight %
Gangue
FeS2
Fe2O3
0.150
0.029
0.547
20.66
3.99
75.35
Total
0.726
100.00
\ FeS2 reacted = (0.85 – 0.029) = 0.821 kg = 6.85 ´ 10–3 kmole
Fe2O3 formed = 0.547 kg = 3.424 ´ 10–3 kmole
512 Û
Ë
= 1.068 kg = 0.016688 kmole
SO2 formed = Ì 0.821 –
479.4 ÜÝ
Í
Heat of reaction = –[(0.016688 ´ 296.9) + (3.424 ´ 10–3 ´ 822.3)
–(6.85 ´ 10–3 ´ 177.9)]
= –6.55 kJ
ENERGY BALANCE
215
9.30 For the following reaction, estimate the heat of reaction at 298 K.
A+B®C+D
Compound
DH°f (kcal/g mole)
A
B
C
D
–269.8
–195.2
–337.3
–29.05
DH° at 25 °C = SDH°f, Products – SDH°f, reactants
= [–337.3 – 29.05] – [–269.8 – 195.2]
= 98.65 kcal
9.31 Estimate the standard heat of reaction DH°298 for the reaction.
A+B®C
Standard heats of combustion are:
DHc, 298 for A = –328000 cal/g
DHc, 298 for B = –212000 cal/g
DHc, 298 for C = –542000 cal/g
DHc, 298 = SDH°c, reactants – SDH°c, products
= [–328000 – 212000] – [–542000]
= 2000 cal
9.32 Calculate the heat of formation of CHCl3 from the following data:
1
CHCl3 + O2 + H2O ® CO2 + 3HCl, DH = –509.93 kJ
(1)
2
1
H2 + O2 ® H2O; DH = –296 kJ
(2)
2
(3)
C + O2 ® CO2; DH = –393.78 kJ
1
1
H2 + Cl2 ® HCl; DH = –167.5 kJ
(4)
2
2
1
CO2 + 3HCl ® CHCl3 + O2 + H2O; DH = –509.93 kJ
2
1
H2 + O2 ® H2O; DH = –296 kJ
2
C + O2CO2; DH = –393.78 kJ
Eq. (4) × 3 + Eq. (3)/2, gives
1
1
1
3
H2 + Cl2 + C + O2 ® 3HCl + CO2
2
2
2
2
216
PROCESS CALCULATIONS
9.33 Standard heat of reaction accompanying any chemical change is equal
to the algebraic sum of the standard heat of formation of the products
minus the algebraic sum. Calculate the standard heat of reaction,
DH°f, 298 for the reaction
2FeS2 + 1.5O2 ® Fe2O3 + 4SO2
The standard heats of formation are:
FeS2 = –42,520 cal/g mole
Fe2O3 = –1,96,500 cal/g mole
SO2
= –70,960 cal/g mole
From the reaction,
DH°r, 298 = DHFe2O3 + 4 [DHSO2] – 2[DHFeS2]
= –19,6500 + 4[–70,960] – 2[42,520]
= –39,5300 cal
9.34 The heat of reaction at 300 K and 1 atm pressure for the reaction
A + 3B ® C is 30,000 cal/mole A converted.
Cp data is as follows:
A = –0.4 + 0.1 T (T in K)
B = 10
C = 30
Calculate the heat of reaction at 600 K and 1 atm
A + 3B ® C
600
DH600 = DH300 + ± ' C p .dT
300
600
= –30,000 + ± [{30 3 × 10 + 0.4} 0.1 T ] dT
300
= –30,000 + 0.4T – 0.1
T2
2
= –30,000 + 0.4[600 – 300] – [0.05 × (6002 – 3002)]
= –43,380 cal/g mole of A
9.35 Calculate the theoretical flame temperature of a gas containing 20%
CO and 80% N2 when burnt with 150% excess air, with both air and
gas being at 25 °C.
ENERGY BALANCE
217
DH°f Cpm
CO2 = –393.137 kJ/g mole CO2 = 50.16 kJ/kg K
CO = –110.402 kJ/g mole O2 = 33.02 kJ/ kmole
H2 = 31.56 kJ/kmole K
Basis: 100 g moles of feed
CO = 20 g moles
1
CO + O2 Æ CO2
2
1
O2 needed =
× 20 = 10 g moles
2
O2 supplied = 2.5 × 10 = 25 g moles
N2 supplied = 25 × 79/21 = 94.05 g moles
Gases leaving
CO2 = 20 g moles
O2 = 25 – 10 = 15 g moles
N2 = 94.05 + 80 = 174.05 g moles
Atmospheric temperature = 25 °C
Heat in reactants + DHR = Heat in products
Standard heat of reaction,
DHf°products – DHf°reactants = –393.137 – [110.402 +
1
× 0]
2
= –282.835 kJ/g mole
Heat of reactants is zero (Reference temperature)
Heat produced when 20 g moles of CO is burnt
= 282.835 × 20 = 5654.700 kJ
Heat in outgoing gas,
{20 [50.16] + 15 [33.02] + 174.05 [31.56]} (T – 25) = 5654700
Solving, T = 834 °C
EXERCISES
9.1
Determine the theoretical flame temperature that can be attained by
the combustion of methane with 20% excess air. Air and methane
enter at 298 K and a pressure of 1 atm. The reaction is complete.
–DHr = 1,91,760 cal/g mole
218
PROCESS CALCULATIONS
Mean heat capacities (cal/g mole K)
Component
9.2
Temperature
2000 °C
1800 °C
CO2
13.1
12.95
H2O
10.4
10.25
O2
8.4
8.3
N2
8.0
7.9
Determine the heat of reaction at 720 K and 1 atm for the reaction
SO2 + 0.5O2 ® SO3
Mean molar specific heats of
SO2 : 51.5 kJ/kmole K
O2 : 45.67 kJ/kmole K
SO3 : 30.98 kJ/kmole K
Standard heat of formation for
SO2 : –2,97,000 kJ/kmole
SO3 : –3,95,000 kJ/kmole
9.3
Calculate the enthalpy change in J/kmole that takes place in raising
the temperature of 1 kmole of the gas mixture of 80 mole %. Methane
and rest ethane from 323 K to 873 K
Heat capacity equation, Cp = R (A + BT + CT2) cal/g mole
where, R is Gas constant, T is temperature in K. A, B and C are
constants and Cp is the heat capacity at constant pressure.
