INTRODUCTORY
CONCEPTS IN
CHEMICAL
ENGINEERING
Featuring a new strategy in teaching
mass and energy balance calculations
in steady-state situation
W. I. JOSE, Ph.D.
University of the Philippines
TABLE OF CONTENTS
CHAPTER 1 Introduction
Section 1 What is Chemical Engineering?
History of chemical engineering
Duties and functions of a chemical engineer
Research and Development
Modeling and Simulation
Design
Plant Operation
Sales and Marketing
Management and Administration
Jobs Chemical Engineers hold
The chemical engineer and the environment
Pollution prevention
Rationale for pollution prevention
ISO 9000 and 14000 criteria
Globalization
Section 2 Mass balance calculations
Calculation techniques
Unit operations and equipment in mass and balance
Section 3 Unit Operations and Equipment
Unit Operation of Heat Transfer
Unit Operations of Mass Transfer
Simultaneous Heat and Mass Transfers
Momentum Transfer
Other Unit Operations
References
Problems
CHAPTER 2 ESSENTIAL BASIC PRINCIPLES
Dimension and system of units
System of units
SI Units (International System of Units)
Rules in using SI
Customary units and their bases
Conversion of Units
Converting mass to mole units
Dimensional Consistency
Dimensionless groups
Accuracy, precision, and significant digits
Curve Fitting and Graphical Presentation of Data
Some Process Variables
Temperature
Density and Specific Gravity
Pressure
Common Units of Pressure
Absolute and Relative Pressures
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Vacuum, head, and Draft
Composition
Flow Rates
Behavior of gases, liquids, and solids
Boyle's Law
Charles' Law
Other Correlations
The Ideal Gas Law
Limitations of the Ideal Gas Law
Proof of the Ideal Gas Law
Calculations Involving the Ideal Gas Law
Using the ideal gas law indirectly
Ideal gas mixtures
Dalton's Law of Partial Pressures
Amagat's Law of Partial Volumes
Useful Relations for Ideal Gas Mixtures
Real Gases
Liquids
Solids
Problems
CHAPTER 3 MASS BALANCE I: NO CHEMICAL
REACTIONS INVOLVED
The mass balance equation
Steady and unsteady state operations
Overall and Component Balances
Tie components
Use of Mass or Mole Units
Mass Balance Setup for batch processes
Mass Balance Setup for Continuous Processes
Basis of Computation
Solution of Mass Balance Problems
Methods of Solution
Arithmetic Method
Algebraic Method
Choice of Method
The Ratio Method
Single Unit Systems with Boundaries Within
Multiunit Systems
Recycle Streams
Bypass
Purge
Problems
CHAPTER 4 MASS BALANCE II: CHEMICAL
REACTIONS INVOLVED
Stoichiometry
The Balanced Chemical Equation
Chemical or Gravimetric Factors
Crystallization
Chemical Reactions in Industrial Process
Excess and limiting Reactants
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Conversion and Yield
Comparison of ideal and industrial reactions
Combustion
Problems
CHAPTER 5 HEAT BALANCE
The energy balance equation
Heat
The heat balance equation
Internal energy and enthalpy
Datum or Reference Temperature
Heat quantities in heat balance without reaction:
sensible heat and latent heat
Heat Capacity of Solids and Liquids
Heat capacity of gases
Heat Capacity of Mixtures of Gases
Mean Heat Capacities
Phase Changes and Latent Heats
Heat balance with chemical reaction; heat of reaction
Standard Heat of Reaction
Hess law
Latent heat in heat of reaction
Standard Heat of formation
Standard heat of combustion
Heating value of Fuels
CHAPTER 6 GRAPHICAL SOLUTION
Graphical Addition
Graphical Subtraction
Heat of Mixing and Heat of Solution
Enthalpy-Concentration Diagram
Problems
CHAPTER 7 EQUILIBRIUM AND MASS AND HEAT
BALANCES
Some Terms and Definitions in Equilibrium
Gibb's Phase Rule
Phase Rule for one component systems
Equilibrium Vapor Pressure
Some terms in the equilibrium of pure substances
solid-Vapor Equilibria
Mixture of Volatile Liquids; Raoult's Law
Henry's Law
Mixture of Immiscible Liquids
Equilibrium Involving a Condensable and a
Permanent Gas; Humidity and Saturation
Humidity and Saturation
Wet and dry bulb temperatures
The Humidity or Psychrometric Chart for the AirWater System
Adiabatic humidification and dehumidification
Liquid-Liquid Equilibrium
Solid-Liquid Equilibrium
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Problems
CHAPTER 8 ENERGY BALANCE
Units of Energy
Definition of Some Terms
Types of Energy
Kinetic Energy
Potential Energy
Heat
Work
Work in Pushing Fluids In and Out of the System
Point or State Function
Thermodynamic Processes
The Energy Balance Equation
Energy Balance for Flow or Open Systems
Closed or Non-flow Systems
Problems
APPENDICES
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—Chapter 1—
Introduction
DRAFT
SECTION 1 WHAT IS CHEMICAL ENGINEERING?
About four decades ago, chemical engineering was defined as
a field concerned with the design, construction, and operation
of industrial plants in which matter undergoes physical or
chemical changes. Of course, the major objective was to obtain
maximum profit. Today this definition only partly applies.
What is missing? Nothing is mentioned about the
environment. Back then, a criterion in choosing a plant
location was that the plant should be near a body of water
such as a river. The wastes were thrown directly to receiving
waters without treatment. The environment degraded as
expected. By the seventies, all countries around the world
were concerned with protecting the environment. This of
course came at some cost. Gradually, pollution prevention,
waste management, and resource recovery became part of the
manufacturing cost of a product.
Today, chemical engineering impacts society in so many
ways that we cannot imagine the world devoid of commercial
output of antibiotics, fertilizers, agricultural chemicals, special
polymers for biomedical devices, high-strength polymer
composites, and synthetic fibers and fabrics, among others. In
manufacturing these commodities, the major objective is to
produce them economically. An equally important concern is
to protect the environment from any adverse impact. We
achieve a balance in economics and environmental protection
by using especially designed chemicals, materials, and equipment. Chemical engineering deals about developing this ability
and implementing it on a practical scale. Chemical engineers
synthesize, design, test, scale up, operate, control, and
optimize processes in which materials undergo physical or
chemical changes.
HISTORY OF CHEMICAL ENGINEERING
Although chemical engineering is a relatively new field, some
principles and practices have been used for almost 5,000
years. The early Egyptians filtered grape juice through cloth
bags. During the 16th century, saltmakers recovered the
product from sea water using the heat energy supplied by the
sun and the combined operations of fluid flow, evaporation,
crystallization, and sedimentation.
Initially, manufacturers thought that the technology of each
particular chemical industry was a special kind of knowledge
Introduction 1
DRAFT
distinct from the others. Chemical engineering evolved from
the continuous addition and revision of techniques resulting
from dynamic changes in the chemical process industries.
Chemists and engineers (mechanical and civil) undertook
the fields of industrial chemistry and chemical manufacture in
the 18th and 19th century. The term "chemical engineering"
originated in the United Kingdom. George E. Davis tried to
form a society for chemical engineers in England but met
stern opposition. In 1888, he gave a series of 12 lectures on
chemical engineering, later published as a handbook. Near the
turn of the century, four universities started to offer chemical
engineering in the United States. Industrial chemistry and
basic engineering subjects were the bases of the first courses.
The American Institute of Chemical Engineers was founded in
1908.
We could trace the development of chemical engineering
through the metamorphosis of the curriculum. Since chemists
and mechanical engineers were the chief personnel of industrial plants, early courses in chemical engineering were heavily
based on chemistry (industrial chemistry and chemical
manufacture) and engineering (machine design, surveying,
hydraulics, mechanics).
In 1915, A. D. Little introduced the concept of unit operations, which was considered a major breakthrough. The basis
of the concept is that many steps or physical operations in
totally different industries have some similarities. Similar
steps or operations could be classified into a distinct "unit
operation". For example, the evaporation of water from a sugar
solution was no different from the evaporation from a salt
solution or from evaporation that involved solvents other than
water. Other unit operations that evolved were drying, heat
exchange, absorption, humidification, and crystallization,
among others. Soon after, a limit was reached on what could
be done with unit operations. New concepts were needed.
Further scrutiny of the unit operations revealed that they
could still be classified into major groups that use similar
principles: fluid flow or momentum transfer, mass transfer,
heat transfer, and simultaneous heat and mass transfers. The
areas of momentum, heat, and mass transfers were always
considered three different subjects. In 1960, the principles of
momentum, mass, and heat transfer were combined to form
the transport processes. This development is of major significance in chemical engineering. Through the years, the other
specialized branches of chemical engineering evolved.
Essential to the unit operations are the concepts of material
and energy balances. We have to account for materials being
processed while passing through all the unit equipment.
As important as the unit operations were the chemical
reactions called “unit processes” or the unit chemical conversions. In this book we will refer to a unit process as unit
process technology. Typical unit process technologies are
oxidation, neutralization, polymerization, and calcination,
among others. The equipment for these processes is the
chemical reactor. Production of more desirable products and
less waste or by-products (increased yields) is the major objective in carrying out a process. Chemical engineers applied the
principles of chemical kinetics, which deal with the rate of
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chemical reactions. Equilibrium in a chemically reacting
system is inevitable, hence a need for a chemical-engineeringoriented thermodynamics. This is particularly useful in
considering the energy balances around a system.
As the processes and systems became more diversified and
complex, the need for proper control of process variables and
conditions arose. Thus, process dynamics and control
emerged, which was concerned with automatic controls and
instrumentation of chemical plants.
The integration of the above principles is the basis of
process design. The introduction of computer technology and
high speed electronic computers opened up new approaches
and solutions impossible through manual computation.
Gradually, exact solutions superseded empiricism. Table 1-1
gives the specialized branches of chemical engineering and
their years of prominent appearances. Presently, more and
more fields of study find applications for chemical engineering.
DRAFT
Table 1-1
Specialized Branches of Chemical
Engineering
Industrial Chemistry
Unit Operations
Material and Energy Balances
Thermodynamics and Unit
Processes
Transport Phenomena and
Computer Technology
Interdisciplinary Technology
Year of Prominent Appearance
1898
1923
1926
1947
1960
1965
Since the last decade, new products and materials have come
to the market mainly from biotechnology, electronics, and high
performance materials industry. They are dependent on structure and design at the molecular level. The manufacturing
processes require precise control of chemical composition and
molecular level.
The traditional essence of chemical engineering was carried
at a mesoscale level -- that is, at medium level size and
complexity. This is typically the unit operation equipment and
chemical reactors -- and the combination of the unit operations and unit process technologies. The current researches
extend to phenomena at the molecular level -- the microscale,
and with exceedingly complex systems -- the macroscale.
In the future, chemical engineers will be able to manufacture specialty products on gram or kilogram scales rather than
in megagram. This will require strict process control and
chemical purity on a level that is not yet practical at present.
This will result in improved process productivity, quality,
efficiency, and environmental friendliness which are keys to
global competitiveness. With the traditional mesoscale and
macroscale processing, the field of molecular science is
required.
DUTIES AND FUNCTIONS OF A CHEMICAL ENGINEER
E. L. Brown has pointed out the general function of an
engineer: "Engineering is primarily the art of applying the
Introduction 3
resources of materials and power in nature to the use and
convenience of man. Engineers translate the discoveries of
scientists into structures, machines, and processes."** The
chemical engineer performs this general function too. The
traditional duties of the chemical engineer are: research and
development, modeling/pilot plant operation, design, plant
operation, sales and marketing, and management and
administration.
Research and Development
DRAFT
A chemical engineer is responsible for transferring a new
chemical reaction into large scale production. He works in
close coordination with a chemist when a reaction is promising. While the reaction is still only on a small scale, he applies
chemical engineering principles to make a preliminary evaluation of the project. He tries to gear the work towards an
economic venture. The chemist may think only of the technical
difficulties but the chemical engineer also has the profitability
of the project in mind.
Research is not confined to chemical reactions alone,
however. The chemical engineer can always do research in
improving the unit operations, reaction kinetics, and process
control. The transport processes, for example, are the subjects
of many researches.
The chemical engineer always aims for the most efficient
operation. A program for process improvement and development is not only applied at the start of a project, but is an
internal and ongoing part of any project.
Modeling and Simulation
The success of a new chemical process on a commercial
scale is not always assured. Thus, instead of designing the
actual plant at once, chemical engineers first model and
simulate processes usually using computers. This is a development stage that provides an inexpensive basis for the final
design and operation of the plant and equipment. Before the
advent of computers, chemical engineers first try the process
on a pilot plant scale -- a scale-down model of the commercial
plant.
Design
The chemical engineer works with other engineers in all
phases of the design of a chemical plant. Most important is
the design of chemical equipment. He applies his training in
designing equipment for unit operations and unit processes.
In plant design, economics is the prime consideration and
the chemical engineer has to choose the most appropriate
from several possible processes. He also uses foresight in
determining the market potential of the products. The plant
design also includes the selection of site in terms of market
and the raw material source.
The chemical engineer works in close coordination with
other engineers in the design of process control and instrumentation, piping and electrical systems building layout, and
others. Before a final design is reached, the team of engineers
makes several preliminary designs to avoid errors.
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The construction of the plant may not be a major function of
a chemical engineer but his presence is essential in locating
flow lines, instrumentation, and auxiliaries.
Plant Operation
Once a chemical plant is constructed, the chemical engineer is
responsible for its startup, efficient operation, and proper
maintenance. The qualifications needed are similar to those
for design but are more complex since the chemical engineer
must contend with the different problems encountered such
as equipment malfunction and product quality maintenance.
Aside from the initial plant startup, the engineer encounters
startups from time to time since a plant has to be shut down
for repairs and maintenance. In everyday operation, the
chemical engineer always tries to attain conditions for full
production and operation in chemical reaction, product
separation, and equipment performance.
DRAFT
Sales and Marketing
A company may manufacture a product in the most efficient
way but they might not be able to sell it. They have no point in
manufacturing such a product. Sales and marketing are so
important that without the proper techniques, a product may
not get a good share of the market. A chemical engineer can be
a good salesman because of his background. He helps the
customer best by giving information about the product. Moreover, his duties do not end in selling. He tries to help the
customer in problems encountered in using the product as
well as related products. He does this to maintain the market.
A chemical engineer in general gets involved in the strategy for
the marketing of a product.
Management and Administration
From any of the previous functions, a chemical engineer gets
enough experience to qualify him for an administrative or
executive position resulting from his technical background
and a knowledge of total company economics. Starting from
purely technical activities, he integrates all considerations for
efficient company operation. Since the technological changes
require broad training, many plants are developing plant
superintendents and managers from among their chemical
engineers. An important asset of an engineer is his creativity
which results from basic training in analysis and synthesis
from both the school and actual job. The manager runs the
plant efficiently in terms of two things - man and machine - to
produce marketable products at a good profit.
JOBS CHEMICAL ENGINEERS HOLD
Chemical engineers do not always hold jobs described above.
Surveys indicate that many of them have jobs that are not
engineering in nature but which make use of their chemical
engineering background. A chemical engineer may question
his status but he should not worry. He can carry out any job
because his chemical engineering training prepared him. He is
versatile and flexible. There should be no remorse but instead
Introduction 5
DRAFT
pride in being able to carry out a job of a different nature. In
financial management, a chemical engineer uses his knowledge to reinforce his performance in this field. He is better
than his non-chemical engineering counterpart because he
understands the technical aspects and seamlessly includes
them in his analysis. Some chemical engineers are in the
government and directly help in the development of their
country. Many chemical engineers become successful entrepreneurs. While research is a duty of the chemical engineer,
some chemical engineers have used invention, creativity, and
innovation as a passport to success. A chemical engineer can
easily adapt to the intricacies of technology transfer. By studying technology intelligence, they can easily take advantage of
the fact that only about 5% of a technology is secret.
A chemical engineer can take a second course (such as the
other engineering professions, medicine, dentistry, microbiology, agriculture, or practically any course) and be good at it
provided his interest is keen. If possible, he should take
advanced studies in his second degree. The emerging fields
such as biotechnology, material science, nanotechnology, DNA
computation, and surface science, among others are fields
that can easily capture his interest. The arts, humanities, and
social sciences are fields some chemical engineers enter.
A listing and description of successful chemical engineers in
different jobs would be an interesting compilation.
THE CHEMICAL ENGINEER AND THE ENVIRONMENT
Human activities cause environmental degradation in form of
global warming, acid rain, thinning of the ozone layer, defoliation, desert formation, soil and water pollution, and marine
pollution. These environmental problems are interrelated and
cannot be addressed singularly. Multidisciplinary approach is
necessary. Everyone must not only recognize his role in
solving problems directly concerning his field but he must also
realize that some environmental problems require expertise
from other disciplines. Chemical engineers are versatile. As
expected they can help in the solution of the problems.
The relationship between engineering and the environment
was accelerated by the United States National Environment
Protection Act of 1969 and the requirement for environmental
impact assessment. Initially, the industries adapted the
“end-of-pipe” treatment of wastes. This meant added expense
to the manufacturing cost that ultimately affected the profit
margin. Methods to avert treatment costs led to concepts that
include pollution prevention and waste minimization, “cleaner
production”, life cycle analysis, and redesign of plants for
higher process efficiency, among others. Ultimately, we expect
the development of new highly efficient processes with
minimal waste emission.
While the problems concerning the environment require
concerted effort by people from different disciplines, the
chemical engineer has advantage over other professionals
because he is flexible and versatile. As the field concerning the
environment is broad and some aspects require specialization
such as in the design of waste treatment plants, a background
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in pollution prevention and waste minimization is most useful
to the chemical engineer.
DRAFT
POLLUTION PREVENTION
Pollution prevention is a combination of reuse, recovery, and
minimization of waste sources. To start a pollution prevention
program is to realize that reducing waste does not only
concern the environment, but also gives economic advantage
in production. The goal is to minimize unwanted impact to the
environment and at the same time, to lower the production
cost. We can see the compatibility of pollution prevention with
the definition of chemical engineering given earlier.
Instituting a pollution prevention plan is one of the surest
ways to improve productivity. This instills a new culture and
pride in attaining set goals. More profits, less environmental
problems, and increased employee morale are the major incentives that can be achieved by companies.
While pollution prevention is good business, there are other
support rationales for its implementation. According to
Nemerow (1995), these are mainly concerned with economics
and include the following: reduced waste disposal costs
(because less waste is produced), reduced raw materials costs
(because manufacturing processes more efficiently use and
reuse these materials), reduced damage costs (because
lawsuits and insurance charges are eliminated), and improved
public relations (because both the local and national public
image of the industry is elevated as a result of environmental
improvement). In addition, the paperwork associated with
national, regional, and local permitting regulations is greatly
reduced, providing savings of both time and money.
THE ISO 9000 AND 14000 CRITERIA
Pollution prevention ultimately leads to quality, productivity,
and efficiency. The criteria of the ISO 9000 and ISO 14000
complement pollution prevention.
The ISO 9000 series, established by the International
Organization for Standardization, is a five-part series of
standards. Companies that comply with the set of procedures
are certified and become basis of product acceptance worldwide. It was originally written as a generic guideline for
negotiations between buyers and suppliers. At present, registrars can certify that the documented quality systems and
practices of a firm comply strictly with the selected standards
of the ISO 9000 series. The standards deal with how firms
design, produce, install, inspect, package, and market
products as well as services. A pollution prevention plan would
be a good preparation for the ISO 9000. Being certified under
ISO 9000 series “is fast becoming a de facto prerequisite for
competing in business worldwide” (Hockman et al. 1993).
More directly related to pollution prevention is the ISO
14000 series that deals with environmental management
system (EMS). The EMS provides order and consistency for
organizations to address environmental concerns through
allocation of resources, assignment of responsibilities, and
Introduction 7
ongoing evaluation of practices, procedures and processes
(ISO 1993). Again a company certified with ISO 14000
standard will be accepted worldwide.
The two ISO series put pollution prevention as an essential
practice. We can see that this has some implications for
chemical engineering products and services. Studying the ISO
series is outside the scope of this book but this is a good basis
for the current definition of chemical engineering -- increased
profitability through high productivity, quality, and environmental friendliness.
DRAFT
GLOBALIZATION
Today, we talk of the world as a “global village”. Distance
and language barriers are rapidly diminishing. We can
communicate instantaneously and travel faster. This emerging
concept of the entire globe as a single entity is what we refer to
as “globalization”. While this idea applies to almost any aspect,
the term globalization commonly relates to world economics -defined as the process of creating an integrated worldwide
financial market. During the last three decades, companies
have felt the pressure to build their plants and sell their
products in many countries aside from their home base. They
have to think in global terms in order to be successful.
The chemical engineers in common with other engineers
should now have a global perspective. This they can obtain
from their education and work experience.
Going back to the definition of chemical engineering, we see
that the profession is becoming global in its perspective.
SECTION 2
Mass Balance Calculations
Chemical engineering work always starts with a mass balance.
Thus, a chemical engineering student must first study the
methods and techniques in mass balance calculations.
For a student to be confident in carrying out such calculations, he must have a good background in mathematics,
chemistry, and physics, as well as a practical knowledge of the
properties of various materials. The second chapter presents a
review of some necessary fundamentals.
Let us recall our algebra courses in high school and in
college. Think also about those courses that used arithmetical
solutions. Some of the verbal problems we solved then deal
with mixing (for example, mix 10% acid solution with 30% acid
solution), compounding (given a formula, find the quantities of
components necessary), simple processes (manufacture
low-fat milk from regular milk), etc. We consider them
examples from real life. We were able to solve those verbal
problems (with the help of our teachers). It may surprise you
to know that those problems are mass balance problems. Even
without any chemical engineering background, you have
solved mass balance problems before. You may have used an
algebraic or arithmetical solution, or a combination of both. In
this book, we will use the same technique -- we consider this
subject as an extension of algebra and arithmetic courses. We
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just need to study the appropriate relationships necessary in
the calculations. This book is so structured that only a few
relationships are given in each chapter. Students are not
overloaded.
The basis of mass balance calculations is the law of conservation of mass: Mass cannot be created nor destroyed in an
ordinary chemical reaction. For a chemical engineer, the mass
balance is the accounting for materials entering and leaving
the process, whether or not a chemical reaction occurs. The
following is the mass balance equation:
Eq. 1-1
mass inputs= mass outputs + accumulation
Obviously, anything that enters a process must come out at
some point. If a liquid mass enters an empty container, some
mass accumulates initially and the outgoing mass is zero.
Eventually the liquid overflows out of the container and the
accumulation becomes zero. If the accumulation is zero,
Eq. 1-2
DRAFT
mass inputs= mass outputs
If the mass output is larger than the mass input, then depletion (or negative accumulation) occurs. We normally set up the
mass balance around a process (a single unit of equipment or
a system consisting of several components). Figure 1.1 shows
this schematically.
M1
M'1
M2
M'2
Process
M3
..
.
.
Mn
M'3
..
.
M'm
Figure 1-1
The mass balance equation corresponding to Figure 1-1 is
M 1 + M 2 + M 3 + ... + M n = M ∏1 + M ∏2 + M ∏3 + ... + M ∏m
Eq. 1-3
where
M1, M2, M3 … Mn
= masses of input streams 1, 2, 3, … n,
respectively
M’1, M’2, M’3 … M’m = masses of output streams 1, 2, 3, … m,
respectively
Eq. 1-3 is an algebraic equation with several mass quantities,
which may be known or unknown. We know that we need the
same number of equations, as there are many unknowns. The
Introduction 9
more the unknowns there are, the more difficult it is to solve
the simultaneous equations.
In setting up a mass balance we should be aware of boundaries of domain in which we are setting up a mass balance
equation. This could be a single equipment, a section of an
equipment, a whole plant, or a section of a plant. We can draw
an imaginary boundary using a broken line.
Let us consider a simple example. Water is flowing through
a pipe. Fig. 1-2 represents a section of the pipe.
M1
M2
Figure 1-2
The mass balance equation is
M 1 = M 2 + Accumulation
Eq. 1-4
DRAFT
Since water is flowing for some time, the accumulation is
zero (that is, the pipe section is full of water).
M1 = M2
Eq. 1-5
We can see that the amount of water that comes in (M1)
must be the same as the amount that goes out (M2). We need
extra information to determine the quantities. Normally we
need an instrument to register the mass of a fluid such as
water that flows through a pipe. An example is the water
meter used by water utility companies. If we provide a water
meter at the point where water enters, then we can determine
M1. If the water meter indicates 10,000 liters of fluid then M1 =
10,000. Substituting in Eq. 1-5, M2 = 10,000.
Now, let us say that a leak is present at a point in the pipe,
and some water is lost. Figure 1-3 shows this, with ML as the
water that leaks out.
M1
M2
ML
Figure 1-3
In this case,
M1 = M2 + ML
Eq. 1-6
We have a single equation. If again M1 = 10,000 liters,
10, 000 = M 2 + M L
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Eq. 1-7
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DRAFT
We have two unknowns, M2 and ML. From algebra we know
that we need another equation to solve for the two unknown
quantities. Physically, we can collect the leakage (if the leak is
in the form of droplets that can be collected). We can then
solve for M2. If ML is 1 liter, then M2 = 10,000 - 1 = 9,999 liters.
Installing a water meter to determine M2 is a possibility. ML can
then be solved. However the meter may not be sensitive
enough to give the true value of ML.
In chemical engineering, not all unknowns are measurable
physically. Sometimes we have to estimate them indirectly
using previously known methods or techniques. For example
the tracer technique is used to determine the amount of
leakage. We tie the amount to the flow of a known amount of
injected material that can easily be tracked such as a radioactive compound.
The example discussed above involves the loss of 1 kg of
process water to the environment. What is the impact? The
cost of the lost water is negligible. Since the water is clean, it
will not have an adverse effect on the environment. Plant and
animal life forms would welcome it. How about if the temperature is higher, say 60ºC? In this case, some plants and
animals will be affected because of the high temperature and
will definitely have an impact.
Suppose the piping system transports another liquid like for
example, benzene. In a similar manner, we set up a mass
balance equation in the system under consideration.
Now let us consider the impact of benzene if a small amount
is released to the environment. The cost would be several
times that of water. This is still negligible and a company
would not even look at it. How about its impact to the environment? If we look at the specifications of benzene we see the
following:
“melting point 5.4°C boiling point 80°C. Insoluble with water,
miscible with organic solvents...vapor inhalation may cause
depression of the bone marrow, convulsions, and
paralysis......”
Well from the data, would you allow benzene to contaminate
the environment? Of course not. Benzene is a carcinogen. In
one instance, a bottled mineral water company discovered that
its underground source of mineral water was contaminated
with benzene. You can just imagine the impact of this
phenomenon. Thousands of the bottled product had to be
recalled. It meant the loss of a large amount of business. This
may have been the fault of another company that inadvertently released the benzene contaminant.
If a company contaminates the environment, the owners are
liable to many litigations that could cost the company large
sums of money. Moreover, the negative publicity can affect
product sales. Recently the sales of Coca-cola in Europe
decreased when several thousand batches of the canned
softdrink were contaminated with a toxic chemical. Likewise,
dairy products from Belgium suffered rejection when they were
found to be contaminated with dioxin.
Example 1-1
Introduction 11
900 kg of cement is available to prepare a concrete mixture.
The ratio cement:sand:gravel is 1:2:4. The water to cement
ratio to be used is 0.4. How much gravel, sand, and water is
needed? How much concrete mixture can be prepared from
the available quantity of cement?
Given:
900 kg of cement
cement:sand:gravel = 1:2:4
water:cement = 0.4
Required:
kg of concrete prepared
Basis:
900 kg of cement
Solution:
Arithmetic method using ratio and proportion
cement available = 900 kg
DRAFT
cement:sand = 1:2
900 = 1
sand 2
sand = 1800 kg
900 = 1
gravel 4
gravel = 3600 kg
900
water = 0.4
water = 360 kg
concrete prepared = 900 + 1800 + 3600 + 360 = 6660 kg
Example 1-2
Water flows through a pipe that branches to three smaller
pipes. The first pipe allows water to flow two times greater
than the second pipe and 1.5 times that of the third pipe. If 50
m3 flows through the main pipe, how much water flows
through each of the branched pipes?
Given: 50 m3 water flows through the main pipe and splits to a
branch of three pipes. The flow rate in pipe 1 is twice that in
pipe 2. The flow rate in pipe 1 is 1.5 times that of pipe 3.
Required: Amount of water that flows through each branch.
Solution: Algebraic method
Basis:
50 m3 of water flowing through the main pipe
Let
x = be the volume of water flowing through pipe 1
y = the volume water flowing through pipe 2
z = volume of water flowing through pipe 3
Water balance:
water flowing through the main pipe = water flowing
through pipe 1 + water flowing through pipe 2 + water
flowing through pipe 3
50 = x + y + z
12
Eq. 1
National Engineering Center CEEC Seminar, June 2 -4, 2004
y= x
2
Eq. 2
z= x
3
Eq. 3
Substituting Eq. 2 and Eq. 3 in Eq. 1,
50 = x + x + x
2 3
Solving for x and substituting the values for y and z,
x=
50 = 23.08 m3, water flowing through pipe 1
2.167
y = x = 11.54 m3, water flowing through pipe 2
2
z = x = 15.38 m3, water flowing through pipe 3
1.5
DRAFT
We used three unknowns in solving the problem. We could
even simplify the algebraic solution by using only one
unknown such as x and then expressing the unknown quantities in terms of x at once. Hence,
Let
x
= the volume of water flowing through pipe 1
x
2
= the volume water flowing through pipe 2
x
1.5
= the volume of water flowing through pipe 3
In this case we have only one unknown. Implicitly, we used
some arithmetic in getting the unknown quantities. This
solution is therefore a combination of arithmetic and algebraic
method. We will use this technique later on.
CALCULATION TECHNIQUES
It may seem that the technique boils down to setting up and
solving a system of simultaneous algebraic equations, or
analyzing and applying arithmetic. This is not the case in
actual practice. Finding all the equations, the constants,
measurable quantities, etc., may not be possible. The data
may be lacking (such as in designing a process) or more are
available than necessary (as in plant operation). In this book
however, we start from principles taken up in mathematics,
physics and chemistry. We then apply chemical-engineeringoriented calculations
While all the calculations can be performed in a computer,
the student still has to go through the suggested methods to
solidify his background.
Mass balance is indispensable in every phase of chemical
engineering work, such as in the design, construction, and
operation of industrial plants. Most chemical engineers
perform mass balance calculations in their job. They use the
Introduction 13
same techniques that beginning chemical engineering
students study. The only difference is that they know how to
find the data necessary for calculations. They use their knowledge of chemical engineering.
A chemical engineer who designs a process has to find all
the data necessary, which are usually lacking. He has to use
empirical, theoretical, or experimental data. He can make
assumptions based on his experience or previous knowledge.
A chemical engineer who operates a plant gathers data and
finds out he has more data than is necessary. He has to decide
which ones to use and use the rest to check for consistency.
His problem is how to make use of the extraneous data he has
collected in the process. A beginning chemical engineer with
minimal background will be given most of the data as well as
some techniques. Practicing chemical engineers and students
both use the same techniques in solving mass and energy
balance problems.
SECTION 3 Unit Operations and Equipment
DRAFT
The discussion in books on mass and energy balance calculations sometimes puts pressure on the students to cope with
new information. This includes the unit operations and the
respective equipment. Mass balance calculations are usually
carried around unit operations or the corresponding equipment. A beginning student may not totally understand the
function and the operation of an equipment if inadequate
information or explanation is provided by the author who is
familiar with the entire chemical engineering knowledge. We
are aware that the beginning student is building his understanding of the profession and in a way still requires careful
guidance regarding some concepts. This is because the discussion on some principles still lies in years ahead in his
schooling.
This section introduces the student to the various unit
operations -- with the underlying principles in a qualitative
manner accompanied by the corresponding equipment. We
need this in the study of mass and energy balances involved in
an equipment or a combination of different equipment. The
detailed theoretical treatment of these operations is covered in
books on unit operations.
The unit operations provided the original basis for classifying various physical operations in chemical process plants.
Now these operations are grouped under the different transport processes: heat transfer, mass transfer, mass and heat
transfers, and momentum transfer. The rest is classified in a
special group.
UNIT OPERATION OF HEAT TRANSFER
We encounter heat transmission in most chemical engineering
operations. Whenever a temperature difference exists, the
transfer of heat occurs. Heat travels from a hot area or volume
to a cold one until the two temperatures become equal. In a
plant we want to maximize the heat transfer rate and
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National Engineering Center CEEC Seminar, June 2 -4, 2004
DRAFT
minimize heat losses at the lowest possible cost. The equipment for heat transfer is the heat exchanger. The heat energy
of a hot fluid is usually transferred to a cold one without direct
contact. Several types of heat exchange equipment are available. The design and the name depend on the particular application. Some examples of heat transfer equipment are: direct
heaters, steam generators, furnaces, preheaters, shell-andtube heat exchangers, coolers, evaporators, and condensers.
In energy conservation, heat transfer is important in salvaging
heat energy with equipment such as exchangers, recuperators,
and regenerators.
(a)
(b)
Figure 1-4
(b) shell-and-tube heat exchanger (Ref. McCabe and Smith, Unit
Operations, McGraw-Hill 2000 Permission being sought.)
Double-Pipe Heat Exchanger
The simplest type of heat exchange equipment is the
double-pipe heat exchanger. It consists of two concentric
pipes. One fluid flows inside the inner pipe while the other in
the annulus. The flow may either be parallel or
countercurrent. As long as a difference in temperature exists
between the two fluids, the transfer of heat from the hot to the
cold fluid occurs. This heat exchanger is normally applicable
for experimental or small scale operations
Shell-and-Tube Heat Exchanger
This equipment is similar to the double pipe heat exchanger
and is the most common type of industrial heat exchanger.
Introduction 15
Several tubes are placed inside a large tube known as the shell
to increase the heat transfer area. One fluid flows inside the
tubes (the tube side) and the other fluid, outside of the tubes
(the shell side). It can easily be assembled and disassembled
for quick cleaning and repairs. Some heat exchangers have
special names, depending on their functions. Some examples
are reboilers (function: to boil a liquid), vaporizers (one fluid
heats up the other fluid until it vaporizes), and condensers (a
cold fluid absorbs heat from the vapor that subsequently
condenses).
UNIT OPERATIONS OF MASS TRANSFER
DRAFT
Most physical separations are based on differences in density
or sizes of particles. Instead, mass transfer operations depend
on concentration differences between two homogeneous
phases. A component transfers from a phase where its concentration is higher to a phase where its concentration is lower.
For example, water transfers from a wet cloth to a dry cloth
until both cloths have the same water content. Place a raisin
in a glass of water. The water will diffuse into the raisin. The
concentration difference can be in terms of solubility or vapor
pressure differences and involves the diffusion of components.
In the production of chemicals from raw materials, the direct
product from the reaction oftentimes contains impurities in
the form of unreacted reactants, by-products, and other
contaminants. The most suitable processes for separation and
purification are the mass transfer operations. Distillation,
extraction or leaching, absorption and desorption, adsorption,
and ion-exchange are some examples of these operations.
Distillation
When a solution containing a volatile component is heated,
the composition of the vapor is different from the composition
of the liquid that remains. The vapor is richer in the more
volatile component. Condense this vapor, then heat again. The
resulting vapor will be even richer in the volatile component.
Repeat the process until you get a pure substance. This is the
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National Engineering Center CEEC Seminar, June 2 -4, 2004
principle used in distillation. The equipment is the distillation
Column
Reflux
Condenser
Feed
Distillate
or overhead
product
DRAFT
Reboiler
Bottoms
product
Conventional distillation column
liquid-phase flow
downcomer
Plan view of a bubble cap plate
vapor-phase flow
Figure 1-5 Equipment for distillation
column. This is similar to the laboratory fractional distillation
apparatus which uses a packing material usually made of
glass beads. A distillation column consists of several stages
Introduction 17
DRAFT
called plates.
Figure 1-6 Equipment for absorption and desorption. (Ref.:
McCabe, et al., Unit operations of Chemical Engineering,
6th ed. McGraw-Hill, 1978 p. 547. Permission being
sought.)
The plate at the bottom is the reboiler where heat (usually
from steam) is applied to boil the solution. The vapor from this
solution in turn heats and boils the solution in the next higher
plate. The resulting vapor, likewise, boils the liquid in the next
plate, and so on. Since the vapor contains more of the volatile
component than the corresponding liquid, the vapor mixture
becomes richer in the volatile component as it ascends from
plate to plate. In a similar manner, the liquid flowing down the
column, becomes richer of the less volatile component. The
product withdrawn at the top, is the distillate, while from the
bottom plate, the bottoms. Side products can be withdrawn
from any other plate. The column operates continuously with
fresh feed input near the middle of the column. Complex liquid
mixtures composed of several components of different volatilities can be separated into suitable fractions by withdrawing
the products from different plates in a distillation column.
Petroleum or crude oil is processed in this manner. Different
fractions withdrawn from the top downwards are LPG,
gasoline, naphtha, kerosene, diesel fuel, fuel oil, lubricating
oil, and others. Each of these fractions is still composed of
several components which can be further separated in another
distillation column. Alcohol from fermentation is recovered by
18
National Engineering Center CEEC Seminar, June 2 -4, 2004
distillation. Oxygen and nitrogen are separated from liquefied
air also by distillation. Vacuum flash distillation is a one-stage
distillation process where a liquid mixture is subjected to
vacuum at a specified temperature. A vapor mixture with a
composition different from that of the liquid results. The vapor
and the liquid portions are obtained as the products.
DRAFT
Absorption and Desorption
In gas absorption, a constituent of a gas mixture is removed
by contacting with a liquid. Ammonia in an air-ammonia
mixture, when placed in contact with liquid water, will transfer to the liquid. In desorption (also known as stripping), the
opposite happens. Here, the component of a liquid phase
transfers to the gas phase. The benzene in a benzene-oil
mixture is recovered by treating the mixture with steam. The
benzene transfers to the steam. Upon cooling, the benzene
separates easily from the steam. For more efficient transfer,
the equipment should provide a large surface of contact
between the gas and the liquid phases. The equipment is the
packed tower. It consists of a vertical cylinder filled with a
packing material. The packing material wetted by the liquid
provides the necessary contact surface. Some typical packing
materials are stoneware shapes, wood slats, broken rocks, and
coke. Chemical stoneware finds applications when the fluids
treated are corrosive.
To operate the tower, the liquid phase is distributed evenly
at the top of the tower while the gas is fed at the bottom.
Contact between phases occurs as they flow countercurrently.
Figure 1-7 Common tower packings: (a) Raschig rings; (b) metal
Pall ring; (c) plastic Pall ring; (d) Berl saddle; (e) ceramic Intalox
saddle; (f) plastic Super Intalox saddle; (g) metal Intalox saddle
Reference (Ref.: McCabe, et al., Unit operations of Chemical
Engineering, 6th ed. McGraw-Hill, 1978 p. 547. Permission being
sought.)
Introduction 19
Solid-Liquid Extraction
DRAFT
The valuable component (solute) in a solid, (e.g., oil in copra),
can be separated by soaking the solid in a solvent (e.g.,
hexane) where the solute is soluble and allowing the solute to
diffuse out. This is the unit operation of solid-liquid extraction.
Some of the other terms equivalent to solid-liquid extraction
are washing, leaching, lixiviation, elution, elutriation,
diffusion, percolation, decoction, decantation, infusion,
maceration, digestion, steeping, and dissolution. These terms
mean the transfer of the soluble material from the solids to the
solvent. Solid-liquid extraction finds application in the recovery of oil from oil seeds, fish oil from fish livers, copper from
oxidized copper ore, etc., using suitable solvents. Extraction
involves two major steps: (1) contact of the solvent with the
solid to effect a transfer of the solute to the solvent, and (2)
separation of the solute from the solution. The transfer of the
solute from the solids to the solvent occurs by diffusion and
takes some time. A recovery of 85% is considered economical.
An open tank extraction unit is the simplest equipment for
extraction. After the solids are placed in the tanks, the solvent
is pumped in until the solids are soaked. Contact is allowed
until most of the solute is dissolved. The liquid is then drained
from the bottom. The solute is recovered by evaporating off the
solvent which is condensed and recycled.
In practice, several extractors in series are used which
allows more efficient operation. The extracted solids from one
extraction unit (known as a stage) is sent to the next stage
where it is treated again. In extraction, the raffinate or the
underflow refers to the extracted solids, while, the extract
phase or the overflow, to the solution containing the solute.
Liquid-Liquid Extraction
This is a unit operation in which a solute dissolved in a
liquid phase is transferred to another liquid phase. Ideally, the
two liquids should not be miscible. Otherwise, the two fluids
mix and no separation takes place. Two major steps comprise
the process: (1) the intimate contacting of the two phases, and
(2) the separation of the two phases.
An example is in the recovery of the acetic acid from acetic
acid and water solution. The separation of these two
substances using distillation is quite difficult. Instead, a
solvent such as isopropyl ether can extract the acetic acid.
When isopropyl ether comes in contact with the solution, most
of the acetic acid transfers to the ether layer. We can easily
separate the acetic acid in the isopropyl ether by distillation.
Liquid-liquid extraction is widely used in lubricating oil processing, vegetable oil technology, plastics manufacture, and
recovery of penicillin, among others.
The equipment for liquid-liquid extraction provides
maximum contact between two phases. Some of these are the
spray column, the packed column, the plate column, the sieve
plate column, and the bubble cap column. The spray column
consists of a cylindrical vessel and spray nozzles at the top
and at the bottom. The nozzles at the top spray the dense
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National Engineering Center CEEC Seminar, June 2 -4, 2004
DRAFT
Figure 1-8 Some equipment in solid-liquid extraction
fluid while at the bottom, the light fluid. Intimate contact
between small droplets of the two fluids results. The packed
column on the other hand, contains packing materials that
allows contact between phases. The plate column consists of a
series of perforated plates. The dense fluid is drawn out at the
bottom by gravity while the lighter fluid, at the top.
Introduction 21
DRAFT
Figure 1-9
al, Unit Operations, John Wiley, 1950. pp 298, 300, 301. Permission
being sought)
Adsorption
Adsorption is a unit operation in which a component of a fluid
mixture adheres to the surface of the solid (called the adsorbent), resulting in a change of composition in the unadsorbed
fluid. The adsorbed substance could be an impurity (such as
odor or color) or a substance to be recovered (such as
acetone). Activated carbon is an example of an adsorbent used
to remove undesired color, odor, or other impurities from
fluids. Other examples are Fuller's earth (used in refining oils,
fats, and waxes), acid-treated clays (used in refining petroleum
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National Engineering Center CEEC Seminar, June 2 -4, 2004
DRAFT
fractions), bone char or bone black (used in sugar refining),
alumina (used in drying air, gases, and liquids), and silica gel
(used in drying and purification of gases).
Adsorbents have large surface-to-volume ratio and preferential affinity for specific substances. The equipment for adsorption consists of a bed of absorbents arranged to provide
intimate contact with the fluid for treatment. The spent
adsorbent is regenerated and used for many cycles.
Figure 1-10 Equipment for humidification and dehumidification. (Ref.:
Treybal, Mass Transfer Operations, McGraw-Hill, 1980. pp.
261-26211 Permission being sought.)
SIMULTANEOUS HEAT AND MASS TRANSFERS
Introduction 23
This is a group of unit operations where mass transfer is
equally important as heat transfer. When a gas phase component transfers to a liquid phase, latent heat of condensation
evolves. The reverse, transfer from the liquid phase to the gas
phase, requires heat. Some mass transfer operations such as
absorption and extraction involves heat effects but this is less
important in design. In the unit operations of evaporation,
crystallization, drying, humidification, and dehumidification,
heat and mass transfers are equally important.
Humidification
DRAFT
Humidity is the amount of moisture in the air. Humidification
is the addition of water to air to increase its humidity. As the
water evaporates, the remaining liquid provides the latent heat
of vaporization. This results in the cooling of the remaining
liquid water. Humidification therefore serves two purposes:
increasing the humidity of the air and cooling of the liquid
water. An example of a humidification equipment is the spray
chamber. Spray nozzles distribute the liquid droplets to the air
stream. This is usually applicable for small-scale operations.
An equipment used to cool water is the spray pond. Nozzles
spray the water into the air and the evaporation of some of the
water cools the remaining water. For large-scale cooling,
cooling towers are the proper equipment. The cooling tower is
usually of wood construction. It consists of several wood-slat
decks. A distributor dispenses liquid evenly at the top of the
tower. The liquid flows down countercurrently to the air fed at
the bottom. The cooled water flows out at the bottom. A
simpler way to increase the humidity of air or any gas is by
plain injection of steam.
The term humidification originally referred to air-water
systems. However, it is likewise applicable for processes
involving a permanent gas (air, O2, CO2, N2, CH4, H2, etc.) and
a condensable vapor (H2O, ether, acetone, alcohol, etc.).
Dehumidification
Dehumidification (the opposite of humidification) is the
process of condensing out the water vapor from the air to
decrease its humidity. The cooling tower, an equipment for
humidification can also be used for dehumidification. By using
a cold liquid (a temperature much lower than that of the
entering gas--the wet bulb temperature) some of the water
vapor in the air will condense out.
Dehumidification is important in air conditioning where the
chiller is the corresponding equipment. The chiller cools a
vapor-gas mixture and condenses out some of the water vapor.
By using the same principle, dehumidification finds application in the recovery of solvents such as benzene, methanol,
acetone, and carbon disulfide, among others.
Evaporation
Evaporation is a unit operation where part of the solvent in a
solution is vaporized to concentrate the solution. Either the
concentrated solution or the evaporated solvent is the desired
product. The most common heating medium is steam delivered through heating coils, tubes, or jackets. In some
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National Engineering Center CEEC Seminar, June 2 -4, 2004
DRAFT
Figure 1-11 Typical evaporators. (Ref.: Brown et al, Unit
Operations, John Wiley, 1950. p. 475. Permission being
sought.)
Introduction 25
instances, hot gas may be injected directly. For smaller units,
the electric heaters are common.
The standard equipment uses vertical or horizontal tubes
with steam as the heating medium. The evaporated solvent
from a unit can be used as the heating medium in the next
unit. The process is known as "multi-effect" evaporation,
where a unit is an "effect".
Some examples of mixtures separated by evaporation are
sugar-water solution, NaCl-water solution, glycerine-water
solution, etc. For heat sensitive substances, the operation
under vacuum is resorted to. In the production of salt, water
is evaporated from sea water by solar energy.
Crystallization
DRAFT
Crystallization is a unit operation in which a solution is
brought to a concentration where it cannot hold all the soluble
component (the "solute") in solution. We can attain this condition by cooling a hot solution and/or by evaporating some of
the solvent. Crystallization can take place either by addition of
a solvent, or by evaporation, or by simple cooling of the
solution.
The solute precipitates out as crystals. These crystals separate
out as pure substance. The impurities left in solution may
adhere to the crystals. Washing of the crystals is necessary to
obtain a purer product.
The simplest type of equipment is a tank where natural
cooling and evaporation occurs. Usually, the crystallizer is
integral with an evaporator. The Swenson-Walker crystallizer
is provided with a cooling system. The vacuum crystallizer
removes part of the solvent by evaporation under vacuum,
with the crystals precipitating out of solution.
Drying
Drying involves the removal of water or another liquid from a
solid, slurry, or liquid by vaporization. Hence, the definition of
drying is not clear-cut from that of evaporation. Drying is the
operation that usually follows evaporation, crystallization, or
filtration for the final removal of water.
In drying, a gas medium picks up the vaporized liquid from
the solid. By raising the temperature and/or lowering the
humidity of the gas medium, we can enhance its drying capacity. In such cases a heater and a dehumidifier are integral
with a drier.
The design of drying equipment depends on the type of
materials handled. Different types of materials such as sheets,
granules, pastes, sludge, slurries, caking crystals, or solutions
require specific drier designs. The operation of experimental or
small-scale driers are usually carried out in a batchwise
manner. The tray drier is an example of a batch drier. For
continuous operation, tunnel and rotary driers find application. For materials in solution, drum driers or spray driers are
preferable. For heat sensitive materials, vacuum operation is
required.
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National Engineering Center CEEC Seminar, June 2 -4, 2004
DRAFT
MOMENTUM TRANSFER
Figure 1-12 Some crystallizers. (Ref.: 13Brown et al, Unit Operations,
John Wiley, 1950. pp 499-500. Permission being sought.)
In a process plant, materials are constantly flowing. This
involves transfer of momentum. Processes that involve
momentum transfer require study of material and energy
relationships. Flow of fluids requires application of fluid
mechanics. Fluid handling, filtration, and fluidization are
some unit operations that involve momentum transfer.
Fluid Flow and Handling
The transportation and movement of fluids in a process is the
unit operation of fluid flow. The equipment consist of the fluid
movers (pumps, compressors, blowers) and conduits
Introduction 27
DRAFT
(
Figure 1-13 Typical drying equipment. (Ref. 15Brown et al, Unit
Operations, John Wiley, 1950. p. 560. Permission being sought.)
normally pipes and fittings). Pumps, compressors, or blowers
provide energy to move fluids through the process equipment.