Component
CH4
C2 H 6
9.4
Value of constant in Cp equation
A
B ´ 103
C ´ 106
1.702
1.131
9.081
19.225
–2.164
–5.561
Chlorine is produced by the reaction
4HCl(g) + O2(g) ® 2H2O(g) + 2Cl2(g)
The feed stream to the reactor consists of 67 mole % HCl, 30 mole %
O2 and 3 mole % N2. If the conversion of HCl is 75% and the process
is isothermal, how much heat is transferred per mole of entering gas
mixture.
ENERGY BALANCE
219
Data:
(a) Standard heat of formation at 25 °C, J/g mole
HCl(g) : –92,307 J/g mole
H2O(g) : –2,41,818 J/g mole
(b) Mean heat capacities: (cal/g mole K)
HCl(g) : 7.06
O2(g) : 8.54
H2O : 7.52
Cl2(g) : 8.61
N2(g) : 7.16
9.5
Calculate the number of joules required to calcine completely 100 kg
of limestone containing 80% CaCO3, 11% MgCO3 and 9% water. The
lime is withdrawn at 900 °C and the gases leave at 200 °C. The lime
stone is charged at 25 °C.
Data: Standard heat of formation at 25 °C and 1 atm, cal/g mole
CaCO3 : – 2,88,450 cal/g mole
MgCO3 : – 2,66,000 cal/g mole
CaO
: – 1,51,900 cal/g mole
MgO
: – 1,43,840 cal/g mole
: – 94,050 cal/g mole
CO2
Mean molal heat capacity, cal/g mole K
: 8.2
H2O
: 10.5
CO2
CaCO3 : 25.0
MgCO3 : 23.0
CaO
: 14.0
MgO
: 10.0
9.6
In the reaction
4FeS2(s) + 11O2 (g) ® 2Fe2O3(s) + 8SO2(g)
the conversion from FeS2 to Fe2O3 is only 80% complete.
If the standard heat of formation for the above is calculated to be
–197.7 kcal/g mole, what –DH°reaction should be used in energy
balance per kg of FeS2 fed.
9.7
Calculate the theoretical flame temperature for CO burnt at constant
pressure with 20% excess air. The reactants enter at 366 K.
CO + ½O2 ® CO2
220
PROCESS CALCULATIONS
The heat capacities are 29.23 for CO, 29.28 for air, 54.18 for CO2,
34.5 for O2 and 33.1 for N2 in J/g mole K. Standard heat of reaction:
–283.13 kJ/g mole.
9.8
A gaseous mixture of 1000 m3 containing 60% hydrogen and 40%
ammonia is cooled from 773 K to 313 K at 1 atm pressure. Calculate
the heat removed. The Cp values in kcal/kmole K, and T in K are:
for hydrogen Cp = 6.9 – 0.2 ´ 10–3 T + 0.48 ´ 10–6 T2
for ammonia Cp = 6.08 + 8.81 ´ 10–3 T – 1.5 ´ 10–6 T2
9.9
Determine the theoretical flame temperature that can be obtained by
the combustion of methane with 25% excess air. Air and methane
enter at 298 K and a pressure of 1 atm. The reaction is complete.
Standard heats of formation at 298 K in kJ/kmole are:
Methane (g) = –74,520
Carbon dioxide (g) = –3,93,500
Water vapour (g) = –2,41,813
9.10 Calculate the amount of heat given off when 1 m3 of air at standard
condition cools from 600 °C to 100 °C at constant pressure
Cp air = 6.386 + 1.762 ´ 10–3 T – 0.2656 ´ 10–6 T2
Cp is in kcal/kmole K and T is in K.
9.11 CO is burnt under atmospheric pressure with dry air at 773 K with
20% excess air. The products leave at 1223 K. Calculate the heat
involved in the reaction chamber in kcal/kmole of CO burnt, assuming
complete combustion.
Data: – DH298 K = – 67,636 kcal
Mean specific heats are: 7.017 for CO, 7.225 for air, 11.92 for CO2,
7.941 for O2 and 7.507 for N2 in kcal/kmole K.
10
Problems on Unsteady
State Operations
The term unsteady state refers to chemical processes in which the operating
conditions generally fluctuate with time. Although unsteady-state processes
are difficult to formulate, the general formula used to represent the total
amount of material and energy in the process is given as
Rate of input + Rate of generation = Rate of output + Rate of accumulation
This is the guiding principle in solving problems on the unsteady state
operations.
WORKED EXAMPLES
10.1 A storage tank contains 10,000 kg of a solution containing 5% acetic
acid by weight. A fresh feed of 500 kg/min of pure water is entering
the tank and dilutes the solution in the tank. The mixture is stirred
well and the product leaves the tank at a rate of 500 kg/min. At what
instant of time the acid concentration in the tank will drop to 1%
acetic acid by weight? After one hour of operation, what will be the
concentration in the tank?
F iX i
M, X
F o, X o = X
221
222
PROCESS CALCULATIONS
Here inlet flow rate, Fi = 500 kg/min
Outlet flow rate, Fo = 500 kg/min
Initial mass = 10,000 kg
The total mass M, at any time in the tank
= Initial mass + (Inflow rate – Outflow rate)(time)
= 10,000 + (500 – 500)t
= 10,000 kg
We know that,
rate of input + rate of generation = rate of output
+ rate of accumulation
Here, rate of generation is zero.
(1)
Let X be the concentration of acid and M the mass of solution of acid
at any time.
Hence, rate of accumulation = rate of input – rate of output
or,
M
d ( MX )
= Fi X i – F o X
dt
(2)
dX
dM
X
= Fi X i – F o X
dt
dt
(3)
dM
= 0 (Since inlet and outlet flow rates are the same)
dt
dX
Therefore, M
= FiXi – Fo X(4)
dt
È dX Ø
Now substituting values, 10,000 É Ù = 500 ´ 0.0 – 500X
Ê dt Ú
Here,
(i.e.)
dX
= – 0.05X
dt
(i.e.)
dX
= – 0.05dt
X
(5)
Integrating, we get
X
t
Xo
t0
dX
± X 0.05 ± dt
È XØ
ln É Ù = –0.05(t – 0)
Ê Xo Ú
= –0.05t
(6)
PROBLEMS ON UNSTEADY STATE OPERATIONS
Therefore,
X
= e–0.05t
Xo
223
(7)
Time taken to reach a concentration of 1% is given by,
0.01 Ø
= –0.05t
ln ÈÉ
Ê 0.05 ÙÚ
i.e.
t = 32.19 minutes
(b) Substituting for t as 60 minutes in Eq. (7), we get
X = 0.05 e–0.05 (60) = 0.00249 = 0.249%
i.e. after one hour of operation, the concentration in the tank will be
0.249%
10.2 A tank contains 10 kg of a salt solution at a concentration of 2% by
weight. Fresh solution enters the tank at a rate of 2 kg/min at a salt
concentration of 3% by weight. The contents are stirred well and the
mixture leaves the tank at a rate of 1.5 kg/min.
(a) Express the salt concentration as a function of time and
(b) At what instant of time the salt concentration in the tank will
reach 2.5% by weight?
Here inlet flow rate, Fi = 2 kg/min
Outlet flow rate, Fo = 1.5 kg/min
Initial mass = 10 kg
The total mass M, at any time in the tank
= Initial mass + (Inflow rate – Outflow rate)(time)
= 10 + (2 – 1.5)t
or,
M = 10 + 0.5t
Differentiating, we get
dM
= 0.5
dt
We know that
rate of input + rate of generation
= rate of output + rate of accumulation
Here, rate of generation is zero.
(1)
Let X be the concentration of acid and M the mass of solution of acid
at any time.
Hence, rate of accumulation = rate of input – rate of output
d ( MX )
= F i Xi – Fo X
i.e.
dt
dM
dX
+X
= F i Xi – Fo X
or,
M
dt
dt
(2)
(3)
224
PROCESS CALCULATIONS
dM
= 0.5
dt
Therefore, by substituting values, we get
Here,
dX
+ 0.5X = 2 ´ 0.03 – 1.5X
dt
dX
= 0.06 – 2X
(10 + 0.5t)
dt
dX
dt
0.06 2 X 10 0.5t
(10 + 0.5t)
(i.e.)
or,
(4)
(5)
Integrating, we get
X
dX
©
¸
± ª« 0.06 2 X ¹º Xo
X
also,
0.5
±
X o 0.02
± 10
t0
dt
0.5t
t
dX
dt
2 ±
20 t
X 0.03
Ë X 0.03 Û
ln Ì
Ü
Í 0.02 0.03 Ý
X 0.03
0.01
or,
t
t0
Ë (20 t ) Û
4 ln Ì
Ü
Í 20 Ý
Ë 20 Û
Ì 20 t Ü
Í
Ý
4
Ë 20 Û
X = 0.03 – 0.01 Ì
Ü
Í 20 t Ý
\
(6)
(7)
4
(8)
Time taken to reach a concentration of 2.5% is given by substituting
X = 0.025 in Eq. (7).
Hence, we have,
(20 + t)4 = 204 ´
0.01
0.03 X
Hence, t = 3.784 minutes
Aliter
We shall go back to Eq. (4), which is
(10 + 0.5t)
or,
dX
+ 0.5X = 2 ´ 0.03 – 1.5X
dt
Û
dX Ë
2
X
Ì
dt Í 10 0.5t ÜÝ
Ë 0.06 Û
Ì 10 0.5t Ü
Í
Ý
(9)
PROBLEMS ON UNSTEADY STATE OPERATIONS
dy
+ Py = Q
dx
Where, P and Q are either functions of x or constants
This equation is of the form
225
(10)
The solution for this differential equation is
ye òPdx = òQe òPdx dx + constant
(11)
Using the same analogy we can solve Eq. (9) in the following manner
P=
2
10 0.5t
4
20 t
Q
0.06
10 0.5t
0.12
20 t
and
Substituting for P and Q in Eq. (11), we get
4
Ô
X e 20 t
4
dt
Ô
0.12 Ô 20 t dt
dt + constant
e
20 t
(12)
X ´ 4 ´ exp[ln(20 + t)]
=
X(20 + t)4 =
Ë 0.12 Û
Ô ÌÍ 20 t ÜÝ ´ 4 ´ exp[ln(20 + t)] dt + constant
Ë 0.12 Û
Ô ÌÍ 20 t ÜÝ [20 + t]4dt + constant
or, X(20 + t)4 =
Ô [0.12][20 + t]3dt + constant
X(20 + t)4 = 0.12
[20 t ]4
+ constant
4
X = 0.03 +
constant
(13)
(14)
(20 t )4
Initial conditions are:
t = 0, X = 0.02
Substituting in Eq. (14), we get,
Constant = – 0.01 ´ (20)4 = –1600
Equation (14) thus becomes,
X = 0.03 –
1600
(20 t )4
and
(15)
226
PROCESS CALCULATIONS
Ë 20 Û
X = 0.03 – 0.01 Ì
Ü
Í (20 t ) Ý
4
(16)
Comparing Eq. (16) with Eq. (8) we find both are same and hence the
time taken to reach a concentration of 2.5% is 3.784 minutes
10.3 A tank contains 10 litre of a salt solution at a concentration of 2 g/litre
Another salt solution enters the tank at a rate of 1.5 litres/min at a
salt concentration of 1 g/litre. The contents are stirred well and the
mixture leaves the tank at a rate of 1.0 litre/min.
Estimate (a) the time at which the concentration in the tank will be
1.6 g/litre and (b) the contents in the tank will be 18 litres
Here, Inlet flow rate, Fi = 1.5 litres/min at a salt concentration of 1 g/litres
Outlet flow rate, Fo = 1.0 litre/min
Initial volume = 10 litres.
The total volume V, at any time
= Initial volume + (Inflow rate – Out flow rate) (time)
= 10 + (1.5 – 1.0)t
or,
V = 10 + 0.5t
Differentiating, we get
dV
= 0.5
dt
We know that,
rate of input + rate of generation
= rate of output + rate of accumulation
(1)
Here rate of generation is zero.
Let C be the concentration of salt and V the volume of solution at any
time.
Hence, rate of accumulation = rate of input – rate of output
i.e.