The types of pumps are many and depend on the nature of
materials handled. Materials flow through pipes and fittings.
Mild steel is the common material of construction but stainless steel, nickel, plastics, rubber, ceramics, brass, copper,
concrete, and glass, among others, find application when they
can resist corrosion from fluids being handled. The fittings
(which join the pipes) involve either threaded, bell-and-spigot,
flanged, or welded connections. Valves control the flow while
different flow measuring devices indicate the flowrate.
Filtration
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National Engineering Center CEEC Seminar, June 2 -4, 2004
Filtration is a unit operation in which a fluid dispersed with
solids particles passes through a filter medium. The fluid flows
through but the filter medium retains the particles that could
not pass through the pores. In filtration, we call the deposited
solid the "filter cake" while the separated liquid, the "filtrate".
For flow to occur, a driving force is necessary. This force can
be gravity, positive pressure, vacuum, or centrifugal force.
This is also a basis of classifying filtration equipment. For
example, the gravity filter consists of a tank filled with sand
filters. The feed is at the top. The liquid drains at the bottom
while the solid remains entrapped within the filter medium.
Periodic backwashing cleans the filter medium. The plate-andframe filter press consists of alternating frame and plate with
cloth as the filter medium. With a positive pressure supplied
by a pump, the solids remain as a cake bounded by the cloth
filters. The leaf filter, on the other hand, needs vacuum. We
immerse the "leaf" in the material to be filtered. We apply
vacuum within the "leaf" and the solids deposits outside the
leaf while the applied vacuum sucks the liquid. The drum filter
operates continuously and needs vacuum. With part of a
rotating drum immersed in the fluid, the solids deposits on the
drum. A knife scrapes the filter cake on the drum.
DRAFT
Fluidization
Fluidization refers to the movement of solid particles in a
packed bed when a fluid passes through the bed. The particles
move up and down and the bed appears to expand. The bed
height (expanded) depends on the fluid velocity. We can
increase the velocity to a point that the solid particles flow
with the fluid. Whether as expanded bed or flowing with the
fluid, the solid particles are "fluidized". With water as the
medium, we can transport sand. Fluidization with air is a
means of transporting grain materials. A pump or blower
supplies the energy.
OTHER UNIT OPERATIONS
This section considers the unit operations not included in the
first four groups.
Centrifugation is a unit operation that separates solids from
solids and liquids from other liquids by centrifugal force. The
equipment, batch or continuous, involves rotation at a high
speed to cause the separation of phases.
Sedimentation involves the separation of a suspension of fine
solids in a liquid by gravity settling. The fine particles settle to
form a denser slurry. Clear fluid issue out from the top. The
operation is either batch or continuous. Waste treatment
plants use large sedimentation tanks operated continuously.
The solids settle at the bottom and an arm continuously
scrapes it. The clarified solution overflows at the top.
Mixing and agitation is a method for combining liquids,
solids, and gases to form intimate mixtures or to maintain
Introduction 29
DRAFT
i
Figure 1-14 Plate and frame filter press. (Ref.: Foust, et al, Principles
of Unit Operations, John Wiley and Sons, p. 48617. Permission being
sought.)
ntimate contact between the components. The equipment
consist of a tank provided with propellers run by motors.
Size reduction involves subdividing solids by application of
force to fracture or shatter large masses. Examples are crushers for coarse size reduction and grinders and pulverizers for
intermediate and fine size reduction.
Screening is the separation of solid particles into different
sizes by passage through several screens with different sizes of
openings. Each screen retains the particles that cannot pass
through its opening. The size fraction depends on the size of
the opening of the screen.
Flotation is a unit operation where solids separate by floating
to the surface of a fluid. Air is the common flotation agent.
Solids that readily take air bubbles on the surface float and
separate from those without adhering air bubbles.
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National Engineering Center CEEC Seminar, June 2 -4, 2004
DRAFT
Figure 1-15
Foust, et al, Principles of Unit Operations, John Wiley and Sons, p.
48820. Permission being sought.)21
Introduction 31
REFERENCES
1. Commission on Physical Sciences, Mathematics, and Applications
2.
3.
4.
5.
6.
(1988), ‘Chemical Engineering Research Frontiers: Needs and Opportunities’, Final report of the National Research Council Committee,
The National Academy Press, Washington D.C. 9 – 16.
Hougen, O.A. (1977) ‘Seven Decades of Chemical Engineering’,
Chemical Engineering Progress, 73 (1), 89-104.
Judson King, C. (1976) ‘The Expanding Domain of Chemical
Engineering’, Chemical Engineering Progress, 72 (3) 34-37.
Nemerow NL. Zero pollution for industry; waste minimization through
industrial complexes. John Wiley and Sons, Inc., New York, 1995.
Hockman, K.K. and D.A. Erdman (1993) “Gearing Up for ISO 9000
Registration, Chemical Engineering, April Issue, pp. 128-134.
International Organization for Standardization (ISO) (1993) Environmental Management Systems -- General Guidelines on Principles,
Systems and Supporting Techniques.
DRAFT
PROBLEMS
1-1
How does the profession of chemical engineering differ from
that of chemistry? Of physics? Of mathematics?
1-2
Interview some chemical engineers you know. Ask them if
they feel versatile and how did it affect their work and
profession.
1-3
Identify chemical engineers that are known locally and their
contribution to the Philippines.
1-4
Identify internationally known chemical engineers and their
breakthroughs.
1-5
Identify some chemical process industries within your city or
town.
1-6
Cite additional fields where you believe chemical engineering
will be increasingly important.
1-7
What characteristic of the chemical engineer is necessary in
globalization?
1-8
A building has a common water meter and a cistern for the
households. In the last month, tenant #1 consumed 40 m3;
tenant #2, 35.5 m3; tenant #3, 54.3 m3; tenant # 4, 36 m3;
and tenant #5, 55 m3. If the reading of the main meter for
that month is 250 m3, how much is the leakage?
1-9
A batch of 50 m3 of wastewater contains 5% oil by volume.
An oil separator is used to recover the oil. If the separator is
98% efficient, how much oil is recovered?
1-10 s an irrigation system delivers water from the source to the
point of usage, the following losses are encountered: 9% to
evaporation, 7% to run-off, and 15% to infiltration. If 4,000
liters was finally used what was the volume of the water at
the source?
32
National Engineering Center CEEC Seminar, June 2 -4, 2004
1-11 Biomass materials normally contain an average of 50%
lignin, 30% cellulose, and 20% hemicellulose. How many
grams of each component does 5 kilograms of the biomass
contain?
1-12 To make a low fat milk containing 1% butterfat, how much
butterfat must be removed from 5,000 liters of milk containing 5.1% butterfat?
1-13 We want to mix coconut oil with corn oil at a ratio of 2:1 (by
mass). How many grams of each is necessary to prepare 1.5
kg mixture?
1-14 A laboratory maintains 35% (by mass) hydrochloric acid and
10% (by mass) hydrochloric acid solutions. The process
needs 100 kg 15% solution. How many kg each of 35% and
10% acid solutions are needed?
1-15 An alloy contains 90% (by mass) of gold while another
contains 20%. How many grams of each alloy should be
combined to produce an alloy containing 60% gold?
DRAFT
1-16 A glycerin-water solution contains 60% glycerin (by mass).
How much water must be removed from the solution to
produce a 90% glycerin?
1-17 A 100-gram mixture of salt and sugar contains 50% of each.
How many grams of sugar must be added to the mixture to
produce a mixture containing 60% sugar?
1-18 Three pipes having different constant rates of water flowing
are used to fill up a tank. The first pipe can fill the tank
alone in half the time that it takes the second pipe to fill the
tank alone. The third pipe takes 1 ½ times the time it take
the second pipe to fill the tank alone. If all the pipes are
used, the tank will be full in 6 minutes. How long will it take
for the second pipe to fill the tank alone?
1-19 The composition of the medium for nata de coco production
is 54 liters of water, 4 kg of sugar, 2 kg of grated fresh
coconut, 1.2 liters of glacial acetic acid, and 12 liters of
mother liquor. The total volume is 72 liters. How much of
each component is needed to produce a 10-liter solution?
Introduction 33
DRAFT
34
National Engineering Center CEEC Seminar, June 2 -4, 2004
DRAFT
—Chapter 2—
Essential
Basic Principles
Chemical Engineering deals with different substances either
as raw materials, intermediate stuff, final products, byproducts, or waste. We need chemistry to understand chemical
transformation and chemical properties. With physics, we
learn about the physical changes, interaction, and properties.
The two sciences share some common ground. The principles
in physics and chemistry use both chemical and physical
properties. They are some of the bases of techniques chemical
engineers devise. The properties are presented in the form of
either experimental data, theoretical equations or empirical
relationships.
Mathematics is important to the chemical engineer. It serves
as a useful tool for communicating as well as manipulating
available scientific and engineering data. While these data can
be stored in tabular form, they would require immense space if
stored in printed form. Access to them is difficult. (Of course
storing them in computer databases is now a practical alternative.) By applying mathematics, we can store theories, data,
and procedures in more compact forms. Mathematical formulas can represent tabular or experimental data. Sets of mathematical equations can represent different phenomena. The set
of equations is also called a mathematical model. While the
solutions of some equations can be solved analytically, the
majority are impossible or too tedious to solve. Numerical
techniques can approximate the analytical solution to the
desired degree of accuracy. Solution by hand is practical for
simple equations but powerful computers do unwieldy computations in a shorter time.
This chapter presents familiar principles, data, and procedures for the beginning chemical engineering students. More
often than not, they have encountered them in chemistry,
physics, and mathematics subjects. We present the procedures with a distinct chemical engineering flavor. We are
trying to integrate some knowledge from chemistry, physics,
mathematics, as well as other relevant courses to fit into the
chemical engineering knowledge.
You may be familiar with the principles and theories
discussed. However, the presentation and techniques are now
Essential Basic Principles 35
in a chemical engineering orientation. If you find that the
difference is not perceptible, that is a good sign. It means that
you are adapting to the chemical engineering procedures.
In this chapter, we will discuss mainly the principles necessary in mass and energy balance calculations to keep our
focus. To facilitate your computations, those principles should
be of second nature to you. Remember, some of them may be
identical to what you already know. We start with the dimensions and system of units.
DIMENSION AND SYSTEM OF UNITS
DRAFT
Chemical engineering uses quantities that have to be qualified. Therefore, we should recognize the importance of dimension and units. A dimension describes a particular kind of
extent while a unit gives the measure of the same extent.
Mass, length, and time are examples of dimensions. The gram
is a unit of mass; the kilometer, a unit of length; and the
second, a unit of time. Table 2-1 shows four fundamental
dimensions. Secondary dimensions result when we combine
fundamental dimensions. Table 2-2 shows some secondary
dimensions. Dimension and systems of units are necessary in
qualifying the variables, parameters, and other quantities in
any technical work.
Table 2-1 Fundamental Dimensions
Dimension
Length
Mass
Time
Temperature
Symbol
L
M
θ
T
SYSTEM OF UNITS
The standard system of units today is the SI Units (System
Internationale d'Unites), which is based on the metric system.
However, in some parts of the world especially the United
States, units in the old system such as the English system
still persist. We will refer to units in use other than the SI as
“customary units.”
When the SI units were introduced, the thinking was that
the transition to SI would be achieved in relatively shorter
time. However, full conversion today has not yet been
attained. For example, older machines still in use are not in
SI. Older books are in their original systems of units. More
significantly, the old systems of units are imbedded in the
knowledge of all professions including chemical engineering.
Chemical engineers who graduated before the implementation
36
National Engineering Center CEEC Seminar, June 2 -4, 2004
of SI think in terms of the old systems. Usually they picture,
sense, compute, and estimate in terms of the original systems
they are familiar with. Older teachers explain better using the
old system. Specifically, some examples and techniques are
based on the old system. Generations of chemical engineers
were honed in these techniques. We will discuss them in this
chapter not just for historical significance but for their role in
computing techniques and better appreciation of physical
counterparts.
This book uses both SI units and some customary units and
provides a discussion on conversion of units.
DRAFT
Table 2-2 Combination of Fundamental Dimensions
Quantity
Area
Volume
Density
Force
Velocity
Acceleration
Weight
Definition
Measure of extent of surface
Measure of extent of space
Mass in a unit volume
Mass × acceleration
Distance covered per unit time
Rate of change of velocity
Force resulting from the gravitational
attraction of a body
Work
Power
Pressure
Specific volume
Surface tension
Viscosity
Force x distance
Rate of performing work
Force per unit area
Reciprocal of density
Force exerted by a surface of length
Resistance to flow
Dimension
L2
L3
M/L3
ML/θ2
L/θ
L/θ2
ML/θ2
ML2/θ2
ML2/θ3
F/L2
L3/M
F/L
M/Lθ
SI UNITS (INTERNATIONAL SYSTEM OF UNITS)
The SI units, adopted by the General Conference on Weights
and Measures, is the modernized metric system - a gradual
improvement of the system first proposed in 1670. Table 2-3
shows the SI base units. The SI units has many advantages,
being a uniform system.
Table 2-4 shows some SI derived units such as velocity,
m/s; acceleration, m/s2; and others. Mass density is in kg/m3.
The numerical value of this quantity is 1000 times its equivalent in g/cm3.
Table 2-5 gives SI derived units with special names such as
hertz, Hz for frequency; pascal, Pa for pressure; joule, J for
energy; and others. Take note that all unabbreviated units are
not capitalized (e.g. Hz for hertz, J for joule, N for newton,
etc.).
Essential Basic Principles 37
Table 2-3 SI Base Units
Quantity
length
mass
time
electric current
thermodynamic temperature
amount of substance
luminous intensity
Name
meter
kilogram
second
ampere
kelvin
mole
candela
Symbol
m
kg
s
A
K
mol
cd
Table 2-4 Some SI Derived Units Expressed in terms of Base Units
DRAFT
Quantity
area
volume
speed, velocity
acceleration
kinematic viscosity
mass density
concentration
Name
square meter
cubic meter
meter per second
meter per second sq.
square meter per second
kilogram per cubic meter
mole per cubic meter
Symbol
m2
m3
m/s
m/s2
m2/s
kg/m3
mol/m3
Table 2-5 SI Derived Units with Special Names
Quantity
Name
Symbol
frequency
force
pressure, stress
energy, work, quantity
of heat
power
quantity of electricity
hertz
newton
pascal
joule
Hz
N
Pa
J
Expression in Terms of
Other Units
1/s
kg.m/s2
N/m2
N.m
watt
coulomb
W
C
J/s
A.s
Quantity
dynamic viscosity
moment of force
surface tension
heat capacity
specific heat capacity, specific
entropy
specific energy
molar energy
molar entropy, molar heat
capacity
Name
pascal-second
meter-newton
newton/meter
joule per kelvin
joule per kilogram kelvin
Symbol
Pa·s
m·N
N/m
J/K
J/(kg.K)
joule per kilogram
joule per mole
joule per mole·kelvin
J/kg
J/mol
J/(mol·K)
Table 2-6 shows some SI derived units expressed by means
of special names. In forming derived unit abbreviations, use a
dot to indicate multiplication and a slash for division, e.g.,
kilogram meter per second square = kg⋅m/s2. Use the
38
National Engineering Center CEEC Seminar, June 2 -4, 2004
parenthesis to show the hierarchy of arithmetic operation as
in J/(mol⋅K) for joule per mol-kelvin.
Table 2-7 shows the other units not strictly SI which are still
in use with the system. Note that the symbol for liter is L.
Avoid using kg-force/m2, calorie, kilowatt-hour, poise, and
dyne and kilogram for force.
Table 2-7 Units in Use with the International system
Units
minute
hour
day
liter
ton
angstrom
hectare
bar
standard atmosphere
Symbol
min
h
d
L
t
A
ha
bar
atm
Value in SI units
1 min = 60 s
1 h = 60 min = 3600 s
1 d = 24 h = 86,400 s
1 L = 1 dm3 = 10-3 m3
1 t = 103 kg
1 A = 0.1 nm = 10-10 m
1 ha = 1 hm2 = 104 m2
1 bar = 0.1 MPa = 105 Pa
1 atm = 101325 Pa
DRAFT
Another feature of the SI is the use of powers of 10. Table
2-8 shows the symbols used for different powers of 10. We
express quantities larger than 103 (kilo) or smaller than 10-3
(milli) in third order prefix increment (103, 106, 109, ..., and
10-3, 10-6, 10-9, ..., etc.). E, P, T, G, and M are in the upper
case.
Table 2-8 SI Prefixes
Factor
1018
1015
1012
109
106
103
102
101
Prefix
era
peta
tera
giga
mega
kilo
hecto
deka
Symbol
E
P
T
G
M
k
h
da
Factor
10-1
10-2
10-3
10-6
10-9
10-12
10-15
10-18
Prefix
deci
centi
milli
micro
nano
pico
femto
atto
Symbol
d
c
m
µ
n
p
f
a
The main advantage of the SI is its coherence. No conversion
factors are necessary when using basic or derived units.
Some problems may occur due to errors in abbreviation. An
example is an error in using mg and Mg. The first is milligram
and the second, megagram, which are different by a factor of
109. The prefixes are quite difficult to remember and oral
miscommunication could lead to disastrous results. Therefore,
when using SI units, observe utmost care.
RULES IN USING SI1
Here is a summary of some points to remember when using SI
units:
1. Periods. Never use a period after a symbol of an SI unit
unless it is at the end of a sentence.
2. Capitalization. Do not capitalize the first letter of units
written out in full except at the beginning of a sentence.
1
Adapted from Peter and Timmerhaus, Plant Design and Economics for Chemi-
cal Engineers, 2nd ed., McGraw-Hill Book Co., New York, pp. 860-863, 1983.
Essential Basic Principles 39
However, capitalize the first letter of the symbols of units
derived from the name of a person (for example, 10 joules or
10 J; 10 pascals or 10 Pa; 200 kelvins or 200 K). Be careful
in the use of single letter symbols having an uppercase and
a lowercase usage such as:
g for gram
G for giga
k for kilo
K for kelvin
m for milli
M for mega
n for nano N for newton
t for ton
T for tera
3.
Plurals. Write the plural of units in words in the
normal grammatical sense. Never use plurals with symbols.
Units in words with numerical values equal to 0, greater than
1, or less than -1 are plural. All other values are singular.
For example:
300 kelvins or 300 K
120 pascals or 120 Pa
0.9 degree celcius or 0.9oC
0.5 kelvin or 0.5 K
0.2 centimeter or 0.2 cm
-.91 joule or -0.91 J
DRAFT
4. Grouping of numbers and decimal points. In SI, numbers
are placed in groups of three from both sides of the decimal
point. Examples:
234 358 298 568. 232
1. 400 324 098
23 344 675. 000
5.
Prefixes. If possible, keep the numerical values of the
units between 0.1 and 1000 by using approximate prefixes
with the unit symbol.
6 767 000 Pa = 6.77 MPa
0.0062 g = 6.2 mg
Do not use two or more prefixes in succession before a unit
symbol, e.g., use 1 pF instead of 1 µµF.
6.Spacing. A space is left between a number and a unit
symbol.
40 mJ
5 ms
6.4 mg
7. Hyphens. Use a hyphen between a number and a unit
symbol used as an adjective.
Example:
He bought a 10-m yacht.
The yacht is 10 m long.
8. Combination of units. Avoid prefixes in the denominator of
compound units (except for the base unit, kg).
Use kJ/m instead of J/mm.
Use kg/s instead of g/ms.
40
National Engineering Center CEEC Seminar, June 2 -4, 2004
Use the slash (/) as a division indicator. Avoid the use of
double slashes (//). For example, use g/(cm⋅s) or g cm-1 s-1
instead of g/cm/s. Use parenthesis in compound units to
avoid confusion, as in J/(kg⋅oC).
CUSTOMARY UNITS AND THEIR BASES
Due to their place in the chemical engineering , we have to
consider the customary units. Books printed before the use of
SI are still relevant. Some chemical engineering techniques
require their use. We consider the metric and the English
systems.
DRAFT
The MKS and the CGS Systems
The metric system originated in France and provided simplicity in measurements. Based on the decimal unit, the linear
measures, weights, and volume measures correspond well in
this system.
In the MKS (meter-kg-second) system, fundamental units
are meter, kilogram, and second. Force is a secondary
quantity. The system uses newton instead of kilogram-force to
avoid dependence of force on g. A newton is the force needed
to give a 1 kilogram mass an acceleration of a m/sec2. Following Newton's Law,
F = kma
where
Eq. 2-1
F = force, newton
m = mass, kg
a = acceleration, m/s2
with k = 1, dimensionless
1 newton = 1 kg × 1 m/s2.
In the cgs system, the fundamental units are centimeter,
gram, and second. Force is in dynes. A dyne is the force
needed to give a 1 gram mass an acceleration of 1 cm/s2.
Following Newton’s first law,
F = kma
with k = 1, dimensionless
1 dyne = 1 g × 1 cm/s2.
English Absolute System
In this system, the fundamental units are: foot for length,
pound mass (denoted as lbm) for mass, and second for time.
Force, a derived unit, is in poundal. A poundal is the force
that gives a pound mass an acceleration of 1 ft/s2. Following
Newton's first law, F = kma, with k = 1, 1 poundal = 1 lbm × 1
ft/s2.
Essential Basic Principles 41
English Gravitational System
In this system, the fundamental dimensions are length in feet,
time in seconds, and force in pound force (denoted as lbf). The
mass in this case is derived and is called a slug. A mass of one
slug will accelerate 1 ft/s2 when a force of 1 pound force is
applied to it. Applying Newton's first law, 1 lbf = 1 slug × 1
ft/s2 and 1 slug = 1 lbf × s2/ft.
English Engineering System
The English engineering system uses both mass and force as
fundamental dimensions. Length is in feet, time is in seconds,
mass is the pound mass, and force is in pound force. A pound
force is the force exerted by the earth's gravity on a mass of 1
pound mass at a standard acceleration of gravity of 32.2 ft/s2
(at sea level and 45o latitude) We designate this quantity as gc.
Substituting the values in Newton's first law, F = kma
F
k = ma
k=
DRAFT
or
1lbf
lbm % 32.2 ft/s 2
gc = 1 = 32.2 lbm $ ft
lbf $ s2
k
k in this case is not equal to 1 and possesses dimensions. gc
acts as a conversion factor between lb-force and lb-mass. So,
at a place where g = gc (at sea level and 45o latitude).
F = m
numerically.
CONVERSION OF UNITS
Every engineer must be proficient in converting from one
system of units to another. Conversion ratios with corresponding dimensions are useful in converting units. (See Table 2-9
for some ratios). We treat dimensions as though they are
mathematical quantities, that is, we can divide or multiply
similar quantities.
For example, consider the ratio 2.2 lb/kg. This shows that
there are 2.2 pounds in one kilogram. To find out how many
kg there are in 10 lb, we follow this method:
Write the conversion ratio such that lb is in the denominator
while kg is in the numerator. Knowing that a kilogram is
larger than a pound, place the coefficient 2.2 in the denominator. The lb units cancel out, leaving kg as the unit of the
desired quantity.
10 lb %
42
1 kg
= 10 kg = 4.54 kg
2.2 lb 2.2
National Engineering Center CEEC Seminar, June 2 -4, 2004
The next three examples illustrate the simple technique
used above: Select the appropriate conversion ratio such that
the unit we are converting cancels out.
Table 2-9 Common Conversions
Mass:
454 g/lb
2240 lb/long ton
2.2 lb/kg
1000 kg/metric ton
1000 g/kg
16 oz/lb
2000 lb/short ton
Time:
60 s/min
60 min/h
Length:
2.54 cm/in
3 ft/yd
100 cm/m
5280 ft/mile
1000 mm/m
106 µm/m
12 in/ft
1010 8/m
43560 ft2/acre
Area:
DRAFT
Volume:
1000.0217 cm3/liter
231 in3/US gal
7.48 U.S. gal/ft
3
Example 2-1
Convert 60 miles/h to cm/min.
Solution
h
60 miles % 5280 ft % 12 in % 2.54 cm %
mile
ft
in
60 min
h
= 1.61 % 10 5 cm
min
Example 2-2
Change a viscosity of 9 cP (centipoise) to viscosity in BVU
(British Viscosity Unit) where
g
1 poise = 1 P = 1 cm $ s
1 P = 100 cP
1BVU = 1 lbm
ft $ h
Essential Basic Principles 43
Solution:
viscosity in BVU =
g
cm
$ s % lbm % 2.54 cm % 12 in % 3600 s % BVU
P
9 cP%
%
P
454 g
100 cP
in
ft
h
lbm
ft $ h
= 21.75 BVU
Example 2-3
What is the equivalent of 0.229 cm2/s in ft2/h? (This quantity
is called diffusivity.)