V
Here,
dV
= 0.5
dt
d (VC )
= Fi C i – F o C
dt
(2)
dC
dV
+C
= Fi C i – F o C
dt
dt
(3)
PROBLEMS ON UNSTEADY STATE OPERATIONS
Therefore, by substituting values,
dC
(10 + 0.5t)
+ 0.5C = 1.5 ¥ 1.0 – 1.0C
dt
dC
= 1.5 – 1.5C = 1.5(1 – C)
i.e.
(10 + 0.5t)
dt
dC
dt
=
or,
1.5(1 - C ) 10 + 10.5t
Integrating, we get
C
227
(4)
(5)
t
dt
⎛ 1 ⎞ ⎛ dC ⎞
⎜ 1.5 ⎟ ⎜ 1 − C ⎟ =
+
(10
0.5t )
⎝
⎠C ⎝
⎠ t=0
∫
∫
o
C
t
dt
⎛ 1 ⎞
⎛ dC ⎞
⎜ − 1.5 ⎟
⎜C −1⎟ = 2
(20
+ t)
⎝
⎠C = 2⎝
⎠
t=0
∫
∫
o
È C - 1˘
È 20 + t ˘
= - 3 ln Í
ln Í
˙
˙
Î 2 -1˚
Î 20 ˚
C - 1 È 20 ˘
=Í
˙
1
Î 20 + t ˚
(6)
3
(7)
È 20 ˘
\ C=1+ Í
˙
Î 20 + t ˚
3
(8)
Time taken to reach a concentration of 1.6 g/litres is given by
substituting C = 1.6 in Eq. (8):
È 20 ˘
0.6 = + Í
˙
Î 20 + t ˚
solving, t = 3.71 minutes.
3
(b) Final volume
= Initial volume + (volume flowing in – volume flowing out)(time)
or,
18 = 10 + (1.5 – 1.0)t
Therefore, time taken for the water in the tank to reach 18 litres is =
16 minutes.
EXERCISES
10.1 A tank contains 500 kg of a 10% salt solution. A stream containing
salt at 20% concentration enters the tank at 10 kg/h and the mixture
leaves the tank after thorough mixing at a rate of 5 kg/h. Obtain an
expression for the salt concentration in the tank as a function of time
and the salt concentration in the tank after 3 hours.
228
PROCESS CALCULATIONS
10.2 A tank contains 1000 kg of a 10% salt solution. A stream containing
salt at 20% concentration enters the tank at 20 kg/min and the mixture
leaves the tank after complete mixing at a rate of 10 kg/min. Obtain
an expression for the salt concentration in the outlet as a function of
time and the salt concentration in the tank after 1 hour. What will be
the time at which the salt concentration in the tank will be 15%?
10.3 A tank contains 50 litres of a salt solution at a concentration of
2.5 g/litre. Another salt solution enters the tank at a rate of 2.0 litres/min.
at a salt concentration of 1 g/litre. The contents are stirred well and the
mixture leaves the tank at a rate of 2.5 litres/min.
Estimate (a) the time at which the concentration in the tank will be
1.25 g/litre and (b) the contents in the tank will be 20 litres.
10.4 A tank contains 20 kg of a salt solution at a concentration of 4% by
weight. Fresh solution enters the tank at a rate of 2.5 kg/min at a salt
concentration of 3% by weight. The contents are stirred well and the
mixture leaves the tank at a rate of 2.0 kg/min. (a) Express the salt
concentration as a function of time and (b) At what instant of time the
salt concentration in the tank will reach 3.75% by weight?
10.5 A storage tank contains 5000 kg of a 1% sugar solution by weight. A
fresh feed of 400 kg/min of pure water is entering the tank and dilutes
the solution in the tank. The mixture is stirred well and the product
leaves the tank at a rate of 400 kg/min. At what instant of time the
sugar concentration in the tank will drop to 1% sugar by weight?
After one hour of operation, what will be the concentration in the
tank?
10.6 A 15% Na2SO4 solution is fed at the rate of 12 kg/min into a mixer
that initially holds 100 kg of a 50 : 50 mixture of Na2SO4 and water.
The exit solution leaves at the rate of 9 kg/min. Assuming uniform
mixing, what is the concentration of Na2SO4 in the mixer at the end
of 10 minutes. Volume change during mixing can be neglected.
10.7 A cylindrical tank of cross-sectional area A is filled with liquid up to a
height Ho. A hole of diameter d at the bottom of the tank, which was
plugged initially is opened to let the liquid drain through it. Set up an
unsteady state balance equation to calculate the time for the liquid
level to fall to a new height H1.
10.8 A solution is having solute A at a concentration CAf is fed
continuously into a mixing vessel of constant volume V, into which
water is added continuously and diluted. The outlet concentration of
the solution is Cao. Write the unsteady state solute balance equation.
Discuss about the solution of the equation.
Tables
Important Conversion Factors
TABLE I
Quantity
Length
Area
To convert
to
in (¢¢)
m
0.0254
ft (¢)
m
0.3048
cm
m
0.01
Angstrom (Å)
m
10–10
microns (m)
m
10–6
in2
m2
6.452 ´ 10–4
ft2
m2
0.0929
cm
Volume
Density
2
2
10–4
m3
0.02832
3
m
3
10–6
litre
m3
10–3
Gallons (UK)
m3
4.546 ´ 10–3
Gallons (US)
m3
3.285 ´ 10–3
Pound (lb)
kg
0.4536
Gram (g, gm)
kg
10–3
lb/ft3
kg/m3
g/litre
3
kg/m
1.0
3
3
kg/m
1000
lbf
N
4.448
kgf
N
9.807
Pa
N
980.7
dyne
N
10–5
g/cm
Force
m
ft3
cm
Mass
Multiply by
from
16.019
(Contd.)
229
230
TABLES
Important Conversion Factors (contd.)