Solution:
2
0.229 cm
s %
3600 s = 0.887 ft 2
ft 2
in 2
%
%
(2.54) 2 cm 2 (12) 2 in 2
h
h
DRAFT
The above examples show one line solutions. In some cases,
we can solve mass balance problems with the same technique.
We treat the description of streams and other quantities as
though they are units. This identification makes our solution
clearer and more systematic. In chemical engineering, we do
not solve problems blindly. We do it systematically and with
practice, it becomes like a mental solution. Sometimes, we do
not have to remember formulas. Of course, the technique
would not apply to complex problems. In those cases, we can
always use the computer.
CONVERTING MASS TO MOLE UNITS
We can express the dimension of mass in either mass or its
corresponding mole unit for molecules or atom unit for
elements. We form the mole units by adding the suffix mol to
the mass units, e. g., kg-mol for kg, g-mol for g, lb-mol for lb,
etc. In the SI, the mole unit is kmol, mol (= g-mol), and mmol
(= mg-mol). A mol of a compound (or a g-atom of an element)
contains 6.02 × 1023 molecules (or atoms for an element). This
quantity is called the Avogadro's Number, represented by the
symbol N with a corresponding unit of molecules/mol or
atoms/g-atom.
Each element has an atomic mass, which is the average
value for its constituent elements. The basis of the atomic
mass is that of isotope C-12 which has a value 12.000 atomic
mass units (amu). For example, the hydrogen isotope H-1 has
1.00 amu; H-2, 2.00; and H-3 3.00. The average value for
hydrogen is 1.01 amu. The periodic table gives the average
value for each element depending on their respective isotopic
composition. The atomic mass of carbon is 12. For molecules,
the sum of the atomic masses of the atomic components is the
molecular mass. We refer to this quantity more commonly as
molecular weight (MW).
44
National Engineering Center CEEC Seminar, June 2 -4, 2004
One mol of a pure compound has a mass numerically equal
to MW g of that compound. For example, one mol of hydrogen
(MW = 2.02) contains 2.02 g.
One kmol of a pure compound has a mass numerically equal
to MW kg of that compound. For example, two kmol of water
(MW = 18.02) has a mass of 36.04 kg.
The MW becomes a conversion factor between the mass and
mol units such as the following values:
kg
mg
g
ton , etc.
,
, lbm ,
,
mol kmol lbmol mmol tonmol
and is effectively dimensionless since the dimension is M .
M
DIMENSIONAL CONSISTENCY
DRAFT
The terms in an equation should agree dimensionally. Terms
added should have the same dimensions and exactly the same
units. The dimensions and units of the left side of the equation
should be the same as that of the right side.
Example 2-4
Verify the consistency of the energy equation for mass flow:
m
g
(P 2 − P 1 )
v 22 − v 21
+ m g (z 2 − z 1 ) + m
= −f − w
!
2g c
c
where
m:
v:
gc :
g:
z:
ρ:
P:
f:
w:
Subscript 1 and 2
mass, M
fluid velocity, L/θ
dimensional constant, ML/(F θ2)
acceleration due to gravity, L/θ2
vertical height, L
Density, M/L3
Pressure, F/L2
frictional loss, FL
work done, FL
Conditions at points 1 and 2, resp.
Solution:
2
2
M( L2 − L2 ) M L2 (L − L) M F2
L = - FL - FL
+ M L
+ M
M L
F 2
F 2
L3
Essential Basic Principles 45
2
M( L2 )
M L
F 2
FL
M L2 (L)
+ M L
F 2
M F2
L =
+ M
L3
− FL - FL
+
+
- FL - FL
FL
FL =
Each term of the equation has the unit FL. We can add the
terms together since they have the same dimensions, viz., they
are dimensionally consistent.
Example 2-5
Which of the following equations is correct?
dT =
dP
T( dV ) − V
dT
cP
or
dV − V
dT = T( dT
cP )
dP
DRAFT
The symbol definition and respective dimensions are
T = temperature, T
P = pressure, F/L2
V = specific volume, L3/M
cP = heat capacity, FL/MT
Solution:
Substitute the dimensions in the right-hand side of the two
equations. Both sides of the equations should have the dimension of the left side, T/(F/L2) corresponding to dT/dP
First Equation:
Second Equation:
T =
F
L2
L3
T M
T
3
−L
M
FL
MT
L3
2
= M = L T
FL
F
MT
L3
M − L3
T ! T( T
M ) = MT ( L 3 − TL 3 )
F
FL M
FL
M
MT
L2
We see that the first equation is correct because the dimension of the left-hand side of the equation is the same as the
dimension of the right-hand side. For the second equation, the
dimension of the right side does not simplify to the dimension
of dT/dP.
46
National Engineering Center CEEC Seminar, June 2 -4, 2004
Example 2-6
We can estimate the critical viscosities of diatomic gases by
using the following equation:
c =
where
(7.40)M1/2(Pc ) 2/3
(Tc ) 1/6
µc = viscosity, micropoise
Pc = pressure in atmospheres
Tc = temperature in K
M = molecular weight
We desire to transform the equation to a new form
∏c =
DRAFT
where
M 1/2 (P ∏c ) 2/3
(T c ) 1/6
µ'c = viscosity, lbm/(ft⋅h)
P'c = pressure in lbf/in2
Tc = temperature in K
M = molecular weight.
Find the numerical value of α in the above equation.
Given: 1 P = 1 g/cm⋅s
Solution:
The dimensions of both sides of the equation
c =
(7.40)M 1/2 (P c ) 2/3
(T c ) 1/6
should be the same. To maintain the dimensional
homogeneity, the constant 7.40 should possess appropriate
dimensions:
1/6
7.4 micropoise K 2/3
atm
which means
that
when this constant is multiplied by the
2
1
factor (Pc) 3 /(Tc) 6 with Pc in atm and Tc in K, µc is in
micropoise. In the new equation, the dimension of α is
lbm $ K 1/6 .
h $ ft $ psi 2/3
We can find the numerical value of α by converting the
dimensions of the original constant. Thus,
1g
1/6
poise
cm
$ s % 2.2 lbm
K
= 7.40 micropoise
%
%
1000 g
10 6 micropoise poise
atm2/3
Essential Basic Principles 47
(12)(2.54) cm
1
%
ft
14.7 psi 2/3
atm
1/6
−4 lbm K
= 2.98 % 10
h $ ft psi 2/3
% 3600 s %
h
We can check the consistency easily.
DIMENSIONLESS GROUPS
In engineering, quantities with units are grouped together to
result in a single quantity with no net unit or dimension.
These groupings are based on theoretical, experimental, or
empirical relationships. In future subjects, you will encounter
the importance of these “dimensionless groups”, which are
conveniently introduced to simplify and systematize engineering calculations. The groups are used as variables or parameters. The Reynolds number, NRe, that is used in fluid mechanics is an example.
Dv!
N Re = DRAFT
where
D = diameter, L
v = velocity,
L
ρ = density, M3
L
µ = viscosity,
M
L$
We can verify that the group is dimensionless by substituting the dimensions of the individual symbols. When using
units, they should be consistent.
ACCURACY, PRECISION, AND SIGNIFICANT DIGITS
Measurements are necessary in scientific or technological
fields such as chemical engineering. The improvement in
measuring techniques and instruments reflects the advances
in technologies. The accuracy of a measurement depends on
the instrument itself and the measuring technique. The
associated significant figure becomes important as it indicates
how accurate or inaccurate a measurement is.
Let us say that we weighed a plastic container on a weighing
balance. We got a reading of 24.5 g. In this reading, we are
sure that the tens digit is 2 and not 1 or 3. We are also sure
that the ones digit is 4 and not 2 or 3. When it comes to the
tenths digit, we are not sure whether it is 4, 5, or 6. We say
that the reading is 24.5±0.1. The reading, therefore, has three
significant figures. The significant figure of a numerical data is
48
National Engineering Center CEEC Seminar, June 2 -4, 2004
the first non-zero digit at the left of the quantity up to the
number which is doubtful. If we express the mass in terms of
kg, the reading will be 0.0245 kg. The significant figures is still
three. In mg, the value is 24,500 mg. This implies that the
number of significant figures is five. To keep the number of
significant digits, we can use the scientific notation 2.45×104
mg.
In our calculations, we should bear in mind the number of
significant figures of our data. The final answer should not
contain more significant figures than that of the datum which
has the least significant figures. Let us say we use the value
24.5 g, then our final answer in the calculation should not
contain more than three significant digits.
We should bear in mind that some measurements in chemical engineering are in very small quantities compared to the
major components of the system. Some may come in mg or
less. They could have an effect on the purity of the product.
Some may be toxic and will have a negative impact if they are
released to the environment.
Errors in calculations are important in statistics and
computer applications. Error propagation and related studies
are not covered in this book.
DRAFT
CURVE FITTING AND GRAPHICAL PRESENTATION OF DATA
Graphs are important since most of the data used by
engineers appear in graphical forms. They are powerful tools
and offer clear representations of processes. In graphical
methods, the size of the
graduations
limits
accuracy. The discussion that follows shows
the
student
some
graphical applications.
Many
computer
c
programs on presentation of data and curve
y
d
fitting are available. We
can
readily
apply
slope = c/d
computers
once
we
know the principles
b
involved.
Plotting a set of data
on a rectangular coordix
nate system shows the
Fig. 2-1
nature
of
the
variations. The curve
can be a straight line, a parabola, a circle, or any
non-standard form. A straight line is the simplest curve we
can use. We can perform mathematical manipulations easily
with this curve.
The equation
y = mx + b
Eq. 2-2
Essential Basic Principles 49
represents a straight line. Here, m is the slope and b is the
y-intercept. Consider the plot in Fig. 2-1.
We can see that the trend is a straight line. Let us draw a
straight line, even if it does not pass exactly through all the
points. Visually, we determine the line. Then we easily calculate the value of m and b from the graph. Hence, we also find
the equation that represents the variation of the data. The
spread of the points cannot be avoided because actual experimental data have inherent errors and will not be exact.
Example 2-7
The following table shows the specific heat of ethane as a
function of temperature:
38
0.43
Temp.,°C
Specific heat, cal/(g⋅°C)
66
0.46
93
0.49
121
0.51
149
0.55
204
0.61
Given that the variation is linear, find the equation that represents the data.
DRAFT
Solution:
In the above set of data, specific heat is the dependent variable
so we assign it on the ordinate. Since we know that the variation is linear, we expect the plot to be a straight line, as Fig.
2-2 shows.
0.65
Specific Heat, cal/(g)(oC)
0.60
0.55
0.50
0.45
0.40
0.35
0.30
0.25
0.20
0
20
40
60
80
100 120
140 160 180
200 220
Temp., oC
Fig. 2-2
At x = 0, the y-intercept is equal to 0.385. Considering two
points A and B on the line, the slope is
0.58 − 0.40 = 0.00108
180 − 13
The equation that relates Cp to T is C p = 0.00108T + 0.385
50
National Engineering Center CEEC Seminar, June 2 -4, 2004
Whenever the variation of some test data with a variable is
known to be linear, the constants m and b in the equation are
specific. To maintain the linear variation, we can rearrange or
regroup terms of the equation to substitute for the variables x
and y in the equation y = mx + b to maintain the straight line
relationship.
Letting y = f(y) and x = f(x), the equation becomes
f(y) = mf(x) + b
Eq. 2-3
Plotting f(y) versus f(x) still results in a straight line with m
as slope and b as the intercept at f(x) = 0. For example, if f(y)
is log y and f(x) is log x, then the equation is written as
log y = m (logx) + b
Eq. 2-4
The problem then boils down to plotting the data set and
evaluating m and b. Table 2-10 shows some techniques in
plotting non-linear data as straight lines. Forms other than
the straight line can be used but the manipulation is more
difficult.
Table 2-10 Methods of Plotting To Give Straight Lines
DRAFT
Equation
1. y = a + bx
2. y = axn
3. y = c + axn
4. y = aebx
5. y = abx
6. y = a + b/x
7. y = x/(a + bx)
Method
Plot y vs x.
Plot log y vs log x on rectangular coordinates or y vs
x on semi-logarithmic coordinates
First obtain c as intercept on plot of y vs. x then
plot log (y - c) vs log x, or y vs. xn, or (y - c) vs x on
logarithmic coordinates.
Plot log y vs. x on rectangular coordinates or y vs. x
on semilogarithmic coordinates.
Plot log y vs. x on rectangular coordinates or y vs. x
on semilogarithmic coordinates.
Plot y vs 1/x.
Plot x/y vs x or 1/y vs 1/x.
1
Mickley, H. S., T. K. Sherwood, and C.E. Reed, Applied Mathematics in Chemical
Engineering, McGraw-Hill Book Co., Inc., New York, 1957, p. 11.
In graphical curve-fitting, the straight line methods are the
only simple ones that can be considered. The rule to follow
when the variation of a data is not known is to plot them first
on rectangular coordinates, then on semi-logarithmic, and
then on a log-log graph. The one which gives a straight line
represents the suitable equation.
Example 2-8
In an electric generating station, the steam silica should be
maintained at a prescribed level. Excessive silica in the steam
will cause silica deposition in the turbine blades which will
considerably reduce turbine efficiency or even cause a plant
shutdown.
The following tabulation shows the maximum boiler silica
that can be tolerated to maintain the steam silica at a satisfactory level:
Essential Basic Principles 51
Silica, 7
5
4.2 3.5
3
2.5 2.1 1.7 1.5 1.2
1
ppm
Pres., 800 1,000 1,100 1,200 1,300 1,400 1,500 1,600 1,700 1,800 2,000
psig
If the above values will form a straight line on a semi-log
scale, find the appropriate equation that will represent the
data.
Solution:
Let
C SiO 2 = be aP
Taking the natural logarithm of both sides
ln C SiO 2 = aP + lnb
With the above form of equation, we can plot the data in the
form of a straight line. Refer to Fig. 2-3.
100
Silica concentration, ppm
DRAFT
intercept = 29
C
10
D
1
0.1
0
200
400
600 800
1000
1200 1400 1600 1800 2000 2200
P, psig
Fig. 2-3
From the graph, the y intercept at x = 0 is b = 29
Considering two points C and D on the line,
at C, C SiO 2 = 10
P = 600
at D, C SiO 2 = 1.2
P = 1800
Therefore
a = ln 10 − ln 1.2 = −1.77 % 10 −3
600 − 1800
Thus, the equation representing the data is
C SiO 2 = 29 e −0.00177P
Example 2-9
52
National Engineering Center CEEC Seminar, June 2 -4, 2004
Two substances A and B react according to the following
equation 4A + B d C + D + 4E + 2G . It is possible to measure the
concentration of A at anytime and also the rate at which it is
being depleted. This rate also represents the rate of the chemical reaction.
It is known that
−r A = k(C A ) a
where
-rA is the rate of depletion of A, mol/h
CA is the concentration of A, mol/liter.
k and a are constants.
From the following data, evaluate the values of k and a
graphically.
CA,
0.02
0.05
0.07
0.09
0.49
0.74
0.98
2.48
3.97
mol/L
1.90E-0 3.00E-0 3.42E-0 4.20E-0 8.60E-0 1.16E-0 1.25E-0 1.80E-0 2.30E-0
-rA,
8
8
8
8
8
8
8
8
8
mol/h
A log-log graph is used with (-rA) as the ordinate and CA as the
abscissa. All the points are plotted and a straight line that
represents the data is visually approximated. Refer to Fig. 2-4.
Two points P1 and P2 on the line drawn are located for the
calculation of the slope. The coordinate of P1 is (0.018, 2 × 10-8)
while that of P2, (2.8, 2 × 10-7) The slope is equal to
a=
log (2 % 10 −7 ) − log (2 % 10 −8 )
= 0.46
log(2.8) − log(0.018)
P2
b = 1.2
10-7
(-rA), mol/h
DRAFT
Solution:
P1
10-8
0.01
Fig. 2-4
1
0.1
CA, mols/liter
k is equal to the (-ra) intercept at CA = 1.0 or at logCA = 0.
From the graph, k = 1.2 x 10-7. The equation therefore is
Essential Basic Principles 53
−r A = (1.2 % 10 −7 )(C A ) 0.46
SOME PROCESS VARIABLES
Variables are necessary to describe chemical systems and
processes. These variables are properties which we can
measure and record. They define the conditions of a system at
any time. The variables in an entering stream can be classified
as disturbances (inputs that we cannot control) or manipulative variables (inputs that we can change). Likewise, the output
variables can either be controlled or uncontrolled. This makes
process variables indispensable elements in process control.
The present trend in chemical engineering is the application
of process modeling, simulation, and control. To derive mathematical models of chemical engineering systems, process
variables are necessary. Temperature, pressure, flow rate,
composition, density, humidity, liquid level, pH, specific
gravity, viscosity, consistency, color, and moisture are
examples of variables. Some of these are still awaiting scientific breakthroughs in the refinement of methods of detection
and measurement.
DRAFT
TEMPERATURE
Temperature is the measure of the kinetic energy of a molecule
or an atom. Molecules and atoms are in constant motion. They
move in translational, vibrational, and rotational manner. We
cannot measure the kinetic energy of the molecules and atoms
directly. However, temperature readings can indicate the
relative magnitude of this molecular energy.
A thermometer is the instrument that indicates temperature. Several types exist. The most common type is the filledsystem thermometers which functions on the basis of the
expansion of fluids. An example of the filled-system
thermometer is a tube containing a thin column of mercury
which expands proportionally to the hotness. Based on the
effect of heat on the current produced at the junction of two
dissimilar metals, we have the thermocouples. The resistance
thermometers are based on the variation of electrical resistance with temperature. For high temperature measurements,
we use the radiation pyrometers. Here, the measurement
depends on the intensity of emitted radiation of a material.
The temperature scales are either absolute or relative. A
postulate states the existence of a temperature at which
molecules or atoms cease to move. This is the absolute zero
temperature. The absolute temperature increases as the
energy of the molecules increases. In the SI units, K (Kelvin) is
the unit for the absolute scale, while in the English system, oR
(degree Rankine). The unit graduations of these two scales are
based on relative scales first used. In the SI system, oC,
degrees Celsius is unit graduation. The freezing point and the
normal boiling point of water (at 1 atm) are the two base
points of the Centigrade scale. The span is divided into 100
graduations with the freezing point of water as 0oC and the
54
National Engineering Center CEEC Seminar, June 2 -4, 2004
boiling point, 100oC. The scale extends from both ends for
values less than zero and greater than 100.
Graduations are indicated on these devices. One Celsius
degree is the same as one Kelvin degree. OoC corresponds to
273.16 K (rounded to 273 for engineering work) or
T K = T ) C + 273
where
Eq. 2-5
T K is the reading in Kelvin
ToC is the reading in Celsius
The English system employs oFahrenheit, with the freezing
point of water at 32oF and the boiling point at 212oF. One
Fahrenheit degree is the same as a Rankine degree. 0oF corresponds to 459.60oR (rounded to 460 for engineering work) or
T ) R = T ) F + 460
DRAFT
where
Eq. 2-6
ToR is the reading in Rankine
ToF is the reading in Fahrenheit
Based on the boiling and the freezing points of water, 100
centigrade graduations are equivalent to 180 Fahrenheit
graduations .
T ) F − 32 = 1.8( T ) C−0)
Eq. 2-7
We can use Eq. 2-7 as a basis for converting temperature
readings.
T ) F = 1.8( T ) C)+ 32
T ) C = (T ) F − 32)/1.8
Eq. 2-8
Eq. 2-9
For the absolute scales,
T ) R = 1.8T K
Eq. 2-10
C, oF, K, and oR are the units for the dimensions of
temperature, T, as for instance in heat capacity, cal/g-oC or
BTU/#-oF. The oC or oF in this case does not refer to the
reading on the scale but rather to the size of the scale graduations. Referring to the scales, a Celsius degree (∆oC or Co) is
larger than a Fahrenheit degree (∆oF or Fo). In conversion of
units,
o
1.8 F o
Co
should be used. Unfortunately, the symbols ∆oC or Co, ∆oF or
Fo, etc., are not in standard usage as unit temperature
Essential Basic Principles 55
difference (scale size). Be careful in using the units in scale
size and in temperature readings. Remember that they are not
the same.
Example 2-11
Convert 1 cal/(g·°C) to BTU/(lb·°F).
Solution:
1 cal % BTU % 454 g % ) C = 1 BTU
g $ ) C 252 cal
1.8 ) F
lb $ ) F
lb
DENSITY AND SPECIFIC GRAVITY
DRAFT
A substance possesses a definite amount of mass in a specified volume. This is the density, mass per unit volume (M/L3).
In SI, the unit is kg/m3.
In general, temperature affects density. When heated, a
substance expands and its density decreases. Liquids and
gases expand more than solids.
The densities of substances are usually compared to the
density of a reference substance at a specified temperature
and pressure. The ratio of the mass of a substance to the
mass of an equal volume of water at a specified temperature is
the specific gravity.
At 4oC, the density of water is 1.000 g/cm3 so that when
o
4 C is the reference temperature the specific gravity is numerically equal to the density in g/cm3. Dry air at a specified
temperature and pressure can also be used as a reference for
gases. The temperature of the substance and the reference are
frequently included with the specific gravity data. Sp. Gr.
25o/4o means the substance is at a temperature 25oC and the
density is referred to the density of water at 4oC or
Sp. Gr. 25 o /4 o =
density at 25 o C
density of H2 O at 4 o C
Specific gravity is dimensionless.
Some special scales for specific gravity are in use. Examples
are A.P.I., *Baume (oBe) and Twadell. We can find the other
special scale conversions to specific gravity in appropriate
references.
In the SI units, the absolute density is preferred. However,
since most data in the literature are expressed as specific
gravity, the above discussion is included.
PRESSURE
Pressure is the normal force acting on a unit area of
surface. This force can be the result of gravity, the expansion
or contraction of material, compression, tension, and thermal
stresses, among others. In chemical engineering, pressure
commonly refers to the forces exerted by contained fluids
56
National Engineering Center CEEC Seminar, June 2 -4, 2004
(static pressure) or from forces of moving fluids (dynamic
pressure). Fluids consist of molecules or atoms that are freely
moving. In gases, the molecules are far apart from one
another. They move around and collide with one another.
When enclosed in an air-tight vessel, they collide with the
walls. The sum total of all the collisions with the wall result in
a total force. This is equal to
F = mv
2gc
2
Eq. 2-11
DRAFT
where m = mass
v = velocity
gc = acceleration due to gravity
The total force divided by the total area of the vessel gives
the pressure exerted by the gas.
Liquids, on the other hand, can be contained in an open
vessel. The force exerted on the container is due to the force of
gravity on the fluid. The pressure increases from zero at the
top to a maximum at the bottom. A fixed height of a fluid
exerts a definite pressure at the bottom. This pressure
depends on the density of the fluid and the height of the
column.
Example 2-12
What is the pressure exerted by 10 m of a vertical column of
water in newtons per square meter?
Solution:
To solve this problem, the density of water and the value of g
are needed to evaluate the force. At 30oC, the density of water
is 996 kg/m3. Assume g = 9.80 m/s2.
Consider the cross-sectional area of the column to be 1 m2.
The total volume of the column then, is 10 m3. The mass of
water is 10 m3 x 996 kg/m3 = 9,960 kg. The total force exerted
by gravity is F = mg = 9,960 x 9.80 m/s2 = 97,608 N. The
pressure at the bottom is, therefore, 97,608 N/m2.
Example 2-13
What is the pressure exerted by 76 cm of mercury in pascals?
In pound-force per sq. inch? The density of Hg is 13,600
kg/m3 and g = 9.8 m/s2.
Solution:
(a) Consider 1 m2 as the cross-sectional area of the column
F = 1 m 2 % 0.76 m % 13, 600
kg
% 9.8 m/s 2 % 1 N
m3
kg $ m
s2
= 1.013 % 10 5 N
The pressure is 1.013 × 105 Pa.