TABLE I
Quantity
To convert
from
Pressure
lbf/ft2
N/m2 = Pa
47.88
2
N/m = Pa
6895
2
N/m = Pa
3386
in water
2
N/m = Pa
249.1
mm Hg
2
133.3
2
2
lbf/in (psi)
in Hg
Heat or energy
Volumetric flow rate
Molar flow rate
(Molar flux)
Enthalpy
Heat capacity (Holds
good for molal heat
capacity also)
N/m = Pa
atm
N/m = Pa
1.0133 ´ 105
torr
N/m2 = Pa
133.3
bar
2
N/m = Pa
105
kgf/cm2
N/m2 = Pa
9.807 ´ 104
Btu
J = N.m
1055
erg
J = N.m
10–7
cal
J = N.m
4.187
kcal
J = N.m
4187
kW.h
J = N.m
3.6 ´ 106
m3/s
0.02832
ft /h, ft /h
m3/s
7.867 ´ 10–6
cm3/s
m3/s
10–6
lit/h
m3/s
2.777 ´ 10–7
lb/ft2.h
kg/m2s
1.356 ´ 10–3
g/cm2s
kg/m2s
10
lb mole/ft2.h
kmole/m2s
1.356 ´ 10–3
g mole/cm2s
kmole g/m2s
10
Btu/lb
J/kg = N.m/kg
2326
cal/g = kcal/kg
J/kg = N.m/kg
4187
Btu/lb. °F
N.m/kg K = J/kg K
4187
cal/g.°C
N.m/kg K = J/kg K
4187
ft3/s
3
Mass flow rate
(Mass flux)
Multiply by
to
3
TABLES
TABLE II
231
Atomic Weights and Atomic Numbers of Elements
Element
Symbol
Atomic Number
Atomic weight
Actinium
Ac
89
217.00
Aluminium
Al
13
26.98
Americium
Am
95
243.00
Antimony
Sb
51
121.76
Argon
A
18
39.94
Arsenic
As
33
74.91
Astatine
At
85
210.00
Barium
Ba
56
137.36
Berkelium
Bk
97
245.00
Beryllium
Be
4
9.01
Bismuth
Bi
83
209.00
Boron
B
5
10.82
Bromine
Br
35
79.92
Cadmium
Cd
48
112.41
Calcium
Ca
20
40.08
Californium
Cf
98
246.00
Carbon
C
6
12.01
Cerium
Ce
58
140.13
Cesium
Cs
55
132.91
Chlorine
Cl
17
35.46
Chromium
Cr
24
52.01
Cobalt
Co
27
58.94
Columbium
Nb
41
92.91
Copper
Cu
29
63.54
Curium
Cm
96
243.00
Dysprosium
Dy
66
162.46
Erbium
Er
68
167.20
Europium
Eu
63
152.00
Fluorine
F
9
19.00
Francium
Fr
87
223.00
Gadolinium
Gd
64
156.90
Gallium
Ga
31
69.72
Germanium
Ge
32
72.60
(Contd.)
232
TABLES
TABLE II
Atomic Weights and Atomic Numbers of Elements (contd.)
Element
Symbol
Atomic Number
Atomic weight
Gold
Au
79
197.20
Hafnium
Hf
72
178.60
Helium
He
2
4.00
Holmium
Ho
67
164.94
Hydrogen
H
1
1.00
Indium
In
49
114.76
Iodine
I
53
126.91
Iridium
Ir
77
193.10
Iron
Fe
26
55.85
Krypton
Kr
36
83.80
Lanthanum
La
57
138.92
Lead
Pb
82
207.21
Lithium
Li
3
6.94
Lutetium
Lu
71
174.99
Magnesium
Mg
12
24.32
Manganese
Mn
25
54.93
Mercury
Hg
80
200.61
Molybdenum
Mo
42
95.95
Neodymium
Nd
60
144.27
Neptunium
Np
93
237.00
Neon
Ne
10
20.18
Nickel
Ni
28
58.69
Niobium
Nb
41
92.91
Nitrogen
N
7
14.01
Osmium
Os
76
190.20
Oxygen
O
8
16.00
Palladium
Pd
46
106.70
Phosphorus
P
15
30.98
Platinum
Pt
78
195.23
Plutonium
Pu
94
242.00
Polonium
Po
84
210.00
Potassium
K
19
39.10
Praseodymium
Pr
59
140.92
(Contd.)
TABLES
TABLE II
233
Atomic Weights and Atomic Numbers of Elements (contd.)
Element
Symbol
Atomic Number
Atomic weight
Promethium
Pm
61
145.00
Protactinium
Pa
91
231.00
Radium
Ra
88
226.05
Radon
Rn
86
222.00
Rhenium
Re
75
186.31
Rhodium
Rh
45
102.91
Rubidium
Rb
37
85.48
Ruthenium
Ru
44
101.70
Samarium
Sm
62
150.43
Scandium
Sc
21
44.96
Selenium
Se
34
78.96
Silicon
Si
14
28.09
Silver
Ag
47
107.88
Sodium
Na
11
23.00
Strontium
Sr
38
87.63
Sulphur
S
16
32.07
Tantalum
Ta
73
180.88
Technetium
Tc
43
99.00
Tellurium
Te
52
127.61
Terbium
Tb
65
159.20
Thallium
Tl
81
204.39
Thorium
Th
90
232.12
Thulium
Tm
69
169.40
Tin
Sn
50
118.70
Titanium
Ti
22
47.90
Tungsten
W
74
183.92
Uranium
U
92
238.07
Vanadium
V
23
50.95
Xenon
Xe
54
131.30
Ytterbium
Yb
70
173.04
Yttrium
Y
39
88.92
Zinc
Zn
30
65.38
Zirconium
Zr
40
91.22
234
TABLES
TABLE III(a) Empirical Constants for Molal Heat Capacities of Gases at
Constant Pressure
Cp = a + bT + cT 2, where T is in Kelvin; g-cal/(g-mole) (K) – Temperature
range 300 to 1500 K
Gas
a
b ´ 103
c ´ 106
H2
N2
O2
CO
NO
H2O
CO2
SO2
SO3
HCl
C2H6
CH4
C2H4
Cl2
Air
NH3(*)
6.946
6.457
6.117
6.350
6.440
7.136
6.339
6.945
7.454
6.734
2.322
3.204
3.019
7.653
6.386
5.920
– 0.196
1.389
3.167
1.811
2.069
2.640
10.140
10.010
19.130
0.431
38.040
18.410
28.210
2.221
1.762
8.963
0.4757
–0.069
–1.005
–0.2675
–0.4206
–0.0459
–3.415
–3.794
–6.628
+0.3613
–10.970
–4.480
–8.537
–0.8733
–0.2656
–1.764
* Gas
TABLE III(b) Molal Heat Capacities of Hydrocarbon Gases
For 10 to 760 oC, Cp = a + bT + cT 2
For –180 to 95 oC, Cp = 7. 95 + mT n
T in Rankine (oF + 460)
Compound
a
b ´ 103
–c ´ 106
m
n
Methane
Ethylene
Ethane
Propylene
Propane
n-Butane
i-Butane
Pentane
3.42
2.71
1.38
1.97
0.41
2.25
2.30
3.14
9.91
16.20
23.25
27.69
35.95
45.40
45.78
55.85
1.28
2.80
4.27
5.25
6.97
8.83
8.89
10.98
6.4 ´ 10–12
8.13 ´ 10–11
6.20 ´ 10–8
2.57 ´ 10–3
3.97 ´ 10–3
0.93 ´ 10–2
0.93 ´ 10–2
3.9 ´ 10–2
4.00
3.85
1.79
1.26
1.25
1.19
1.19
1.00
Answers to Exercises
CHAPTER 1
1.1
(a) 10.84 cm2/s
(b) 39.67 psia
(c) 0.03929 hp-hr
(d) 0.627 lbf/ft2
(e) 0.01355 cal/s cm2°C
(f) 1163 W/m K
1.2
10.93 [Cp G 0.8/D 0.2]
1.3
4.55 * 10 –3 [P/T 0.5]
1.4
(0.09453) [h2.5/g0.5] tan F
1.5
As long as consistent units are used, the equation remains the same.
CHAPTER 2
2.1
500 g moles
2.2
0.5455 kg of carbon
2.3
(a) 3.572 g of O2
(b) 12.77 g of KClO3
2.4
(a) 2.8%
(b) 0.088
(c) 5.378 g moles/kg of water
2.5
Mole ratio: 0.0425
Mole %: Na2CO3: 4.1%
Water: 95.9%
2.6
Molality: 5.98 g moles/kg of solution
Molarity: 5.98 g moles/litre
Volume of solution: 0.0836 litre
Normality: 5.98
235
236
2.7
ANSWERS TO EXERCISES
Weight %: 39.02%
Volume %: 26%
2.8
Compound
Weight %
Volume %
mole %
NaCl
H2O
23.3
76.7
11
89
8.54
91.46
Total
100.00
100
100.00
Atomic %:
Na: 2.93%; Cl2: 2.93%; H2: 62.76%; and O2: 31.38%
Molality: 5.185 g moles/kg of solution
2.9
AVMWT: 65.02
Chlorine: 54.98%
Bromine: 10.15%
Nitrogen: 34.87%
2.10 (a) 0.1455 kg sugar/kg water
(b) 1075 kg/m3 solution
(c) 136.56 g sugar/litre
2.11 (a) Nitrogen
(b) 16.17% and
(c) 59.97%
2.12 0.945 g of Cr2S3
2.13 9.3445 kg of AgNO3
2.14 (a) Mole fraction of H3PO4: 0.02
Mole fraction of Water: 0.98
(b) 890.8 cc of solution
2.15 54,090 g
2.16 31.75 kg Cu, 104.26 kg of 94% H2SO4
2.17 (a) H2SO4
(b) 79.25%
(c) 89.625
2.18 NaCl:75%, KCl 25% and NaCl: 79.25%, KCl 20.75%
2.19 2094 kg iron and 900 kg water
2.20 (a) 26.94
ANSWERS TO EXERCISES
237
(b) Methane: 24.94%, ethane: 14.47%, ethylene: 25.98%, propane:
9.79%, propylene: 14.03% n-butane 10.76%
(c) 0.93
2.21 (a) Methane: 24.16% and carbon dioxide: 66.44% and nitrogen 9.4%
(b) 29.8
(c) 1.33 kg/m3
CHAPTER 3
3.1
Compound
Weight fraction
mole fraction
mole %
Butane
Pentane
Hexane
0.5
0.3
0.2
0.5701
0.2758
0.1541
57.01
27.58
15.41
Total
1.0
1.0000
100.00
AVMWT: 66.138
3.2
70,748 g
3.3
37.63%
3.4
(a) and (b)
Component
mole %
Weight %
Methane
80
68.45
Ethane
15
24.10
Nitrogen
5
7.45
100
100.00
Total
(c) AVMWT: 18.7
(d) Density: 0.0008 g/cc
3.5
AVMWT: 30.5; Density of gas: 1.36 g/litre
3.6
Methane: 0.008892 kmole/m3
Ethane: 0.02231 kmole/m3
Hydrogen: 0.01338 kmole/m3
Velocity: 30,558 m/h
Density: 0.00405 g/cc
3.7
1.169 g/litre
3.8
0.567 litre
238
3.9
ANSWERS TO EXERCISES
10,719.7 K
3.10 C3H8
3.11 300.37 atm
3.12 (a) and (c)
Component
Volume %
mole fraction
N2
64.75
0.6475
CO2
9.52
0.0952
H2O
18.62
0.1862
O2
6.11
0.0611
CO
1.00
0.0100
Total
100.00
1.0000
(b) Average density: 0.792 kg/m3
3.13 52.923 kg
3.14
Component
Volume % = mole %
Chlorine
Bromine
Oxygen
68.59
12.67
18.74
Total
100.00
3.15 (a) Partial pressure: 0.09 atm
(b) 3 m3
(c) 2.06 g/litre
3.16 (a), (b) and (c)
Component
mole fraction
g mole/cc
Concentration,
mm Hg
Partial pressure
CH4
C2 H 6
H2
0.1
0.3
0.6
1.058 ´ 10–5
3.173 ´ 10–5
6.347 ´ 10–5
200
600
1,200
(d) 1.058 ´ 10–4 g mole/cc
(e) 1,248 g/s
(f) AVMWT: 11.8
ANSWERS TO EXERCISES
239
3.17 (a) Nitrogen,
(b) 7.5% hydrogen
(c) 19.76%
3.18 (a) 0.05 atm
(b) 5 m3
(c) 0.707 g/litre
(d) 17.3
3.19 34.4 s
3.20 (a) Chlorine: 54.98%, bromine: 10.16% nitrogen: 34.85%
(b) 65.05
(c) Density: 2.59 g/litre
3.21 (a) Nitrogen: 67.5%, oxygen: 15.74% water: 1.79%, ammonia:
14.97%
(b) 2.48 kg/m3
3.22 10.45 litre
3.23 0.982 kg
CHAPTER 4
4.1
31,941 J/g mole
4.2
77.25 °C
4.3
x
1.0 0.897 0.773 0.660 0.555 0.459 0.369 0.288
0.212
0.140
0.076 0.013 0
y
1.0 0.958 0.897 0.831 0.758 0678 0.590 0.496
0.393
0.281
0.162 0.030 0
x, y: Mole fraction of benzene in liquid and vapour phase respectively.