Essential Basic Principles 57
(b) Solution (a) could be repeated but with the appropriate
units in the English system. However, conversion of the
answer in (a) is used.
kg $ m
s2
1
% 2.2 lbm %
%
Pressure = 1.013 % 10 5 N2 %
N
m
kg
32.2 lbm $ ft
2
lbf $ s
ft 2
1m %
3.28 ft 144 in 2
= 14.65 psi
COMMON UNITS OF PRESSURE
DRAFT
The common units of pressure (F/L2) are:
y Pascal, Pa (kPa and MPa are more common since Pa is
small)
y pound-force per sq in, psi
y dynes/cm2
In terms of the height of a column of liquid, we have the
following:
y cm of mercury, cm Hg
y inches of mercury, in. Hg
y meters or feet of water
y cm or in of water
We can also express pressure in terms of the force exerted
by the atmosphere around us. This pressure at sea level and
45 degrees latitude is designated as the normal atmospheric
pressure or the standard atmosphere. A normal atmosphere
has a value of 1 atmosphere (1 atm). In terms of the other
units,
1 atm = 1.01325 x 105 Pascal
1 atm = 14.7 lbf/in2
1 atm = 760 mm Hg
1 atm = 29.92 in. Hg
1 atm = 33.9 ft of H2O
1 atm = 1.01325 x 106 dynes/cm2.
For convenience, memorize the above values. You could use
them as conversion factors.3
Example 2-14
How many atmospheres is 27 psi?
Solution:
27 psi % 1 atm = 1.84 atm
14.7 psi
3
For columns of fluids, the effect of temperature (normal variations) on density is
neglected when used for engineering purposes.
58
National Engineering Center CEEC Seminar, June 2 -4, 2004
Example 2-15
If the pressure is 107.6 psi, what is its equivalent in inches of
Hg? In feet of H20?
Solution:
P = 107.6 psi %
29.92 in Hg
= 219.0 in Hg
14.7 psi
P = 107.6 psi %
33.9 ft H 2 O
= 248.1 ft H 2 O
14.7 psi
DRAFT
ABSOLUTE AND RELATIVE PRESSURES
The three major classes of pressure-measuring elements are:
y A liquid column where the density and height of the fluid
determines the pressure. Manometers and barometers are
examples.
y Formed metal elements such as diaphragm, bellows, and
bourdon gages.
y Electronic or electrical devices, thermal conductivity
gages, strain gages, and ionization gages.
The bourdon gage indicates pressure by the deflection of a
pointer. Normally, the zero reading of most gages corresponds
to the pressure exerted by the atmosphere. The indicated
pressure therefore, is the "gage pressure". Gage pressure is
the pressure above that of the atmospheric pressure.
Therefore,
P abs = P atm + P gage
where
Eq. 2-12
Pabs is the absolute pressure
Patm is the atmospheric pressure
Pgage is the gage pressure
All three terms must have the same units to maintain dimensional consistency
We express gage pressure as psig, in. Hg gage, mm Hg gage,
etc. and absolute pressure, as lb-force per square inch
absolute or psia, in. Hg abs., mm Hg abs., etc. In S.I. however,
we only use the absolute pressures.
Example 2-16
If the atmospheric pressure is 1.1 atm, what is the absolute
pressure if the gage indicates 28 psig?
Solution:
P abs = P atm + P gage
Essential Basic Principles 59
P abs = 1.1 atm %
14.7 psi
+ 28 psi = 44.17 psia
1 atm
Note: The value of the atmospheric pressure is dependent on
the actual pressure exerted by the atmosphere.
VACUUM, HEAD, AND DRAFT
Pressures below atmospheric may be expressed in terms of
"vacuum". "Vacuum" refers to the value by which the pressure
is less than atmospheric. The absolute pressure, Pabs is related
to the "vacuum", Pvac, by
P abs = P atm − P vac
Eq. 2-13
At pressures near atmospheric we use the term "head" in
lieu of gage pressure and "draft" instead of vacuum.
"Head" is the quantity by which a pressure is more than
atmospheric while "draft" is the quantity by which a pressure
is less than atmospheric.
Example 2-17
DRAFT
If a vacuum gage indicates 26 in. Hg, what is the pressure if
the atmospheric pressure is 28 in. Hg?
P abs = P atm − P vac
P abs = 28 in Hg − 26 in Hg
= 2 in Hg abs
Example 2-18
The pressure of a gas is reported as 3" water head. If the
atmospheric pressure is 758 mm Hg, what is the absolute
pressure?
Solution:
3" H2O is equivalent to
3 in H2 O %
ft % 760 mm Hg = 5.60 mm Hg
12 in 33.9 ft H2 O
The pressure of the gas is, therefore, 5.60 mm Hg higher
than the atmospheric pressure.
P abs = 758 + 5.60 = 763.6 mm Hg
Example 2-19
What is the equivalent of 5" H2O draft in psia if the atmospheric pressure is 28 in Hg?
Solution:
60
National Engineering Center CEEC Seminar, June 2 -4, 2004
P abs = (28 in Hg %
14.7 psia
14.7 psia
) − (5 in H 2 O % 1 ft %
)
29.92 in Hg
12 in 33.9 ft H 2 O
= 13.58 psia
COMPOSITION
Composition denotes the amount of substance present in a
mixture. We can express composition in various ways depending on the type of substances, convenience, or convention.
Mass Fraction and Mass Per Cent
Mass fraction is the ratio of the mass of a component in a
mixture to the total mass of the mixture.
M
xi = Mi
or
DRAFT
Eq. 2-14
t
M
x 1 = M1
t
M
x 2 = M2
t
M
x 3 = M3
t
and so on.
where
xi = mass fraction of component i
Mi = mass of component i
Mt = total mass of the mixture
Since
M t = M 1 + M 2 + M 3 + ... + M n
Eq. 2-15
x 1 + x 2 + x 3 + ... + x n = 1
Eq. 2-16
then
and mass % = 100 xi.
Mass fraction is usually used for solids and liquids, and less
commonly for gases.
Essential Basic Principles 61
Mole Fraction and Mole Per Cent
Mole fraction xi (or yi) is the ratio of the moles of component
i of a mixture to the total moles of the mixture
xi =
moles component i
total moles of mixture
Also,
x1 + x2 + x3 + ...... + xn = 1 or
n
i=1
xi = 1
mol per cent = 100 % x i
While mole fraction is applicable for all types of substances,
it is more commonly used to express the compositions of
mixtures of gases.
Volume Fraction and Volume Percent
DRAFT
Volume fraction refers to the ratio of the volume occupied
by a component to the volume occupied by the total mixture
under the same conditions. This is commonly used to express
the composition of gases and occasionally of liquids. A limitation in the use of this type of composition is that when we mix
certain substances, the resulting volume is different from the
sum of the volume of the components before mixing. Hence,
corrections must be applied. The volume fraction is applicable
only for "ideal mixtures", i.e., mixtures with components that
do not expand or contract when mixed together.
Volume % = volume fraction × 100.
Concentration
Concentration refers to the amount of component present
per unit volume of the total mixture. The unit is either mass
per unit volume of mixture or moles per unit volume of
mixture. kmol/m3 is the unit in SI. Concentration is dependent on temperature since the volume of a substance changes
with temperature.
Concentration Units in Chemistry
In chemistry, concentration is expressed in terms of molarity, formality, molality, and normality.
Molarity is the number of gram moles of a solute dissolved
in a solvent per liter of the solution.
Formality is the number of formula weights per liter of
solution.
Molality is the number of gram moles of a solute per 1000
grams of solvent.
Normality is the number of gram-equivalent weights of a
solute per liter of solution.
Concentration of Trace Quantities
In reporting trace or small quantities, the mass fraction is
expressed in terms of ppm, parts per million (parts of a
specific substance per 106 parts total mixture), or ppb, parts
per billion (parts of a specific substance per 109 parts total
mixture). In aqueous solutions, where concentrations may be
62
National Engineering Center CEEC Seminar, June 2 -4, 2004
expressed in mg solute/liter solution, ppm is approximately
equal to mg/liter.
FLOW RATES
In chemical engineering, we often encounter transfer of large
amounts of materials that are constantly flowing in a process
plant. Therefore, the flow can be measured per unit of time.
Mass flow rate is the mass of a material flowing per unit
time.
Volumetric flow rate is the volume of material flowing per
unit time.
Q= w
!
Eq. 2-17
or
DRAFT
where
w = Q!
w = mass flow rate, kg/s
Q = volumetric flow rate, m3/s
ρ = density, kg/m3
As ρ is temperature dependent, so is the volumetric flow
rate.
Flow velocity refers to the distance traversed by a material
per unit time. It is designated by v, with m/s as the S.I. unit.
If we know the cross-sectional area, A, of the duct,
v=
Q
A
Eq. 2-18
If Q and A are constant, the flow velocity is also constant.
The velocity of a fluid flowing through a pipe or duct varies
from a value of zero at the wall to a maximum at the center of
the pipe. This is due to the resistance offered by the pipe.
Therefore, the flow velocity we use is the average value across
the cross sectional area of the pipe or duct. Substituting the
value of Q (from Eq. 2-18) in Eq. 2-17
w = Av!
Eq. 2-19
Mass velocity (G) is another useful flow expression and is
equal to mass flow/cross-sectional area of flow.
G = w = v!
A
Eq. 2-20
Example 2-20
A certain process needs 600 kg of hot air per hour at 120oC.
To accomplish this, air at 30oC passes through a heat
exchanger.
Essential Basic Principles 63
The inlet pipe is 20 cm in diameter while the outlet pipe is
30 cm. The heat exchanger consists of 100 tubes (each 2.5 cm
diameter) through which the air passes and is heated by the
steam condensing outside the tubes. Calculate the volumetric
flow rate, the flow velocity, and the mass velocity at
(a) the 20 cm dia. pipe
(b) the end of the heat exchanger pipes where the air goes
out
(c) the 30 cm dia. pipe
20 cm dia.
air in
air out
heat exchanger
30 cm dia.
pump
Fig. 2-5
Data: ρair at 30oC = 1.161 kg/m3
ρair at 120oC = 0.895 kg/m3
Solution:
DRAFT
Setting up an air mass balance,
The mass flow rate at the 20 cm dia. pipe
= mass flow rate through all heat exchanger tubes)
= mass flow rate at the 30 cm dia. pipe
= 600 kg/h.
(a) At the 20 cm dia pipe,
A = (0.20) 2 % = 0.0314 m 2
4
The temperature is at 30°C and the density of air is 1.161
kg/m3.
Q= w
! =
600 kg/h
= 516.8m 3 /h
1.161 kg/m 3
600 kg/h
= 16, 450 m/h
v= w =
!A (1.161 kg/m 3 )(/4)(0.2) 2 m 2
600 kg/h
= 19, 098 kg/(h$m 2 )
G= w =
A (/4)(0.2) 2 m 2
(b) At the end of the heat exchanger the air has been heated to
120oC and ρ = 0.895 kg/m3. The cross-sectional area of flow
through each tube,
A = (2.5) 2 (/4) = 4.9087 cm 2
64
National Engineering Center CEEC Seminar, June 2 -4, 2004
Q= w
! =
600 kg/h
= 670 m 3 /h
0.895 kg/m 3
600 kg/h
= 13, 700m/h
v= w =
!A (0.895 kg/m 3 )(4.9087)(1/100 2 ) m 2 (100)
600 kg/h
= 12, 223 kg/(h $ m 2 )
G= w =
A (4.9087)(1/100 2 ) m 2 (100)
(c) At the 30 cm dia pipe, the air is still at 120°C. The crosssectional area of flow, A = (0.3) 2 (/4) = 0.0707 m 2
Q= w
! =
600 kg/h
= 670 m 3 /h
0.895 kg/m 3
600 kg/h
= 9, 480 m/h
v= w =
!A (0.895 kg/m 3 )(0.0707) m 2
DRAFT
600 kg/h
G= w =
= 8, 490 kg/(m 2 $ h)
A (0.0707) m 2
BEHAVIOR OF GASES, LIQUIDS, AND SOLIDS
All substances may exist, although sometimes theoretically, as
a gas, a liquid, or a solid. Generally, a liquid substance vaporizes when heated to a high temperature and solidifies when
cooled to a low temperature. Some substances, when heated,
decompose before a change in phase occurs. For instance,
sugar chars when heated before it can turn to gas.
Each substance possesses its own definite behavior with
variation in temperature. However, the general trend for all
substances is the same. At a temperature of absolute zero,
where it is postulated that all molecules and atoms cease to
move, all substances are in the solid state. As the temperature
increases (or as energy is added), the molecules start to move
(rotational, vibrational, and translational motion). Depending
on the structure of the molecules, they may attract or repel
one another. When the molecules or atoms are able to move
past one another, the "melting point", the temperature at
which a substance changes to the liquid state is reached.
Further increase in temperature allows the molecules to move
more freely with some being able to escape from the liquid.
Then, a temperature will be reached at which the liquid vaporizes, i.e., the molecules are able to escape. This temperature is
the "boiling point". The gaseous form can be heated further
resulting in an increase in the energy of the molecules.
We can subject a substance at a pressure and temperature
so as to keep it in the liquid state. The maximum pressure and
temperature at which the substance can exist as a liquid is
known as the critical pressure and critical temperature.
Above the critical temperature, a gas cannot be liquefied no
matter how large a pressure is applied. The boiling point, the
melting point, and the critical properties of a particular
substance depend on the structure of its molecules or atoms.
Essential Basic Principles 65
To appreciate how the properties of substances vary, note that
helium boils at -268.94oC while water boils at 100oC (both at 1
atmosphere).
If we can correlate data in terms of empirical equations,
then voluminous data on substances can be reduced. Several
models and empirical equations are available with moderate
success achieved for gases.
BOYLE'S LAW
Robert Boyle derived one of the earliest correlations based on
experimental observations. He demonstrated that for different
gases at the same temperature, the volume varies inversely
with absolute pressure. (He made these observations at
ordinary temperatures and pressures.)
V} 1
P
Eq. 2-21
V = K1 1
P
Eq. 2-22
PV = K 1 = constant
or
DRAFT
P 1 V 1 = P 2 V 2 = P 3 V 3 = .....P n V n
where
Eq. 2-23
V is the volume
P is the absolute pressure
K1 is a constant, with dimension FL
n is a subscript referring to a particular state or
condition.
Pn is the pressure at state n
Vn is the volume at state n
CHARLES' LAW
The correlation observed by Charles states that at constant
pressure, the volume is directly proportional to the absolute
temperature.
V}T
Eq. 2-24
V = K2T
Eq. 2-25
V 1 /T 1 = V 2 /T 2 = V 3 /T 3 = ... = V n /T n
where
Eq. 2-26
T is the absolute temperature
K2 is a constant, with L3/T as the dimension
OTHER CORRELATIONS
66
National Engineering Center CEEC Seminar, June 2 -4, 2004
Another correlation is that, at constant volume, the
absolute pressure varies directly with the absolute
temperature.
P}T
Eq. 2-27
P = K3T
Eq. 2-28
Pressure and temperature do not depend on the quantity of
the gas. (They are intensive properties.) Volume (an extensive
property), on the other hand, depends on the quantity of
matter. The more the number of moles of a gas at a particular
temperature and pressure, the greater is its volume.
V}n
at a fixed P and T
Eq. 2-29
where
n is the number of moles of the gas.
DRAFT
THE IDEAL GAS LAW
We can combine the correlations (as discussed above) for the
variables V, P, T, and n as:
V } nT
P
Eq. 2-30
V = (constant)n T
P
Denoting the constant as R, which is designated as the
universal gas constant
PV = R
nT
Eq. 2-31
PV = nRT
Eq. 2-32
or
This is the well-known ideal gas law.
For different conditions,
P1V1 P2V2 P3V3
P V
=
=
= ... = n n
n1T1
n2T 2
n3T3
nnT n
Eq. 2-33
We can note that if T and n are constant, Boyle's law
results; and if P and n are constant, Charles' law results. The
value and units of R depends on the units of P, V, T, and n.
The dimension of R is
(F/L 2 )(L 3 )
Eq. 2-34
= FL
R = PV =
MT
MT
nT
Essential Basic Principles 67
In SI units, where the respective dimensions of the variables
are P = kPa, V = m3, T = K, and n = kmol, R has a value of
8.314 kPa·m3/(kmol·K) or 8.314 kJ/(kmol·K) (since Pa⋅m3 = J).
Table 2-11 gives the values of R in different systems of units.
Table 2-11
Values of R
8.314 kJ/(kmol⋅K)
0.082 (atm⋅liter)/(mol⋅K)
1.987 BTU/(lbmol⋅°R)
10.73 (psia⋅ft3)/(lbmol⋅°R)
8.314 (kPa⋅m3)/(kmol⋅K)
0.73(atm⋅ft3)/(lbmol⋅°R)
1.987 cal/(mol⋅K)
LIMITATIONS OF THE IDEAL GAS LAW
DRAFT
The ideal gas law is a simple correlation derived under the
following conditions: sufficiently low pressure (about two
atmospheres and below) and high enough temperature (near
or above the critical temperature of the gas). For example, if
we consider the air around us, we can apply the ideal gas law
since its critical temperature is way below room temperature
and as long as the pressure is near or below one atmosphere.
The application of the ideal gas law to gases at high pressures
and low temperatures is inadequate. We can explain this
limitation with the aid of the kinetic theory of gases. The
theory provides that for gases to behave ideally, the forces of
attraction or repulsion between molecules should be
negligible. These forces may either increase or decrease the
pressure exerted by the molecules. Another condition is that
the sum of the individual volumes occupied by the molecules
should be negligible compared to the total volume occupied by
the gas. This means that at low temperatures and high
pressures, the molecules are so near one another that the
force of repulsion or attraction between molecules may
increase. Moreover, the volume occupied by the molecules of
the gas are now substantial. Therefore the volume to be
predicted by the ideal gas cannot be obtained.
PROOF OF THE IDEAL GAS LAW
We can easily show the validity of the ideal gas law experimentally. We could set up an apparatus to observe the variation of
volume and pressure with temperature. The apparatus
consists of a cylinder filled with one mole of a gas and fitted
with a piston of a known area on which a known force could
be applied. We can control the temperature of the gas by
providing a constant temperature bath around the cylinder.
At a constant temperature, we add known masses on the
piston and we observe the volumes at different pressures. We
repeat this for different temperatures. At every temperature,
we calculate the product PV. We then plot PV versus the
68
National Engineering Center CEEC Seminar, June 2 -4, 2004
absolute temperature T on a rectangular coordinate system.
The plot must yield a straight line at low pressures.
PV
slope = m
extrapolated
T, K
Fig. 2-6
The equation of the straight line in Fig. 2-6 is
DRAFT
PV = mT + b
Eq. 2-35
where
m = slope
b = intercept at T = 0
If we set b as zero,
PV = mT
Eq. 2-36
and since we used one mol of the gas,
PV = m = constant = R
T
Eq. 2-37
If the gas is ideal, PV varies linearly with T and when
extrapolated to zero, yields a value equal to 0. (Note that in
actual cases, PV at T = 0 cannot be zero).
If the plot is not a straight line and if the extrapolation does
not lead to T = 0 K, then the gas is not ideal. Most chemistry
books include derivation of the ideal gas law by employing the
kinetic theory of gases.
CALCULATIONS INVOLVING THE IDEAL GAS LAW
The ideal gas law is useful in solving gas problems involving P, V, T, and n. With the value of R obtained from tables, we
can solve for any unknown variable. When using the ideal gas
law it becomes essential to memorize the values of R or to refer
to the table.
Essential Basic Principles 69
Example 2-21
What is the volume in m3 occupied by 1 kmol of an ideal
gas at a temperature of 0oC and a pressure of 101.325 kPa (1
atm)?
Solution:
The appropriate value of the ideal gas constant R to use is
8.314
kPa $ m3 .
kJ
or 8.314
kmol $ K
kmol $ K
n = 1 kmol
V = nRT
P =
T = O + 273 = 273 K
P = 101.325 kPa
(1)(8.314)(273)
= 22.4 m 3
101.325
DRAFT
We do not have to memorize all the values of R. For a given
set of conditions of P, V, and T for one mole of an ideal gas, we
implicitly have the value of R. The "standard conditions" or
"standard temperature and pressure" (STP) are particularly
useful. At the "standard conditions" the temperature (To) is
equal to 273 K (0oC) or 492oR (32oF) while the standard
pressure (Po) is 1 atm or 101.325 kPa or 14.7 psia. At these
conditions, the volume (Vo) occupied by 1 kmol of a gas is 22.4
m3; by 1 mol, 22.4 L and by 1 lbmol, 359 ft3. From these, we
can obtain the values of R for any combination of units. For
instance, if P is in inch Hg, T in oR, V in ft3, and n in lbmol, the
value of R is
R=
(in Hg) $ (ft)
(29.92 in Hg)(359 ft 3 /#mol)
PoVo
=
= 21.83
To
(32 + 460) o R
(#mol)( o R)
USING THE IDEAL GAS LAW INDIRECTLY
By knowing that 1 kmol of a gas occupies 22.4 m3 or 1 #mol
occupies 359 ft3 at STP (standard temperature and pressure)
and by applying Boyle's law (P = k1/V at constant T) and
Charles' law (V = k2T at constant P), we can apply the ideal gas
law indirectly.
For example, let us calculate the volume occupied by 20 kg
of nitrogen at 400oC and 1.5 atm without using the ideal gas
formula.
The procedure is as follows: Twenty kg of nitrogen is 20/28
kmol (The molecular weight of N2 is 28). This amount of gas
occupies (20/28)(22.4) m3 of volume at 0oC and 1 atm (STP).
We can change the condition of the gas to 400 oC and 1.5 in
two steps. By using Charles law (the volume increases proportionately with the absolute temperature at constant pressure)
we know that the volume at 400oC (with the pressure
maintained at 1.5 atm is greater than that at 0oC by a factor
equal to (400 + 273)/(0 + 273). From Boyle's law (the volume
varies inversely with the pressure at constant temperature)
that the volume at 1.5 atm is smaller than that at 1 atm by a
70
National Engineering Center CEEC Seminar, June 2 -4, 2004
factor of 1/1.5 (with the temperature maintained at 400oC).
We can write the solution in one line:
3
+ 273 % 1
V = 20 kg % 1kmol % 22.4 m % 400
0 + 273
1.5
kmol
28 kg
↑
↑
volume at STP
Temperature
correction
DRAFT
↑
Pressure
correction
V = 26.30 m3
The value of the temperature correction factor is greater
than 1 since the volume must increase when the temperature
is raised at constant pressure. Hence the correction factor is
(400 + 273)/(0 + 273). The pressure correction is less than 1
(1/1.5) since the volume decreases when the pressure
increases at constant temperature. Fig. 2-7 is an effective way
of remembering this. The method is very convenient in chemical engineering calculations. This simple procedure is a calculation peculiar to chemical engineering, which all students of
chemical engineering should be familiar with. (While the use of
SI diminishes the application of the method, we use this for
instructional purposes.)
Thus, in the procedure above, we "converted" the "mass" of a
gas to "volume" without using a value of R, the universal gas
constant
What we should remember is that at 0°C (32°F, 273K, or
492°R) and 1 atm (14.7 psi, 101.325 kPa, 29.92 in Hg, 760
mm Hg, etc.), the volume of an ideal gas is equal to 22.4 m3 (if
the quantity is 1 kmol), 359 ft3 (if the quantity is one lb-mol),
or 22.4 L (if the quantity is one mol). By using Boyle's Law and
Charles Law, we can "convert" the volume to other temperatures and pressure.
Now, practice this procedure until this becomes of second
nature to you. Notice the examples below how it becomes a
drill for you in order to remember the procedure:
1 mol of gas at 0°C and 1 atm occupies a volume of 22.4 L.
Cooling contracts
the gas volume in a
flexible closed
container; the
correction factor is
less than 1.
Heating expands the
gas volume in a
flexible closed
container; the
correction factor is
greater than 1.
Force
gas
Increasing the
pressure
decreases the gas
volume; the
correction factor
is less than 1.
Pull
gas
Decreasing the pressure
increases the gas volume;
the correction factor is
greater than 1..
Fig. 2-7
Essential Basic Principles 71
44 g of CO2 gas at 0°C and 1 atm occupies a volume of 22.4 L.
44 g of CO2 gas at 273°C and 0.5 atm occupies a volume of
22.4 L.
22 g of CO2 gas at 273°C and 1 atm occupies a volume of 22.4
L.
1 kmol of gas at 0°C and 1 atm occupies a volume of 22.4 m3.
28 kg of N2 gas at 0°C and 1 atm occupies a volume of 22.4
m3.
14 kg of N2 gas at 273K and 1 atm occupies a volume of 22.4
m3.
1 lb-mol of a gas at 32°F and 14.7 psi occupies a volume of
359 ft3.
44 lb of CO2 gas at 32°F and 14.7 psi occupies a volume of
359 ft3.
DRAFT
What is the relevance of doing this? Notice that by performing this drill, we are at once ingraining into our mind several
important principles. This is how the mind of a chemical
engineer works. By remembering the basic principles and a set
of procedure, we can perform calculations in a manner we
understand better.