4.4
xA
1.0
0.724
0.415
0.134
0.0
yA
1.0
0.819
0.548
0.205
0.0
x, y: Mole fraction of ‘A’ in liquid and vapour phase respectively.
4.5
0.1847 kg/kg of steam, 0.1385 kg/kg of steam
4.6
Total pressure: 2,044.79 mm Hg
Mole fraction of methanol in vapour phase: 0.5052
Mole fraction of ethanol in vapour phase: 0.4948
240
4.7
ANSWERS TO EXERCISES
(a) Total pressure: 4087.9 mm Hg
C2H6 : 0.139, n-C3H8 : 0.639, i-C4H10 : 0.039, n-C4H10 : 0.179
and C5H12 : 0.004
(b) 3589.87 mm Hg
4.8
4.9
Total pressure: 906.9 mm Hg
Component
Liquid phase
composition, x
Vapour phase
composition, y
Benzene
0.500
0.739
Toluene
0.377
0.233
Xylene
0.123
0.028
Partial pressure: 102.6 mm Hg
4.10 (a) Ethyl acetate: 12.22 %, air: 87.78% (By volume)
(b) Ethyl acetate: 29.8 %, air: 70.2% (By weight)
CHAPTER 5
5.1
(a) RH% = 24.94%
(b) 0.01358 mole of toluene/mole of vapour free gas
(c) 0.0433 kg toluene/kg of air
(d) % saturation: 23.93%
(e) Mole % = volume % = 1.34%
5.2
Humidity: 0.019 kg/kg
% saturation: 21%
Humid volume: 0.9394 m3/kg dry air
5.3
2961.6 m3/h
5.4
0.567 mole/mole
1.534 kg/kg
5.5
0.25 mole/mole
0.676 kg/kg
5.6
Humidity: 0.025 kg/kg
Dew point: 28.5 °C
Humid volume: 0.907 m3/kg dry air
Adiabatic saturation temperature: 30 °C
Humid heat: 1.0521 kJ/kg dry air
Enthalpy: 0.951 kJ/kg dry air
ANSWERS TO EXERCISES
5.7
Cool to 18.5 °C and then reheat it to 30 °C
Air needed initially 5114.03 m3/h
5.8
27.5 °C and 0.019 kg/kg
5.9
(a) 96.5%, (b) 115.08 m3/h
5.10 (a) 0.0351 kmole/kmole
(b) 0.00857 kmole/kmole
(c) 1.830 kg of water
(d) 49.858 m3
5.11 46.5 °C; 0.0475 kg/kg dry air
5.12 (a) % RH: 29.6%
(b) Humid volume: 0.974 m3/kg dry air
5.13 (a) 3703.7 kg dry air/h
(b) 0.8586 and 0.9739 m3/kg dry air respectively
(c) 18 and 28.5 °C respectively
(d) 35°C
5.14 (a) 0.008 kg/kg
(b) 12 °C
(c) 2.81 kg of water
(d) 7,405.2 kJ
(e) 32.5 °C
5.15 % Relative saturation: 57.5%
Percentage saturation: 44.59%
5.16 Humidity: 0.027 kg/kg
% saturation: 70%
5.17 Humid volume: 0.9205 m3/kg dry air
Enthalpy: 106.6 kJ/kg dry air
5.18 (a) 809.4 m3/min
(b) 74.04 kg
5.19 (a) 0.0259 kmole/kmole, 0.01616 kg/kg
(b) 0.0677 kmole/kmole 0.00423 kg/kg
(c) 13.48 kg of water
5.20 (a) 29.45
(b) 781.24 mm
(c) 810.69 Hg
(d) 0.0377 mole/mole
241
242
ANSWERS TO EXERCISES
5.21 5.7%
5.22 (a) 86%, (b) 1396.5 m3/h
CHAPTER 6
6.1
Mother liquor: 3,324.64 kg
6.2
Feed: 83,078.34 kg
6.3
Crystals: 6,636 kg
6.4
Crystals: 479.2 kg
6.5
Crystals: 342 kg and Mother liquor: 918 kg
6.6
Water evaporated: 7,916.67 kg and crystals: 2,083.33 kg
6.7
Water evaporated: 720 kg Mother liquor: 6589 kg and
crystals: 1,691 kg
6.8
87.4%
6.9
Water evaporated: 561 kg
6.10 Water evaporated: 82.35 kg; Mother liquor: 1,122.85 kg and
Feed: 2,705.2 kg
6.11 2393.6 kg/h
CHAPTER 7
7.1
H2SO4: 5,310.86 kg, HNO3: 2,811.87 kg and Feed: 1,877.27 kg
7.2
(a) 56.96%;
7.3
50.62%
7.4
CO2: 11.41%, H2O: 14.4% O2: 3.53% and N2:70.66%
7.5
Excess air: 39.06%, 471.06 kg of lime/100 kg coke burnt
7.6
O2: 12.14%, N2: 79.34%, SO2: 7.67% and SO3: 0.85%
7.8
54.22% Excess air
(b) 8.338;
(c) 21.81 kg
7.10 CO2: 4.9%, CO: 9.1% H2O: 17.51% and N2: 68.49%
7.11 CO2: 11.183%, O2: 3.956%, SO2: 0.158%, N2: 75.693%, and
H2O: 9.010%
7.14 Butane: 2.46%, Pentane: 42.25% and Hexane: 55.29%
7.15 (a) 6,600 kg, (b) 4,920 kg and (c) 2,165 kg
7.16 (a) Nitrogen; (b) 16.2% and (c) 60%
7.17 (a) 18,973.2 kg/hr and (b) 1,434.4 kg/h in each evaporator
7.19 (a) 1.25 m3 (b) 6.41 m3
ANSWERS TO EXERCISES
243
(c)
Component
Condition (a)
Condition (b)
CO2
H2O
N2
O2
0.23
0.14
0.63
—
0.14
0.08
0.74
0.04
Total
1.00
1.00
7.20 (a) 4.49%
(b) 3.08
7.21 (a) 28.8
(b) 7.55 m3/kg carbon burnt
(c) 50.285 m3/100 m3
(d) CO2: 19.66%, O2: 1.28%, N2: 79.06%
7.22 (a) 446.7 m3
(b) 0.642
7.23 (a) 18
(b) CO2: 17.5%, O2: 6.3%, H2O: 3.5%, and N2: 72.7%
7.24 (a) 2.965
(b) 19.23%
(c) 5081 kg
7.25 Liquid : 81.789 kg/hr, vapour : 18.211 kg/hr, C5H12 : 31.38, C6H14 : 60.06