Example 2-22
What is the mass rate (kg/h) of air flowing at 17 m3/min at
200oC and 122 kPa?
Solution:
The volume at actual conditions is first converted to the
volume at STP and the corresponding mass is evaluated.
3
29 kg 60 min
%
Mass rate = 17 m % 273 + 0 % 122 % 1kmol3 %
kmol
min 273 + 200 101.325 22.4 m
h
less than 1 since ______↑
the temperature
decreased from
200°C to 0°C
↑




greater than 1 since the pressure
Decreased from 122 kPa to 101.325 kPa
Mass rate = 917.69 kg/h
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National Engineering Center CEEC Seminar, June 2 -4, 2004
IDEAL GAS MIXTURES
The ideal gas law applies to any type of gas. Hence, a mixture
consisting of different molecules can also behave like an ideal
gas if it follows the requirements for ideality. Molecules of
different structures should not exert forces of attraction or
repulsion against one another. Although difficult to satisfy,
some gaseous mixtures do follow the ideal gas law. Two laws
are particularly useful for gas mixtures: Dalton's law of partial
pressures and Amagat's law of partial volumes.
DALTON'S LAW OF PARTIAL PRESSURES
This law states that the total pressure of a gas mixture is
equal to the sum of the individual pressures (partial
pressures) exerted by the components at the temperature and
total volume of the mixture.
If a mixture consists of components a, b, c, ..., n, then the
PT = Pa + Pb + Pc + ... + Pn
DRAFT
where
Eq. 2-37
Pa, Pb, Pc, ..., Pn are the partial pressures of a, b, c, ...,
n, respectively.
PT = total pressure
AMAGAT'S LAW OF PARTIAL VOLUMES
This law states that the total volume of a gas mixture at a
fixed temperature and pressure is equal to the sum of the
volumes (partial volumes) occupied by each component (if it
were possible to separate the gases) at the same temperature
and pressure.
For a mixture consisting of a, b, c, ..., n components,
V T = V a + V b + V c + ... + V n
where
Eq. 2-38
Va, Vb, Vc, ..., Vn are the partial volumes of components a, b, c, ..., n, respectively.
VT = total volume
USEFUL RELATIONS FOR IDEAL GAS MIXTURES
Consider a gas mixture consisting of n components and
occupying a total volume V, at a temperature T. Let na, nb, nc,
..., nn be the number of moles of each component. If the gas is
ideal, then we calculate for the partial pressure of each
component as follows:
Pa =
n aRT
V
Eq. 2-39
Pb =
nb RT
V
Eq. 2-40
Essential Basic Principles 73
Pb =
:
:
Pi =
nb RT
V
Eq. 2-41
n i RT
V
Eq. 2-42
Adding all the equations,
P a + P b + P c + ... + P i = (n a + n b + n c + ... + n i ) RT
V
Eq. 2-43
Using Dalton’s law
P T = P a + P b + P c + ... + P i
with
n T = n a + n b + n c + ...n n , then
PT =
n T RT
V
Eq. 2-44
DRAFT
Dividing Eq. 2-39 by Eq. 2-44, Eq. 2-40 by 2-44, Eq. 2-41 by
Eq. 2-44, and so on,
Pa na
=
= ya
PT nT
Pb nb
PT = nT = yb
Pc
nc
PT = nT = yc
:
:
n
Pi
= i = yi
PT nT
where yi = mole fraction of component i. Now dividing Eq.
2-39 by Eq. 2-40, Eq. 2-40 by Eq. 2-41, or taking ratios of any
2 equations from Eq. 2-39 to Eq. 2-42,
Pa na ya
=
=
Pb nb yb
Pa na ya
= nc = yc
Pc
yc
n
Pc
= n bc = y b
Pb
and so on.
74
National Engineering Center CEEC Seminar, June 2 -4, 2004
Therefore, for an ideal gas, the partial pressure fraction is
equal to the mole fraction while the partial pressure ratio is equal
to the mole ratio.
We can also make a similar analysis for a gas mixture
consisting of a, b, c,..., i components at a total pressure P and
at a temperature T. If na, nb, nc,...,ni are the number of moles of
each component and if the gas mixture is ideal, then the
partial volume of each component is
Va =
n a RT
P
2-45
Vb =
n b RT
P
2-46
n RT
Vc = c
P
:
:
n RT
V i = iP
2-47
2-48
DRAFT
Adding all the equations,
V a + V b + V c + ... + V i = (n a + n b + n c + ... + n i ) RT
P
2-49
By Amagat's law, VT = Va + Vb + Vc + ... Vn
n T = n a + n b + n c + ...n n ,
then
VT =
n T RT
P
2-50
Repeating the same operations done in the previous section
Va na
= nT = ya
VT
Vb
nb
VT = nT = yb
nc
Vc
VT = nT = yc
:
:
n
Vi
= n Ti = y i
VT
likewise
Essential Basic Principles 75
Va na ya
=
=
Vb nb yb
Va na ya
= nc = yc
Vc
yc
Vc
n
= n bc = y b
Vb
and so on. For an ideal gas, therefore, the volume fraction is
equal to the mole fraction while the volume ratio is equal to the
mole ratio.
The above relations are useful in engineering calculations.
Since most of the gases handled are mixtures that behave
ideally, the above relations simplify calculations and data
presentations. We commonly represent gas compositions by
volume percent. From the above equations, it is also equivalent to mole percent and to partial pressure percent. For gas
mixtures, the composition is understood to be in mole percent
(volume or pressure percents) unless otherwise specified.
Example 2-23
DRAFT
Combustion gases contain CO2, CO, N2, O2, and H2O vapor. It
is common practice to report the composition in volume per
cent on a dry-basis (water-free) and to indicate the water
content by its partial pressure.
A typical report of an analysis is as follows:
12% CO2, 3% CO, 7% O2, and 78% N2. (all % by volume)
PH2O = 1.33 kPa
Total pressure = 101.06 kPa
What is the composition on a wet basis (including H2O)?
What is the partial pressure of each component?
Solution:
Combustion gases are often at high temperature and low
pressures and so can be considered as ideal gas mixtures.
The composition given is by volume and is also equal to mole
per cent and pressure per cent.
Basis: 100 mols of water-free gas (dry gas)
mols CO2
mols CO
mols O2
mols N2
=
=
=
=
0.12 (100) = 12
0.03 (100) = 3
0.07 (100) = 7
0.78 (100) = 78
The total pressure is 101.06 kPa and the partial pressure of
H2O is 1.33 kPa.
The partial pressure of the combustion gases is, therefore,
101.06 - 1.33 = 99.73 kPa. The number of moles of H20 in the
gas is
100( 1.33 ) = 1.33 mols
99.73
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National Engineering Center CEEC Seminar, June 2 -4, 2004
1.33 mol H 2 O
1.33 kPa is equivalent to
99.73 mol dry gas
99.73 kPa
The composition on a wet basis is:
12
(100) = 11.84%
100 + 1.33
3
% CO =
(100) = 2.96%
100 + 1.33
7
% O2 =
(100) = 6.91%
100 + 1.33
78
% N2 =
(100) = 76.98%
100 + 1.33
% CO 2 =
DRAFT
Since the partial pressure is equal to the mole fraction of
the component × total pressure,
12
(101.06) = 11.97 kPa
P CO 2 =
100 + 1.33
3
P CO =
(101.06) = 2.99 kPa
100 + 1.33
7
P O2 =
(101.06) = 6.98 kPa
100 + 1.33
78
(101.06) = 77.79 kPa
P N2 =
100 + 1.33
Example 2-24
Producer gas is a fuel that is manufactured from the
reaction of coal, steam, and air. A sample has the following
composition by volume on a dry basis: (Dry basis means the
water is not included in calculating the composition.)
CO
CO2
H2
N2
O2
CH4
%
20.35
11.03
15
51.43
1.98
0.21
The gas is flowing at 6,000 cu feet per minute (cfm). It has a
temperature of 200°C, a pressure of 21 cm H20 head and a
partial pressure of water equal to 50 mm Hg. The atmospheric
pressure is 758 mm Hg. The gas is delivered to the furnace at
50°C and 2 atm pressure. At this condition, calculate
a. the volumetric flow rate
b. the mass flow rate
c. the partial pressure of each component
(including H20)
d. the concentration of each component.
Assume ideal gas behavior.
Solution:
Given:
Essential Basic Principles 77
20.35% C0
11.03% CO2
15.00% H2
51.43% N2
1.98% O2
0.21% CH4
6000 cfm
T = 200°C
PT = 21 cm H20 head
Patm = 758 mm Hg
Partial pressure of H2O = 50 mm Hg
Basis of solution:
6,000 cu ft producer gas at 200°C and 21 cm H2O head.
(a) For the volumetric flow rate at 50°C and 2 atm.
The gas is originally at a temperature of 200°C.
The total pressure =
758 mm Hg + 21 cm H 2 O
760 mm Hg
1 ft H 2 O
%
(12)(2.54) cm H 2 O 33.9 ft H 2 O
= 773.4 mm Hg
The volumetric flowrate =
DRAFT
6, 000 cfm % 273 + 50 % 773.4 = 2085 cfm
273 + 200 2(760)
Here, we assume that the mass of the gas remains
unchanged, i.e., nothing is lost due to leakage.
(b) For the mass flow rate (lbm/min)
Considering a time interval of 1 minute, the mass corresponding to 6000 cfm can be calculated. The 6000 cfm of gas
includes the water vapor.
lbmol wet gas = 6, 000 ft 3 % 0 + 273 % 773.45 % 1 lbmol
200 + 273
760
359 ft 3
= 9.82 lbmol
lbmol dry gas = (9.82) 773.45 − 50 = 9.18 lbmol
773.45
lbmol H2 O = (9.82)
50 = 0.64 lbmol
773.45
By an overall mass balance the above amounts also correspond to the cooled gases.
Components
CO
CO2
H2
N2
O2
CH4
H2 0
mol %
20.35
11.03
15
51.43
1.98
0.21
#mol
1.87
1.01
1.38
4.72
0.18
0.02
0.64
MW
28
44
2
28
32
16
18
Total
Mass
52.36
44.44
2.76
132.16
5.76
0.32
11.52
249.32
The mass flow rate is 249.32 lbm/min.
(c) For the partial pressure of each component.
78
National Engineering Center CEEC Seminar, June 2 -4, 2004
Partial pressure of water = 0.64
(2)(760) = 99.82 mm Hg,
9.82
Partial pressure of dry gas = 9.18 (2)(760) = 1, 421 mm Hg
9.82
For the other components,
PCO
PCO2
PH2
PN2
PO2
PCH4
= (0.2035)(1421) = 289.16 mm Hg
= (0.1103)(1421) = 156.73 mm Hg
= (0.1500)(1421) = 213.14 mm Hg
= (0.5143)(1421) = 730.79 mm Hg
= (0.0198)(1421) = 28.13 mm Hg
= (0.0021)(1421) = 2.98 mm Hg
DRAFT
(d) For the concentration of each component (lbm/ft3)
C co = 52.36 = 0.025 lbm/ft 3
2085
C CO2 = 44.44 = 0.021 lbm/ft 3
2085
2.76
= 0.001lbm/ft 3
C H2 =
2085
C N2 = 132.16 = 0.063 lbm/ft 3
2085
5.76
= 0.003 lbm/ft 3
C O2 =
2085
C CH4 = 0.32 = 0.0001 lbm/ft 3
2085
C H 2O = 11.52 = 0.006 lbm/ft 3
2085
REAL GASES
In actual industrial processes, we often encounter deviations
from ideal gas behavior. Conditions of low temperature and
high pressure do not allow assumptions that gas molecules
have negligible volume and do not interact with each other in
any way. P-V-T relationships like those for ideal gases are
called equations of state. To represent actual gas behavior,
several correlations have been formulated.
Van der Waals Equation of State
Van der Waals developed one of the first equations of state:
P=
nRT − n 2 a
V2
V − nb
for n kmols
2-51
P=
RT − a
v − b v2
for 1 kmol
2-52
v denotes the volume for 1 kmol of the gas. a and b are empirical constants (called Van der Waalsconstants) and depend on
the critical temperature and critical pressure of a gas. The
term a2 accounts for the interaction among the molecules
v
Essential Basic Principles 79
while b is the correction term for the volume occupied by the
molecules.
The values of the constants a and b are:
a =
27R 2 T 2c
64 P c
b =
RT c P c
8
Tc is the critical temperature and Pc is the critical pressure.
Refer to Appendix 1 for the values for some gases.
The usefulness of Van der Waals' equation is quite limited. It
may not be applicable for some conditions of pressure and
temperature. However, we can use other equations of state.
Redlich-Kwong Equation of State
The Redlich-Kwong equation of state performs better than
the Van der Waals equation.
P = RT − 0.5 a
v − b T v(v + b)
DRAFT
where
a= 0.42748
R 2 T 2.5
c
Pc
b= 0.08664
RT c
Pc
2-53
Tc = critical temperature
Pc = critical pressure.
The Redlich-Kwong equation of state is quite robust for
most conditions.
The Benedict-Webb-Rubin (BWR) Equation of State
The BWR is a more accurate equation of state developed
with parameters gotten from comprehensive curvefits to actual
gas behavior. It contains eight empirical constants.
P = RT
v +
RTB o − A o − C o /T 2 RTb − a a
2
+
+ 6 + 3c 2 (1 + 2 )e −/v 2-54
V
V3
v T
v
v2
The constants are available in the literature. In general, the
more robust an equation of state is, the more complicated the
equation becomes. It will contain too many constants.
The Compressibility Factor
The compressibility factor z is a simple way of accounting for
the non-ideality of a gas and is defined as:
2-55
z = PV
nRT
80
National Engineering Center CEEC Seminar, June 2 -4, 2004
The equation becomes the ideal gas law when z is equal to 1.
It may be less or greater than unity depending on the pressure
and temperature. z is a function of the reduced pressure, PR
and reduced temperature. The graph is available in many
books on thermodynamics.
TR = T
TC
PR = P
PC
Example 2-25
Calculate the pressure of 100 kg of methane, CH4, at -40oC
and at a volume of 10 m3 using
a. the ideal gas law
b. the Van der Waals' equation of state
c. the Redlich-Kwong equation of state
Solution:
a. Using the ideal gas law,
DRAFT
mass = 100 kg or 100 = 6.25 kmol
16
T = -40 + 273 = 233K
V = 10 m3
R=
3
(22.4 m 3 )(101.325)
= 8.314 m $ kPa
(1 kmol)(273K)
kmol $ K
P = nRT
V =
(6.25)(8.314)(233)
= 1211 kPa
10
b. Using the Van der Waals' equation of state,
P=
RT − a
v − b v2
From Appendix1, for methane, Tc = -82.5°C and Pc = 45.8 atm
P c = 45.8 % 101.325 kPa = 4641 kPa
1 atm
a =
27R 2 T 2c
= 228.04
64 P c
b =
3
RT c P c
= 0.04266 m
8
kmol
3
R = 8.314 m $ kPa
kmol $ K
v = 10/6.25 = 1.6 m3/kmol
(8.314)(233)
P = RT − va2 =
− 228.04
(1.6 − 0.04266)
(1.6) 2
v−b
Essential Basic Principles 81
P = 1155 kPa
c. Using the Redlich-Kwong equation of state
P = RT − 0.5 a
v − b T v(v + b)
a = 0.42748
(8.314) 2 (190.6) 2.5
R 2 T 2.5
c
= 0.42748
= 3193
Pc
4641
b = 0.08664
8.314(190.6)
RT c
= 0.08664
= 0.02958
Pc
4641
v = 10/6.25 = 1.6
8.314(233)
3193
=
−
P = RT − 0.5 a
v − b T v(v + b) 1.6 − 0.02958 (233) 0.5 1.6(1.6 + 0.02958)
P = 1153 kPa
LIQUIDS
DRAFT
For gas systems, the ideal gas law is good model. However, no
suitable correlation for an "ideal or perfect" liquid is available.
Liquid volumes are more easily measurable than those of
gases and this accounts for the voluminous data available in
the literature. A good reference is the "Handbook of Chemistry
and Physics". Several density-temperature relationships for
some liquids in the form of graphs and nomographs are available. Reid and Sherwood's "The Properties of Gases and
Liquids" (McGraw-Hill, 1966) is an excellent reference for
liquid properties. Most liquid properties are function of critical
property or normal boiling point.
The properties of mixtures are also experimentally determined and data are often tabulated. Some gases and solids
can be dissolved in liquids and gas-liquid and solid-liquid
mixtures and solutions can exist. Some of these are discussed
in Chapter 7.
SOLIDS
Majority of the properties of solids are available in the literature either in tabular or graphical form. These include melting
point, specific gravity, expansion due to temperature, specific
heat, and molecular arrangement, among others. Solid
solutions exist. These are homogeneous solutions and metal
alloys are examples.
PROBLEMS
82
National Engineering Center CEEC Seminar, June 2 -4, 2004
2-1 Convert the following quantities to the units indicated: (Use the
unit-ratio method.)
a. Viscosity of 8 centipoise to pascal-second, Pa⋅s. 1 poise = 1
g/(cm⋅s)
b. 150 ft-lbforce to joule
c. 1000 dynes to Newton
d. 105 barrels to m3 (1 barrel = 42 gal)
e. 55 miles per hour to meters per second
f. 34 pints U.S. liq. to m3
g. 21,000 in3 to m3
h. 900 fluid ounces to m3
i. 2100 lbm/ft3 to kg/m3
j. 7150 Btu/second to watt
2-2 Express the following quantities in standard SI units: (Follow the
rules given in the text.)
a. A woman's vital statistics of 36"-25"-37"
b. The distance of the closest galaxy to our own, the Megallanic
c.
d.
DRAFT
e.
f.
g.
h.
i.
clouds -- 200,000 light years
A plane's velocity of 400 miles per hour
The force exerted by gravity on a 10,000 lbm of vehicle on the
moon
A diffusivity of 3.71 ft2/s
A flow rate of 100,000 cu ft per minute
A mass velocity of 568 lbm/h⋅ft2
Your height and weight
A speed of Mach 3
Mach =
speed in consideration
speed of sound
speed of sound = 1083 ft/second)
j. A radio frequency of 100,300,000 cycles per second
k. A corrosion rate of 0.04 inch per year
l. A viscosity of 2.4 × 10-5 lbm/(ft.s)
m. The mass of one atom of carbon-12 (12 grams of carbon-12
contains 6.02 × 1023 molecules)
n. A fuel mileage of 19 miles per gallon
o. An oil rate of 2,400 bbl per hour.
2-3 In the following equation, what is the appropriate unit of H in the
SI? in the English system?
dH = V − T dV
dT
dP
where
2-4
H: is enthalpy
P: pressure, F/L2
V: volume, L3
T: temperature, T
Find out which of the following formulas is correct:
ØC v
ØV
2
T
= T Ø P2
ØT V
or
T
ØC v
ØV
T
2
= Ø P2
ØT V
Essential Basic Principles 83
2-5
C, temperature, θ
P, pressure, F/L2
V, specific volume, L3/M
Cv, heat capacity, FL/MT
Find the value of the following dimensionless groups for the given
data:
a. The Reynolds No., N Re =
Dv!
where D = diameter, L
v = velocity L/θ
ρ = density, M/L3
µ = viscosity, M/Lθ
for a case where D = 16 in, v = 6 m/s, ρ = 4.0 kg/gal and
#
µ = 300
ft $ hr
b. The Schmidt No., N Sc =
!D v
where
µ = viscosity, M/L
ρ = density, M/L3
Dv = diffusivity, L2/θ
DRAFT
for a case where µ = 12 centipoise, ρ = 30 kg/cu ft, Dv = 745 cm2/hr.
2
c. The Froude No., N Fr = v
Lg
where
v = velocity, L/θ
L = length, L
g = acceleration due to gravity
for a case where L = 6 ft, v = 20 cm/s
2-6 The diffusion coefficient DAB for the interdiffusion of two gases, A
and B may be estimated from the following empirical relationship
(E.R. Gilliland, Ind. Eng. Chem., 26, 681 (1934)
D AB = 0.043
T 2/3
1
3
1
P(V A + V B3 ) 2
1 % 1
MA MB
where
DAB = diffusion coefficient, cm2/s
MA, MB = molecular weights of A and B, respectively
P = total pressure, atm
VA, VB = molecular volumes of A and B, respectively
Find the value of the constant corresponding to 0.0043, for an
empirical formula which gives DAB in ft2/hr, and uses P in psi and T
in °R.
ans.: 0.16553
2-7 The viscosity of a liquid at the critical point is given by formula
c = 61.6
where
84
MT c
(V c ) 2/3
µ = critical viscosity in micropoise
M = molecular weight
National Engineering Center CEEC Seminar, June 2 -4, 2004
Tc = critical temperature, K
Vc = critical volume, cm3/mol
For a similar formula using µc in Pa⋅s, and Vc in m3/kmol, and Tc in
K, find the value of the constant corresponding to 61.6.
2-8 The vapor pressure of n-pentane is represented by the Antoine
equation
log 10 P = A -
B
+t
where
A = 6.85221
B = 1916.33
= 385.60
P = vapor pressure, mm Hg
t = temperature °F
Determine the new coefficients A', B', and θ' of the equation for P to
be given in kPa and t in °C.
Conversion factor
DRAFT
Ans: A’ = 5.9771
t o C= 5 (to F−32)
9
B’ = 1064.6
θ = 232.0
2-9 Given the following formula,
h=
0.0682"3
!0.4
where
h, height in ft
µ, viscosity, centipoise
σ, surface tension, dynes/cm
ρ, density, #/ft3
Find the appropriate value of the constant α' in the equation
h∏ =
∏ ( ∏ ) 2 (" ∏ ) 3
(! ∏ ) 0.4
where
h' is in cm
µ', #m/ft⋅hr
σ', N/m
ρ', g/cm3
2-10 The temperature of a cold water being heated varies linearly with
the distance from one end of the heat exchanger. From the following data find the suitable equation representing the variation
graphically:
4.44
5.56
11.67
19.44
25.00
32.22
Temperature, °C
Distance, m
0
0.24
1.13
2.07
2.99
4.11
Ans.: Temp = 6.896 + 4.275
2-11 The following table gives the relationship between dissolved
methane in a solution and the partial pressure of methane above
the solution. (Chem. Eng. Prog. symp. series, Volume 65, 73
(1969).
Dissolved CH4, ppm
Partial pressure of CH4, kPa
20
138
115
689
225 340 450 560
1379 2068 2758 3447
Essential Basic Principles 85
If the variation is linear, find the suitable relationship.
2-12 The following data follows the equation y = a
x + b. Find the value of
a and b graphically.
2.5
3.8
4.8
y
1.43
2.00
3.33
x
Ans: a = -6.06
b = 6.75
2-13 The following set of data:
6.2
10.00
6.1
6.4
6.9
7.35
8.0
8.45
y
x
1.61
2.19
2.61
2.93
3.44
3.69
is known to follow the relation y = ax2 + b. Find the value of a and b
graphically.
2-14 The following data when plotted on a semi-log graph yield a
straight line:
12.6
6.75
3.86
y
1.25
2.5
3.8
x
Find the equation graphically.
1.95
5.26
2-15 From the following data (Wynkoop and Wilheim, Chem. Eng'g.
Prog. 46, 300 (1950)) find the value of A and E/R in the equation:
DRAFT
E
−
k p = A(e) RT
kp × 106, mol/s⋅cm3⋅atm 33.1
353.0
T, K
28.2
348.0
13.9
337.0
7.0
327.0
3.0
312.5
2-16 The following experimental data from the calibration of a venturimeter is expected to yield a straight line when plotted in a log-log
scale: (Zimmerman and Lavine, Chem. Eng. Lab. Equip., Industrial
Research Service, Inc., New Hampshire, 1955, p. 42)
Velocity, ft/s
Reading, in Hg
42.25 38.5 36.63 34.38 29.63 17.13 15.25 9.73
5.33 5.06 4.92 4.85 4.47 3.41 3.27 2.6
4.25
1.8
2-17 The following table gives experimental data on the absorbance as a
function of time for the reaction of ethylene oxide with n-acetylenemethionine in aqueous solution (Powers and Corcoran, Ind. Eng.
Chem. Fundam., Vol. 16, No. 1, 1977, p. 160):
Time, s
0
5,000 9,000 13,000 22,000 30,000 39,100
Net absorbance
1.430 1.140 0.905 0.719 0.447 0.217 0.108
Find an adequate equation representing the data.
2-18 Convert the following temperature readings
a. 180K to °C, °R, and °F
b. 800°F to K, °C, and °R
c. 100,000°C to K, °R, and °F.