12.24 kg ,
air required theoretically for complete
7.26 Air supplied =
kg of fuel
combustion = 15.3 kg, volumetric analysis = (mol. basis)
N2 = 82.8%
CO = 10.9%
CO2 = 6.5%
7.27 (a) Moles of CO: 5.444 kmoles; moles of CO2: 1.089 kmoles
(b) Moles of air supplied: 18.388 kmoles
(c) Percentage of carbon lost in the ash: 7.407%
7.28 (a) Composition of the flue gas (wt. basis): CO2 = 273.0933 kg,
CO = 9.1467 kg, N2 = 1091.3911 kg, O2 = 118.5901 kg,
H2O = 4.5 kg,
(b) Ash produced = 21.1 kg
(c) Carbon lost per 100 kg of coke burnt = 7.58%
244
ANSWERS TO EXERCISES
7.29 (a) Volume of air being introduced = 21729.54 m3/h
(b) Composition of flue gas on dry basis (mol. basis): CO2: 10.74%,
CO: 0.5373%, N2: 85.487%, O2: 3.2267%
7.30 (a) 3.846%
(b) 70.3%
7.31 Air–fuel ratio =12.59478, flue gas analysis:
Composition
Weight %
Volume %
CO2
17.7382
11.9749
O2
7.2088
6.69
N2
72.5086
76.92
H2O
2.4726
4.08
SO2
0.7167
0.3326
Amount of air supplied = 37.78434 ton
7.32 Air–fuel ratio by mass = 16.7165
Flue gas analysis:
Composition
Weight (kg)
CO2
3.113
O2
0.6408
N2
12.8717
H2O
1.026
SO2
0.064
Total = 17.7155 kg/kg of fuel
7.33 Air requirement = 432.433 kmoles, volume = 15416.56 Nm3/h
Flue gas analysis:
Composition
Weight (kg)
CO2
16.63
O2
1.2
H2O
8.032
N2
74.14
7.34 Air–fuel ratio = 17.51058, % excess air = 16.167%
ANSWERS TO EXERCISES
CHAPTER 8
8.1
(a) 100 moles (b) 0.4286 mole of Ammonia/mole of NO formed
8.2
(a) 1,425.8 kg, (b) Recycle fraction: 0.7179
CHAPTER 9
9.1
1760 °C
9.2
–1,16,295.8 kJ/kmole
9.7
1,874 °C
9.10 51.44 kcal
CHAPTER 10
10.1 C = –0.1 [100/(100 + t)]2 + 0.2
10.2 C = [10,000 + 400t + t2]/(100 + t)
10.3 (a) t = 100 – 100 [(C – 1)/1.5]0.25, 36.1 minutes
(b) 60 minutes
10.4 (a) C = 0.01 [40/(t + 40)]5 + 0.03
(b) 2.369 minutes
10.5 28.78 minutes
10.6 0.2675 kg Na2SO4/kg of solution
245
Index
Gas constant, 35, 36
Adiabatic reaction temperature, 198
Antoine equation, 75
API scale, 12
Atomic numbers of elements, 231–233
Atomic percent, 11
Atomic weights of elements, 231–233
Average molecular weight, 38
Avogadro’s hypothesis, 9
Avogadro’s number, 36
Hausbrand chart, 75
Heat, 196
Heat capacity of gases (empirical
constants), 234
Heat of
combustion, 197
formation (also standard), 196
mixing, 197
reaction, 197
from combustion data, 197
from enthalpy data, 197
Humid heat, 88
Humid volume, 88
Humidity, 87
absolute, 87
percentage absolute humidity, 88
relative, 88
Hydrated salt, 111
Baume’ gravity scale, 12
Brix scale, 13
Bypass, 180
Clausius–Clapeyron equation, 74
Composition of
liquid systems, 11
mixtures, 10
solutions, 10
Conservation of mass, 8
Conversion, 9
Conversion factors, 229–230
Crystal, 111
Crystallization, 111
Ideal gas law, 35
Kopp’s rule, 198
Density, 12, 38
Dew point, 88
Dry bulb temperature, 87
Law
Amagat’s, 37
Dalton’s, 37
Hess’s, 197
Leduc’s, 37
Law of conservation of mass, 8
Energy balance, 196
Enthalpy, 88
247
248
INDEX
Magma, 111
Mass balance, 122
Mass relations, 7
Mole fraction, 11
Mole percent, 11
Mother liquor, 111
Normal boiling point (NBP), 74
Normal temperature and pressure
(NTP), 9
Partial pressure, 36
Percentage saturation, 88
Process, 3
Property, 3
extensive, 3
intensive, 3
Psychrometry, 87
Pure component volume, 36
Purge, 180
Rate of
accumulation, 221
generation, 221
input, 221
output, 221
Reactant, 9
excess, 9
limiting, 9
Reaction
endothermic, 197
exothermic, 197
Recycle, 180
Saturated vapour, 89
Saturation, 89
partial, 89
percentage, 89
relative, 89
Specific gravity, 12
Standard
condition, 35
state, 196
Steam distillation, 75
System, 3
isolated, 3
Table of
atomic numbers, 231–233
atomic weights, 231–233
conversion factors, 229–230
molal heat capacities, 234
Theoretical flame temperature, 198
Twaddell scale, 13
Units and notations, 1
derived units, 2
Unsteady state operations, 221
Vapour pressure, 74
Volume percent, 10
Weight percent, 10
Wet bulb temperature, 87
Yield, 9