2-19 Convert a thermal conductivity of
Btu
to s $ KJ$ m
hr $ o F $ ft
2-20 What is the specific gravity of 65°Be sulfuric acid at 25°C?
145
Sp. Gr. 60/60 =
145 − o Bé
where
Sp.Gr. 60/60 is the specific gravity at 60°F referred to density of
water at 60°F
0.003
86
National Engineering Center CEEC Seminar, June 2 -4, 2004
°Bé is degrees Baumé
2-21 A hot air balloon makes use of the lower density of the hot flue gas
and air to provide lifting power. If a spherical balloon 10 m in
diameter uses hot gas with a specific gravity of 0.70 referred to air
at 0°C and 760 mm Hg (air density = 1.2929 g/liter), what mass, in
kilograms, can the balloon lift if the surrounding air has a specific
gravity of 0.92 (same reference)?
2-22 The pressure gage of a vacuum oven reads 28.2" Hg vacuum. If the
barometric pressure is 751 mm Hg, what is the absolute pressure
in kPa? in psi (lb/in2)? in psf (lb/ft2)?
2-23 Express the following absolute pressures in terms of vacuum or
gage pressures if the barometric pressure is 753 mm Hg.
a.
b.
c.
d.
65 ft of water
3.07× 106 Pa
1500 cm of carbon tetrachloride
7 cm of air, ρ = 1.2928 g/L
2-24 A brine solution has a specific gravity 20°C/4°C of 1.149. How high
is a column of this fluid equivalent to 150 psig? Barometric
pressure = 14.7 psia.
2-25 What is the equivalent of 7 cm of mercury (sp. gr. = 13.6) in meters
DRAFT
of air (density = 1.29 g/L)?
2-26 Convert the following gage pressures to absolute values if the
barometric pressure is 757 mm Hg:
a. 200 psig
b. 4 cm of Hg vacuum
c. 29 mm of H20 head
d. 15 mm of H20 draft
2-27 Convert the following pressures to the indicated absolute
pressures:
a. 20.0 psig to atm abs with barometric pressure = 740 mm Hg
b. 28" Hg vacuum to mm Hg absolute with barometric pressure =
760 mm Hg
c. 4" H2O draft to mm Hg abs with barometric pressure = 760 mm
Hg
d. 5" H2O head to mm Hg abs with barometric pressure = 750 mm
Hg.
2-28 A flue gas has the following composition on a molar basis:
77% N2
12% CO2
2% CO
2% H2O
7% O2
What is the composition in mass percent? What is the average
molecular weight of the mixture?
2-29 Some ionic components of sea water are given in the following
table:
Cation
Na+l
Mg+2
Ca+2
K+1
Sr+2
mass %
1.06
0.13
0.04
0.04
0.01
Anion
Cl-1
SO4-2
HCO3-1
Br-1
F-1
mass %
1.90
0.26
0.01
0.0065
10.0001
Essential Basic Principles 87
If the specific gravity of sea water is 1.03, how much NaC1 could be
recovered from a cubic km of sea water? How much Mg(OH)2?
2-30 The average composition of dry air is as follows:
Gas
Volume %
Nitrogen, N2
78.09
Oxygen, 02
20.05
Argon, Ar
0.93
Carbon Dioxide, CO2 0.03
Neon, Ne
0.002
Helium, He
0.0005
Methane, CH4
0.0002
Others
0.8973
How much argon could be recovered from a cubic kilometer of air?
Density of air = 2.2928 g/l.
2-31 Steel, submerged in running water, corrodes at the rate of 1 inch
per year. If the area of the steel surface is 10 m2 and if running
water at a volumetric rate of 104 m3/hr flows passes by it, what is
the concentration of iron in the water in ppb?
2-32 In a process plant, benzene is evaporating and is leaking into an
air stream at the rate of 0.1 kmol per hour. If the air flow rate is
8000 kg/min, what is the concentration of benzene in the air in
ppm?
DRAFT
2-33 5.7 kg of sucrose is dissolved in 18 liters of water. What is the
sugar concentration of the solution formed in
a.
b.
c.
d.
mass fraction?
mole fraction?
molality?
molarity?
2-34 What is the molarity of a 35% HCl solution if the specific gravity
25°/4° is 1.48?
2-35 Ethane is flowing through a 1-m I.D. pipe at the rate of 6530 kg/h.
If the density of ethane is 1.3567 g/L what is
a
b
c
the volumetric flow rate?
the mass velocity?
the flow velocity?
2-36 A lube oil (sp. gr. = 0.85) is pumped to a header at the rate of 4,000
liters/hr. At the header, the flow branches in two lines. One pipe
has an inside diameter of 7 cm while the other, 15 cm. Assuming
that the flow is directly proportional to the cross sectional area of
flow, calculate for both pipes
a. the flow velocity in m/s
b. the mass velocity in kg/hr·m2
c. the volumetric flow rate in m3/hr.
A
10 cm dia
88
B
5 cm dia
C
15 cm dia
National Engineering Center CEEC Seminar, June 2 -4, 2004
2-37 Air flows through the piping system shown above. If the flow velocity at A is 10 m/s, what is the flow velocity at B and at C?
2-38 Find the volume occupied by 20 kg of C2H6 at a pressure of 2 atm
and at a temperature of 50°C.
2-39 Butane gas is stored in a cylinder at a pressure of 100 psig and at
a temperature of 30°C. Some of the butane gas was used and after
some time, the pressure had gone down to 50 psig. If the temperature is 25°C, what fraction of the butane had been used?
2-40 Hydrogen is stored in a 2 cubic feet cylinder at a pressure of 130
atm and at a temperature of 25°C. What volume can the cylinder
supply at 1 atm and 25°C?
2-41 An airship has a volume of 60824 m3. If the conditions of the gas in
this volume is 100 kPa and 20°C, how many kilograms of helium is
needed?
2-42 A tank contains hydrogen at 7 atm and 35°C. When 1.5 kg of the
gas is removed, the pressure goes down to 6 atm and the temperature to 25°C. What is the volume of the tank?
DRAFT
2-43 Two tanks are connected by pipes provided with a valve. The valve
is initially closed. The first tank, with a volume of 1 m3 contains N2
at 6.8 atm and 60°F. The second tank contains oxygen at 2 atm
and 15°C. Its volume is 0.4 m3. If the valve is left opened and the
final temperature of both tanks becomes 27°C, what is the
pressure inside the two tanks?
2-44 A natural gas has the following analysis: 90% CH4, 3% C2H6, 0.5%
O2, 0.2% CO2, and 6.3% N2. Calculate the density in kg/m3 at 30°C
and at 761 mm Hg pressure.
2-45 The flue gas from a furnace goes out with the following composition
(dry basis): 12% CO2, 3% O2 2% CO, and 83% N2. The partial
pressure of water is 10 mm Hg. The total pressure is 4 cm H20
draft. The barometric pressure is 759 mm Hg. Calculate the
composition on a wet basis (including H20).
2-46 Air contains 79% N2 and 21% 02. Calculate the density of moist air
at 760 mm Hg and at 40°C if the partial pressure of water is equal
to 12 mm Hg. Find the kg of H20 present per kg of dry air.
2-47 A gas mixture contains 20% C2H6, 70% CO2, 1.3% 02, and 8.7% N2.
What is the partial pressure of each component if the total
pressure is 30 atm and the temperature is 35°C?
2-48 Inert gas containing 90% N2 and 10% CO2 is bubbled through
benzene liquid and issues out with a partial pressure of benzene
equal to 30 mm Hg. The gas is at 30°C and at 31" Hg total
pressure. The meter indicates a flow rate of 20 liters per minute.
What is the corresponding flow rate of the benzene-free gas at a
temperature at 25°C and at a pressure of 30" Hg?
2-49 200 kg per hour of chlorine gas is flowing through a pipeline at a
temperature of 80°F and at a pressure of 5 psig. What is the flow
velocity in meters per second if the pipe is 5" in diameter? Use the
ideal gas law indirectly.
2-50 A pipeline is transmitting 6,500,000 cubic feet per hour of natural
gas. Near the outlet, the pressure is 50 psig and the temperature is
Essential Basic Principles 89
40°F. If the composition is the natural gas is 90% CH4, 7% C2H6,
and 3% C3H8, what is the mass flow rate in pounds per minute?
Use the ideal gas law indirectly.
2-51 A sample of wet air is flowing at the rate of 100 kilograms per
minute. If the total pressure is 765 mm Hg and the partial
pressure of water is 30 mm Hg, what is the amount of water that is
flowing with the air in kilograms per second?
2-52 An inert gas mixture has the following composition: 80% N2, 15%
CO2, and 5% Ar. It is bubbled through liquid benzene, C6H6 and
steam is then injected into the gas. The resulting mixture has a
partial pressure of benzene equal to 50 mm Hg and that of water is
equal to 15 mm Hg. If the total pressure is 1.5 atm at a temperature of 200°C, how much benzene and water (kg) are contained per
100 m3 of the gas mixture?
DRAFT
Answers to selected problems:
2-15 A = 6841 E/R = 6745
2-17 NA = 1.59exp(-0.0000638t)
2-20 1.8125
2-22 Pabs = 38.3 mm Hg = 5.169 kPa = 0.750 psi = 108 psf
2.24 h = 230.4 ft
2-27 a) 5928 b) 738 c) 761 d) 757 all in mm Hg
2-28 Composition in mass %: 17.6% CO2, 1.9% CO, 7.5% O2
2-30 29,400,000 kg
2-32 16.25 ppm
2-34 14.21
2-36 c) for 7 cm D; Q = 1.33 m3/h; for 15 cm D; Q = 2.67 m3/h
2-38 V = 8.834 m3
2-40 V = 260 ft3
2-42 V= 23.73 m3
2-44 Density = 0.6973 kg/m3
2-46 Density = 1.116 kg/m3 kg H2O/kg dry air = 0.010
2-48 Q = 19.551 L/min
2-50 w = 23,2891bm/min
2-52 kg benzene = 7035.6 kg water = 487.08
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National Engineering Center CEEC Seminar, June 2 -4, 2004
DRAFT
—Chapter 3—
Mass Balance I:
No Chemical
Reactions Involved
In Chapter 1, we discussed simple mass balance calculations. In this chapter, we consider mass and heat balance
where no chemical reaction is involved. Usually, heat balance
is part of the chapter on energy balance. However, the principles involved are simple enough to be included in this chapter.
Moreover, a heat balance requires the associated mass
balance. Remembering from Chapter 1 that mass balance
calculations boil down to solving algebraic equations, the real
problem is recognizing and collecting the necessary data and
information to set up the equations. As a starting chemical
engineering student, your available stock knowledge comes
from the courses you have taken such as mathematics,
chemistry, and physics. However you have to tailor them to
the chemical engineering techniques. In this chapter, we will
consider some unit operations. To be able to set up the mass
balance calculations, all the unit operations to be discussed
will be explained although mostly in a qualitative manner.
THE MASS BALANCE EQUATION
As introduced in Chapter 1, we can state a mass balance
around any system as:
Mass inputs − Mass outputs = Mass Accumulation
1-1
Normally, we set up mass balance around an equipment, an
enclosure, or a vessel where mass (gas, liquid, solid, or combination) can be retained. Some examples are a pipe section, a
storage tank, a mixing tank, a conveyor, a pot, or a filter bag.
Consider a tank shown in Fig. 3-1 with water flowing in and
out. Q1 is the input flow rate while Q2 is the output flow rate.
At any time t, M is the mass of the fluid within the tank.
Mass Balance I: No Chemical Reactions Involved
91
For a finite time interval ∆t,
Q1
at time t, the flow rates are
Q1|t and Q2|t
M
Q2
Fig. 3-1
at time (t + ∆t), the flow rates are
Q1|t + ∆t and Q2|t + ∆t
The average input rate over the
period ∆t is
Q 1|t + Q 1|t+t
2
and the average output rate is
Q 2|t + Q 2|t+t
2
Applying a mass balance for the time interval, ∆t
Mass input − Mass output = Mass Accumulation
DRAFT
Q 1|t + Q 1|t+t
Q 2|t + Q 2|t+t
t −
t = M
2
2
3-1
where ∆M is mass accumulation and is the change in mass for
the finite time interval, ∆t. Dividing both sides by ∆t,
Q 2|t + Q 2|t+t
Q 1|t + Q 1|t+t
−
= M
2
2
t
3-2
As ∆t approaches 0, the equation becomes
Q1 − Q2 = dM
dt
3-3
The mass balance equation can thus be restated as
time rate of change
mass flowrate − mass flowrate
=
out of the system
of mass in the system
into the system
In general, the mass balance equation applies to systems
with more than one input and output streams and Eq. 3-3
becomes
3-4
Qin −
Qout = dM
dt
STEADY AND UNSTEADY STATE OPERATIONS
In mass balance calculations, if the flow rates and amount of
mass in a system vary with time, unsteady state exists. As an
example, let us consider that a tank is initially empty. Its
capacity is 1 m3. (The density of water is 1,000 kg/m3.) This is
to be filled with water at a constant rate of 10 kg/s. At this
92
National Engineering Center CEEC Seminar June 2 -4, 2004
flow rate, the tank will fill up in 100 s. During this time interval, the output is zero and the water is accumulating. After
100 s, the water starts to overflow. Since the tank is already
full, the accumulation of water is zero. Now if the valve near
the bottom of the tank is opened such that the flow rate is
larger than 10 kg/s, the water level will start to go down. (The
accumulation is negative).
Whenever the accumulation is not zero, unsteady or
transient state exists. We encounter both unsteady and steady
state operations in chemical engineering systems. Unsteady
state exists in a batch process and in the start-up of a
continuous operation.
input
(a)
(b)
(c)
output
DRAFT
Fig. 3-2 Batch and semi-batch processes. (a) The input is added progressively
with no output. (b) No input, no output. (c) The product is gradually withdrawn.
Fig. 3-2 shows batch and semi-batch processes. In Fig.
3-2b, the input and the output streams are zero. We can set
up a mass balance in this set up by considering the components inside the tank. Fig. 3-2a and Fig. 3-2c are semi-batch
processes. Here, unsteady state exists. In a continuous
process as shown in Fig. 3-3, steady state operation (input =
output) is easily attained. Recall the example of the tank
initially empty and to be filled with water. In start-up, the tank
has to be filled up first before attaining steady state operation.
In a similar manner, an industrial process that operates
continuously is in unsteady-state operation during initial
start-up. The plant has to be shutdown for maintenance and
has to be re-started-up. Therefore unsteady-state operations
are always encountered.
input
output
Fig. 3-3
OVERALL AND COMPONENT BALANCES
As almost all processes deal with different components, we
have to consider both the overall and component balances.
Consider a system where components A, B, C, and D are
involved. See Fig. 3-4. As the diagram shows, the block has
three input streams (1, 2, and 3) and two output streams (4
and 5). M represents the masses while x, the composition in
Mass Balance I: No Chemical Reactions Involved
93
mass fraction. We apply mass balances around the system.
1
xa2
xb2
xc2
xb3
xc3
xd3
M1
xa1
xb1
2 M2
4 M4
3 M3
5 M5
xa4
xb4
xc4
xd4
xa5
xb5
xc5
Fig. 3-4
Overall mass balance:
mass inputs
=
mass outputs
M1 + M2 + M3 = M4 + M5
3-5
DRAFT
Component A balance:
A in stream 1 + A in stream 2 = A in stream 4 + A in stream 5
(No A in stream 3)
M 1 x a1 + M 2 x a2 = M 4 x a4 + M 5 x a5
3-6
Component B balance:
B in stream 1 + B in stream 2 +B in stream 3 = B in stream 4 + B in stream 5
M 1 x b1 + M 2 x b2 + M 3 x b3 = M 4 x b4 + M 5 x b5
3.7
Component C balance:
C in stream 2 + C in stream 3 = C in stream 4 + C in stream 5
M 2 x c2 + M 3 x c3 = M 4 x c4 + M 5 x c5
3-8
Component D balance:
D in stream 3 = D in stream 4
M 3 x d3 = M 4 x d4
3-9
The above equations show that in a mass balance setup
involving N components, N+1 relationships are available: one
overall mass balance and N component balances.
TIE COMPONENTS
If we analyze the streams around a unit, we can identify a
component (or components) that enters in one input stream
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National Engineering Center CEEC Seminar June 2 -4, 2004
only and goes out in one output stream only. The quantity of
this material can easily be traced in the process. We call this
component a "tie component.” If we know the quantity of a tie
component, we can use this information to relate the masses
of the two streams, which contain that tie component.
Consider Fig. 3-5. The process involves four components.
Let us see whether we can identify a tie component. Component A enters in stream 1, 2, and 3 and leaves in streams 4
and 5; component B enters in streams 1, 2, and 3 and leaves
in streams 4 and 5; component C enters in streams 2 and 3
and leaves in streams 4 and 5; and component D enters only in
stream 3 and leaves only in stream 4. It is clear that D is the tie
component.
DRAFT
1
M1
xa1
xb1
xa2
xb2
xc2
2 M2
4 M4
xa3
xb3
xc3
xd3 = 0.2
3
5 M5
xa4 = 0.05
xb4 = 0.10
xc4 = 0.20
xd4 = 0.65
xa5
xb5
xc5
M3 = 200
Fig. 3-5
Let us illustrate the use of component D as a tie component.
From Fig. 3-5, M3 = 200 kg and xd3 = 0.2. The composition of
stream 4 (M4), is xa4 = 0.05, xb4 = 0.10, xc4 = 0.20, and xd4 =
0.65. D, as a tie component, can relate stream 3 to stream 4.
Setting up a D component balance,
D in stream 3 = D in stream 4
D in stream 3 = xd3M3 = 0.2 (200) = 40 kg
We can solve for M4. From the data, we know that 1 kg of M4
contains 0.65 kg of D. This corresponds to the ratio 1 kg
M4/0.65 kg D or its reciprocal 0.65 kg D/kg M4. Now recall the
conversion of units as discussed in Chapter 2. We can use the
ratio as a conversion factor -- to "convert" D to M4.
M 4 = 40 kg D %
kg M 4
= 61.5 kg
0.65 kg D
or divide the quantity D by the appropriate ratio
M4 =
40 kg D
= 61.5 kg
0.65 kg D/kg M 4
This technique is useful in mass balance calculations.
Always look for a tie component. In drying wet solids, part of
the moisture is removed to give a product that may still
contain some residual moisture. Here, we consider the waterfree solids (also called the "bone-dry" solids) as a tie
Mass Balance I: No Chemical Reactions Involved
95
component. The water-free solids in the input is the same
water-free solids in the output. The water-free solids can relate
the quantities of these input and output streams.
USE OF MASS OR MOLE UNITS
We can use either mass or mole units in mass balance
problems without chemical reactions. We usually use mass
units for solids and liquids. Mole units are more suitable for
gases and occasionally, for liquids. When chemical reactions
are involved, the number of moles of the input materials may
not equal the number of moles of the output materials.
Chapter 4 presents the cases where chemical reaction occurs.
The equality for mass units, on the other hand, always holds
true for cases with or without chemical reactions.
MASS BALANCE SETUP FOR BATCH PROCESSES
DRAFT
If we consider a batch process as shown in Fig. 3-2b, we do
not see any input and output streams -- we cannot set up a
mass balance equation based on the figure. For a batch
process, what we really have to do is consider a time interval
-- the starting time (to charge the inputs) and the end of the
process. Fig. 3-6 shows this.
100 kg A
100 kg B
Mixing tank
300 kg
mixture
100 kg C
At time 0
100 kg A
100 kg B
Mixing tank
300 kg
mixture
100 kg C
At the time after
inputting and mixing
the components
(a)
(b)
Fig. 3-6
For example, we want to mix 100 kg of A, 100 kg of B, and
100 kg of C in a tank. We consider the starting time (or time =
0) when we start charging the inputs into the tank while
letting the agitator run. This will take a period of time. Assuming that no chemical reaction occurs, at the end of the charging time, the tank contains 300 kg of a mixture that comprises
equal quantities of A, B, and C. We can represent this as in
Fig. 3-6a with a diagonal line drawn across. The left side of the
diagonal line represents the start of mixing showing the inputting of the quantities A, B, and C. The right side of the diagonal represents the total mass after charging and mixing. We
can simply delete the diagonal line and represent the
schematic diagram as in Fig. 3-6b.
If we know that a chemical reaction will occur, then we have
to wait for a reasonable length of time for the reaction to
proceed and end. (What happens in between the start and
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National Engineering Center CEEC Seminar June 2 -4, 2004
completion of a reaction is an unsteady state situation, which
is discussed in a later chapter.)
MASS BALANCE SETUP FOR CONTINUOUS PROCESSES
In a continuous process, we know that the stream flow rates
are constant. In this case, we base our calculation on a fixed
time interval (normally a second, minute, hour, day, etc.,
whichever is convenient). Taking a basis of one minute operation, for example, Fig. 3-7 shows that the amount of A and B
flowing in the process within one minute is equal to the
amount of C flowing out of the process within one minute. The
concept of a basis of computation (introduced in Chapters 1
and 2) is discussed next in detail.
BASIS OF COMPUTATION
In a process, we deal with several streams that have different
flow rates. In mass balance calculations, we should be clear as
DRAFT
10 kg/min A
10 kg/min B
Process
20 kg/min C
Fig. 3-7
to what quantities or streams we are referring to. To be consistent in our calculations we should take a basis of computation.
A basis of computation is a quantity of a specific stream or
component upon which our calculation is based. Depending
on convenience or calculation technique, we can choose the
amount of any of the input or output streams or a corresponding component as the basis. Usually 100 kg, 100 kmol, or 100
units of a material is convenient. If the stream is expressed in
terms of rate, say kg/h, the convenient basis is the number of
kilograms flowing in one hour. In this case, the basis can be a
unit of time, say basis = 1 hour. Frequently, we can deduce
the proper basis from the statement of the problem. Notice the
use of basis in the examples. We cannot underestimate its
importance.
SOLUTION OF MASS BALANCE PROBLEMS
We have to follow some steps to achieve a systematic solution
to mass balance problems. For simple problems, the steps
may seem superfluous. For complicated problems, they
become essential. Proficiency in solving mass balance
problems depends on solving actual exercises. You may
observe a pattern for each type of calculation but the "feel" or
proficiency is attained only by actual practice. Almost every
chemical engineering work starts with a mass balance, hence
you should not take these lessons for granted.
Mass Balance I: No Chemical Reactions Involved
97
These are the steps we have to follow in solving mass
balance problems:
A. Read the problem carefully. Determine what particular
types of process and equipment are involved. For a beginner like you, be sure you understand what the equipment
or the process is all about. You have to rely on your
instructor for information regarding process and equipment. In this book, qualitative descriptions of some unit
operations are given. For example, in drying problems, we
know that water is usually removed from wet materials by
using a medium such as hot air to transport the water
vapor.
B. Draw a simple block diagram (or a schematic diagram)
that represents the system. Arrows pointing to the block
depict the materials entering, while arrows pointing
outward, the output streams. Draw broken lines around
the unit in which the mass balance will be set up. See
Fig. 3-8.
Output 1
Output 2
Input 1
DRAFT
Distillate
Feed
Input
Input 2
Output 3
(a)
(b)
Bottoms
Fig. 3-8
C. From the statement of the problem, identify the given
or known quantities (or streams) together with the
relevant properties, such as temperature, pressure, and
flow rate, among others. Indicate the unknown quantities
or properties required.
D. Take a basis of computation. Be sure that you choose
the most convenient one, which you can deduce from the
statement of the problem. Use only one basis at any
given time. You can change the basis within a problem,
but you should indicate this properly.
E. Setup overall and component balances. Extra relationships can be developed from chemical and physical
principles. Solve the unknowns or the desired quantities
using any of the techniques to be discussed below. Be
practical in your solution. You will be able to sense an
error if you get impractical results such as too large
quantities or negative masses or flow rates.
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National Engineering Center CEEC Seminar June 2 -4, 2004
F. Check your answers by considering the given conditions or by using common sense if applicable.
METHODS OF SOLUTION
DRAFT
We can solve the unknowns in a mass balance problem by
considering stream to stream relationships or by solving
simultaneous equations. We can use any of the following
methods:
a. Arithmetic method
b. Algebraic solution
c. Combination of algebraic and arithmetic method
d. Graphical technique
e. Computer solutions including packaged software
The arithmetic method is most appropriate for beginners.
Anybody who knows algebra can use the algebraic method,
but a beginner will gain little proficiency in mass balance
calculations. (We showed this in Chapter One.) The algebraic
method is recommended for those already adept in mass
balance calculations. A system of simultaneous equations with
more than four unknowns is difficult to solve manually. It is
easily solved using a computer. As much as possible, we
should limit the number of unknowns to less than four. For
learning purposes, the third method is the best.
We will discuss graphical techniques in Chapter 6.
ARITHMETIC METHOD
This method uses stream to stream analysis based on the tie
component. We choose a stream which has a tie component
and start the calculation using this component. The quantity
of the tie component can be used to relate the quantities of the
two streams where it is found. In the arithmetic method, the
technique we used in the conversion of units is useful. (See
Chapter 2.) Retaining the units of the quantities is essential.
From Fig. 3-5, with D as the tie component,
kg D = 200 kg M 3 %
0.2 kg D
= 40 kg
kg M 3
The amount of M4 is calculated as
M 4 = 40 kg D %
kg M 4
= 61.54 kg
0.65 kg D
Note how placing the unit for mass fraction helps in making
the solution more systematic. As we have shown in the
conversion of units in Chap. 2, similar units cancel out.
Example 3-1
A batch dryer can handle 5000 kg feed per hour. 5000 kg of a
material containing 30% moisture is to be dried. How much
Mass Balance I: No Chemical Reactions Involved
99
moisture evaporates per hour if the dried material contains 5%
moisture? Solve using the arithmetic method.
Solution:
First we draw the block diagram. See Fig. 3-9. Next, we
identify the tie component. The wet material consists of the
moisture and the bone-dry solids. (The term "bone-dry solids"
refers to water-free solids.) The dried material consists of the
residual moisture and the bone-dry solids. The bone-dry solids
therefore, is the tie component.
Water evaporated
Wet material in
5,000 kg
30% moisture
Drier
Dried
material out
5% moisture
Fig. 3-9
The amount of entering material as well as its moisture
content is known so that the amount of the bone-dry solids
can be calculated. Now we can identify the most convenient
basis -- the entering material or 5,000 kg input.
DRAFT
Required: kg water evaporated per hour.
Basis: 5,000 kg
(Note: We can instead use "1 hour". This means that our basis is 1
hour of operation.)
Wet material in = 5000 kg
Bone dry solids in
= 5000 kg wet material % (1 − 0.3)
kg bone dry solids
= 3, 500 kg
kg wet material
Bone-dry solids balance:
Bone-dry solids in = Bone-dry solids out = 3500 kg
Dried materials out
= 3, 500 kg bone-dry solids %
kg material with 5% water
= 3, 684 kg
(1 − 0.05) kg bone-dry solids
We can now calculate the water evaporated by using an
overall mass balance.
Overall mass balance:
kg wet material in = kg H2O evaporated + kg dried product out
kg H2O evaporated = 5000 - 3684 = 1316 kg.
ALGEBRAIC METHOD
In the algebraic method, we identify which streams, composition or other quantities are unknown. We assign appropriate
symbols to these unknowns. As we have learned from algebra,
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National Engineering Center CEEC Seminar June 2 -4, 2004
we have to set up independent equations, the number of
which should equal that of the unknown quantities. The
overall and component balances are the primary relationships
available. If the number of equations is not equal to the
number of unknowns, we have to find extra relationships. For
example, this could be the efficiency of the process or physical
relationship. The system of equations is then solved. The
method is simple and similar to solving verbal problems in
algebra.
Example 3-2
Solve Ex. 3-1 using the algebraic method.
Solution:
(Refer to Fig. 3-9.)
Required:
kg water evaporated per hour
Basis:
5000 kg or 1 hour
The unknown streams are the dried material and the water
evaporated.
DRAFT
Let
x = mass of water evaporated
y = mass of dried material.
With two unknowns, we need two equations. Three equations
are available: one overall and two component balances. Let us
use the O. M. balance and a water balance.
Overall material balance:
5, 000 = x + y
3-10
Water balance:
Water in the incoming stream = water evaporated + water in
the outgoing dried material
5, 000(0.30) = x + 0.05y
3-11
From Eq. 3-10,
y = 5, 000 − x
3-12
Substituting Eq. 3-12 Eq. 3-11
1, 500 = x + 0.05(5, 000 − x)
3-13
Therefore,
x = 1, 316 kg water evaporated per hour
Answer
NOTE: Instead of Eq. 3-11, we can set up another equation by
using a bone-dry solids balance:
Bone-dry solids in = Bone-dry solids out
Mass Balance I: No Chemical Reactions Involved
101
5, 000(1 − 0.3) = y(1 − 0.05)
3-14
Therefore, another set of equations is
5, 000 = x + y
5, 000(1 − 0.3)
=
y(1 − 0.05)
We can solve this in the usual manner.
In the algebraic solution of pure mass balance problems
involving N components, the maximum number of independent equations is equal to N. As shown in Ex. 3-2, a total of
three equations are available, but we only used two to solve
the system of equations.
Referring to Fig. 3-4, we have set up the following equations:
Overall mass balance:
M1 + M2 + M3 = M4 + M5
3-15
Component balances:
DRAFT
M 1 x a1 + M 2 x a2 + M 3 x a3 = M 4 x a4 + M 5 x a5
M 1 x b1 + M 2 x b2 + M 3 x b3 = M 4 x b4 + M 5 x b5
3-16
3-17
M 2 x c2 + M 3 x c3 = M 4 x c4 + M 5 x c5
3-18
M 3 x d3 = M 4 x d4
3-19
The sum of the mass fractions of the components of each
stream is one.
x a1 + x b1 = 1
3-20
x a2 + x b2 + x c2 = 1
3-21
x a3 + x b3 + x c3 + x d3 = 1
3-22
x a4 + x b4 + x c4 + x d4 = 1
3-23
x a5 + x b5 + x c5 = 1
3-24
If we know the composition of all the mixtures and we use one
of the streams as a basis, we only need four of the five mass
balance equations available. With unknown compositions,
nine equations are available. We should be careful in determining which equations are independent. Otherwise, we will
not be able to solve the system of equations. We have to refer
to the rules in algebra. In our problems, we will limit our
unknowns to three as much as possible. This is easily
manageable by hand computation. The algebraic method of
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National Engineering Center CEEC Seminar June 2 -4, 2004
solution is useful when we cannot identify a tie component in the
system.
DRAFT
CHOICE OF METHOD
The example discussed is a simple one and any of the two
methods is adequate. For practice and for better understanding of a process, the arithmetic method is preferred over the
algebraic method. The process can be pictured mentally and
the flow path traced easily.
This does not diminish the usefulness of the algebraic
method, however. For some problems, the arithmetic method
may be difficult to apply and we have to use the algebraic
method. Systems of equations containing more than 4
unknowns are difficult to solve manually. If a computer is
available however, the algebraic method should be used.
In solving problems, determine first whether tie components
are available. Their presence indicates that the arithmetic
method can be used. In some instances, a combination of the
algebraic and the arithmetic method is the best method applicable. Here, one strives to limit the unknowns to a maximum
of three, solving for the others by direct arithmetical analysis.
Through this method, you totally understand the process.
Blindly applying the algebraic method makes mass balance
calculations mechanical, that is, you can solve the problem
without entirely understanding the process. When you are a
full-pledged engineer, you can use any method, because you
understand the mechanics. As of now, use the method that
teaches you the basics.
Example 3-3: Distillation
We desire to produce two ethyl alcohol-water mixtures each
containing 90% and 50% alcohol by mole from a dilute
mixture containing 20% mole alcohol by distillation. 98% of
the ethyl alcohol in the feed is to be recovered in these two
products, that is, only 2% of the ethyl alcohol in the feed may
go into the bottoms. If the bottoms contains 0.6% alcohol,
calculate the amount of the different streams per 1000 kmol of
feed stream.
Solution:
Given: 98% of ethyl alcohol in feed goes to the two product
streams.
Examining the different streams, water and alcohol are found
in all of them such that no component can serve as a "tie". The
Mass Balance I: No Chemical Reactions Involved
103
algebraic method is suitable.
Distillate, D
90% alcohol
10% water
Feed, F
Side stream, S
50% alcohol
50% water
20% alcohol
80% water
Bottoms, B
0.6% alcohol
99.4% water
Fig. 3-10
Let D represent the 90% alcohol stream; S, the 50% alcohol; F,
the feed; and B, the bottoms.
Required: D, S, B
Basis: 1000 kmol of feed
DRAFT
Overall mass balance:
1, 000 = D + S + B
3-25
Alcohol balance:
(0.2)((1000) = 0.90D + 0.5S + 0.006B
3-26
Alcohol recovery:
98% of the alcohol in the feed goes to D and S; 2% of the
alcohol in the feed goes to B.
(0.02)(0.20)(1, 000) = (0.006)B
3-27
The resulting system of equation in three unknowns are:
1, 000 = D + S + B
3-28
200 = 0.90D + 0.50S + 0.006B
(0.02)(200) = (0.006)B
3-29
3-30
(Note: A water balance instead of an alcohol balance can
substitute for Eq. 3-26)
[(0.80)(1, 000) = 0.10D + 0.50S + 0.994B]
From Eq. 3-27
B=
104
(0.02)(200)
= 666.7 kmol
0.006
National Engineering Center CEEC Seminar June 2 -4, 2004
Substituting the value of B in Eq. 3-28 and 3-29,
1, 000 = D + S + 666.67
3-31
200 = 0.9D + 0.5S + 0.006(666.67)
3-32
From Eq. 3-31
D = 1, 000 − S − 666.67 = 333.33 − S
3-33
Substituting Eq. 3-33 in Eq. 3-32,
200 = 0.9(333.33 − S) + 0.5S + 4
200 = 300 − 0.9S + 0.5S + 4
0.40S = 104
S = 260 kmol
Substituting the value of S in Eq. 3-33,
DRAFT
D = 1, 000 − 260 − 666.67 = 73.33 kmol
Example 3-4:
Evaporation
An evaporator has a capacity to evaporate 30,000 kg of
water per hour. We want to concentrate a NaCl-water solution
from 10% to 40%. If the evaporator is to operate at its rated
capacity, what must the feed rate be? How much product is
obtained per hour? Solve using the algebraic and the arithmetic methods.
Required:
1. feed rate
2. product rate
Solution:
A. Using the algebraic method
30,000 kg water
evaporated per hour
feed
10% salt
Evaporator
40% salt
Fig. 3-11
Basis: 1 h
Let F = kg feed
P = kg product
Setting up the equations,
Mass Balance I: No Chemical Reactions Involved
105
Overall Material Balance:
F = P + 30, 000
3-34
0.1F = 0.4P
3-35
P = 0.1F
0.4
3-36
Salt Balance:
Substituting Eq. 3-36 in Eq. 3-34,
F = 0.1F + 30, 000
0.4
F = 40, 000
P = 40, 000 − 30, 000 = 10, 000 kg.
The feed rate is : 40,000 kg/h
The product rate is : 10,000 kg/h
DRAFT
b. Using the arithmetic method
In using the algebraic method, no difficulty arose. However, if
we analyze the problem and use the arithmetic method, we
will encounter some problems.
Although the salt is a tie component, we cannot relate the
amount of water evaporated to the other two streams. The tie
component relates two unknown quantities. This is an
instance where the use of another basis is convenient. Instead
of using 30,000 kg of water evaporated or one hour as the
basis, we choose a basis of 100 kg of feed. Then using ratio
and proportion, we can solve for the hourly rate.
Basis: 100 kg feed
Salt balance:
salt in the feed = (0.1)(100) = 10 kg = salt in the product
kg product =
10 kg salt
= 25 kg
kg salt
0.4
kg product
Using an overall mass balance,
H 2 O evaporated = 100 − 25 = 75 kg
Using ratio and proportion,
100 kg feed
F feed rate
=
75 kg H 2 O evaporated 30, 000 kg H 2 O evaporated/h
F = 30, 000 % 100 = 40, 000 kg/h
75
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National Engineering Center CEEC Seminar June 2 -4, 2004
For the product
P product rate
25 kg product
=
75 kg H 2 O evaporated 30, 000 kg H 2 O evaporated/h
P = 30, 000 % 25 = 10, 000 kg/h
75
We should see the merits of each method since in some
instances, one of them will be more convenient to use. In the
next illustrative example, the choice of the method depends on
the conditions imposed in the problem.
Example 3-5: Crystallization
DRAFT
One thousand kilograms of a 64% NaNO3 solution in water is
at a temperature of 100°C. This is sent to a crystallizer where
it is cooled to 30°C. At this temperature the solubility of
NaNO3 is 96 parts per 100 parts of water (the solubility of
NaNO3 is maximum amount of NaNO3 that can stay in
solution; the rests crystallizes out). Calculate the amount of
crystals that precipitates out if
a. No water evaporates during cooling.
b. 5% of the original water evaporates during cooling
Solution:
a. Water is a tie component. We can apply the arithmetic
method.
feed
64% NaNO3
Crystallizer
solution
(mother liquor)
NaNO3 crystals
Fig. 3-12
Basis: 1,000 kg feed
kg NaNO3 in the feed = 0.64(1000) = 640 kg
kg H2O = 1000 - 640 = 360 kg
We set up a water (the tie component) balance:
Water in = water out = 360 kg
NaNO3 in the mother liquor =
96 kg NaNO3
= 345.6 kg
360 kg H 2 O%
100 kg H 2 0
NaNO3 balance:
NaNO3 crystals = NaNO3 in feed - NaNO3 in mother liquor
= 640 - 345.6 = 294.4 kg
b. When some water evaporates, we can no longer consider
water as a tie component. Therefore, we should use the
algebraic method.
Mass Balance I: No Chemical Reactions Involved
107
Basis: 1000 kg feed
E
1000 kg feed
64% NaNO3
water evaporated
mother liquor
Crystallizer
M
C
Fig. 3-13
Let
NaNO3 crystals
C = kg crystals obtained
M = kg mother liquor
E = kg water evaporated.
Overall mass balance:
1, 000 = C + M + E
3-37
NaNO3 balance:
DRAFT
0.64(1, 000) = C + M(
96 )
100 + 96
3-38
Evaporated water balance:
E = 0.05(1, 000)(1 − 0.64) = 18
3-39
Substituting in Eq. 3-39 in Eq. 3-37,
1, 000 = C + M + 18
M = 982 − C
Substituting Eq. 3-40 in Eq. 3-38
640 = C + (982 − C)
3-40
96
100 + 96
C = 311.68 kg crystals obtained
Example 3-6: Solid-Liquid Extraction
A copra batch contains 44% oil and is to be extracted with
n-hexane. The extracted meal contains 0.07 kg oil/kg oil-free
matter, and the mass fraction of n-hexane is 0.05. The extract
contains 49% n-hexane. If 10,000 kg of copra are to be
processed, how many kilograms of n-hexane is required?
Solution:
Required: kilograms of n-hexane for extraction
Basis: 10,000 kg copra feed
In this problem, the oil-free matter is the tie component. It
goes in only in one stream and goes out in only one stream
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National Engineering Center CEEC Seminar June 2 -4, 2004
also.
Extract
49% n-hexane
51% oil
n-hexane
Extractor
Copra feed
Extracted meal
10,000 kg
44% oil
xn-hexane = 0.05
0.07 kg oil
kg oil-free matter
Fig. 3-14
Oil-free matter balance:
Oil-free matter in entering copra = oil-free matter out
Oil-free matter = 10,000 x 0.56 = 5600 kg
Oil balance:
Oil in copra feed = oil in extract + oil in extracted meal
Oil in extracted meal =
DRAFT
5, 600 kg oil-free matter %
0.07 kg oil
kg oil-free matter
= 392 kg
Oil in extract = (10, 000)(0.44) − 392 = 4, 008 kg
n-hexane balance:
n-hexane required = n-hexane in extract + n-hexane in extracted meal
n-hexane in extract = 4, 008 kg oil %
0.49 kg n-hexane
0.51 kg oil in extract
= 3851 kg
n-hexane in extracted meal = 5, 600 kg oil-free matter
%
(1 + 0.07)kg oil and oil-free matter 0.05 kg n-hexane in extracted meal
%
kg oil-free matter
(1 − 0.05) kg oil and oil-free matter
= 315.37 kg
n-hexane required = 3850.82 + 315.37 = 4166.19 kg.
Notice the use of units and dimensions to simplify the arithmetic method.
Example 3-7
Gas Absorption
A gas mixture contains 6.0 mole % acetone, 1.7 mole % water,
and the rest, air. The acetone is to be recovered using an
absorption tower that operates at 30°C and 101.325 kPa.
10,000 kg water per hour will be used to scrub 180 kmol per h
of entering gas. If the recovery is 96%, calculate the mole
fraction of acetone in the outlet gas and in the outlet water.
Mass Balance I: No Chemical Reactions Involved
109
The outgoing gas has a partial pressure of water equal to
4.241 kPa. The water is fed at the top of the tower while the
gas, at the bottom.
Solution:
The arithmetic method is the appropriate method.
Required:
mole fraction acetone in outlet gas, y2
mole fraction acetone in outlet water, x1
Basis:
1 hour
The tie component here is pure air in the incoming gas.
H2O
10,000 kg/h
Outlet gas
2
PH2O = 4.241 kPa
Operating
temperature = 30oC
DRAFT
Operating Pressure
= 101.325kPa
Water + absorbed
acetone
Absorber
1
180 kmol/h
6% acetone
1.7%water
Fig. 3-15
In the entering gas:
kmol entering air = (1 − 0.06 − 0.017)(180) = 166.14
kmol acetone = (0.06)(180) = 10.80
kmol water = (0.017)(180) = 3.06
In the outlet gas:
kmol acetone = (1 − 0.96)(10.80) = 0.432
kmol acetone + pure air = 0.432 + 166.14 = 166.6
Since the outlet gas has a partial pressure of water equal to
4.241 kPa, the partial pressure of acetone and air is (101.325
- 4.241) = 97.084 kPa.
Assuming ideal gas behavior, mole ratio = partial pressure
ratio.
The water vapor in outlet gas
4.241 kmol water
= 7.278 kmol
97.084 kmol air and acetone mixture
kmol of outgoing gas = 166.6 + 7.278 = 174
= (166.6) %
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National Engineering Center CEEC Seminar June 2 -4, 2004
mol fraction acetone in outlet gas, y2 = 0.432 = 0.00248
173.878
kmol water entering at top of tower =
10, 000
= 555.6
18
Water balance:
Water in at the top + water in at the bottom = water out at
the top + water out at the bottom
Water going out at the bottom = 555.6 + 3.06 − 7.28 = 551.34
Acetone in outgoing liquid = (0.96)(10.80) = 10.368 kmol
Mol fraction acetone in outgoing liquid,
x1 =
10.368
= 0.018
10.368 + 551.34
DRAFT
THE RATIO METHOD
This method of solution, is a special form of the arithmetic
method. The tie component is indispensable in this method.
Fig. 3-16 shows the block diagram of process where the
method is generally applicable. Here, A is the tie component.
B removed
A and B
1
M
1
xa in
xb in
Process
A and B
2
M
xa out
xb out
2
Fig. 3-16
Consider the streams containing A and B (streams 1 and 2).
In both the input and output streams, set up the ratio of the
mass of the mixture to the mass of A.
mass of A and B entering
1
= x1a1 =
(1 − x b1 )
mass of A entering
3-41
mass of A and B outgoing
1
= x1a2 =
(1 − x b2 )
mass of A outgoing
3-42
Ratios of the amount of materials with the tie component as
the denominator will have a unit in terms of "per unit amount
of the tie component". The difference between the two ratios,
1
1
x a1 − x a2
must have the unit,
mass of B removed
unit mass of A involved
Mass Balance I: No Chemical Reactions Involved
111
In this set of ratios, note that the basis is made on a unit
mass of A only. A unit mass of entering A has with it, a fixed
amount of B. If B is removed (or added) in the process, the
same unit mass of A will have less (or more) B. The amount of
B removed (or added) therefore is simply the difference
between the 2 ratios.
As an example, given the amount of M1 and the compositions of streams 1 and 2, let us find the amount of B removed.
Here, the solution makes use of the techniques in conversion
of units. Start with M1 (I), find the amount of A in M1 (II), and
find the B removed using the difference of the ratios discussed
(III).
B removed =
M 1 kg stream1 %
I
kg B removed
x a1 kg A
% x1a1 − x1a2
1 kg stream 1
kg A
II
III
DRAFT
Another set of ratios that can be set is
mass of B entering
x
x b1
= x b1
=
a1
(1 − x b1 )
mass of A entering
3-43
mass of B outgoing
x
x b2
= x b2
=
a2
(1 − x b2 )
mass of A outgoing
3-44
and
Again, the difference between the ratios,
x b2
x b1
x a1 − x a2
must have the unit
mass of B removed .
unit mass of A involved
In later chapters, you will appreciate the usefulness of this
special method of solution. While the method looks like a
shortcut, it becomes a powerful technique for the chemical
engineer.
Example 3-8
Solve example 3-1 using the ratio method.
Required: kg H2O evaporated per hour
Solution:
As in the arithmetic solution discussed in Example 3-1, the tie
component is the bone-dry material. Consider 1 kg of wet
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National Engineering Center CEEC Seminar June 2 -4, 2004
material (30% H2O) entering the dryer.
Water evaporated
Wet material in
5,000 kg
30% moisture
70% bone-dry solids
Drier
Dried
material out
5% moisture
95% bone-dry solids
Fig. 3-17
The ratio of the kg wet material to the kg bone-dry material
in the entering stream is
kg wet material
1
(1 − 0.3) kg bone-dry material
3-45
Then consider 1 kg of the dried material (5% moisture) going
out. The ratio of the material to the bone-dry material in the
outgoing stream is
DRAFT
kg dried material
1
(1 − 0.05) kg bone-dry material
3-46
The above two ratios have one thing in common. The unit of
their denominators is the same (kg bone-dry material). Taking
the difference,
kg wet material
kg dried material
1
1
−
(1 − 0.3) kg bone-dry material (1 − 0.05) kg bone-dry material
= 0.3759
kg water evaporated
kg bone-dry material
3-47
Obviously, the difference should be the kg water evaporated
per kg of bone-dry solids. Notice that the calculation is done
implicitly on a basis of 1 kg bone-dry material. Now, making
use of appropriate units and dimension, per 5000 kg of
material,
kg water evaporated
kg bone-dry material
= 5, 000 kg wet material % 0.7
kg wet material
kg water evaporated
% 0.3759
kg bone-dry material
= 1,316 kg water evaporated
Example 3-9
Solve Example 3-4 using the ratio method.
Mass Balance I: No Chemical Reactions Involved
113
Solution:
30,000 kg water
evaporated per hour
feed
10% salt
90% water
Evaporator
product
40% salt
60% water
Fig. 3-18
Here the tie component is the salt.
The solution ratio from the feed composition is 1
0.1
salt
The solution ratio from the product composition is 1
0.4
salt
Therefore
1 − 1 represents kg water evaporated
0.1 0.4
kg salt involved
Using a basis of 1 h,
DRAFT
Feed rate
30, 000 kg H 2 O evaporated
%
h
%
1 kg feed
0.1 kg salt
1
1 − 1 kg H 2 O evap.
0.1 0.4
kg salt
= 40,000 kg/h
Product rate
=
30, 000 kg H 2 O evaporated
%
h
%
1 kg product
0.4 kg salt
1
1 − 1 kg H 2 O evap.
0.1 0.4
kg salt
= 10,000 kg/h
SINGLE UNIT SYSTEMS WITH BOUNDARIES WITHIN
In some equipment, contact and mixing between different
streams are avoided. Suitable boundaries are provided. In the
unit operation of heat transfer, a fluid heats or cools another
fluid while avoiding direct contact or mixing. A specific
example is the double-pipe heat exchanger. Concentric pipes
provide the separation. One fluid flows through the inner pipe,
while the other, in the annular space. In a shell-and-tube heat
exchanger, one fluid passes through the tubes while the other,
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National Engineering Center CEEC Seminar June 2 -4, 2004
outside of the tubes but inside of the shell. In this type of
equipment, we carry out a mass balance on each side of the
boundary.
Boundary
B in
A in
A out
B out
Figure 3-19
DRAFT
As Fig. 3-19 shows,
MA in = MA out
3-48
MB in = MB out
3-49
Another example is the steam boiler. The chemical energy of
a fuel such as diesel is used to boil water to produce steam.
Since uncontaminated steam is desired, a boundary must
separate the boiling water and the burning fuel. Tubes usually
act as the boundary and efficient medium for heat transfer.
See Fig. 3-20.
Liquid water in
Steam out
Combustion
gas
Bank of tubes with
combustion gases
flowing inside while
water is boiling
outside
Fuel
Air
Figure 3-20
The mass balances are:
Steam side:
mass liquid water in = mass steam out
Fuel side:
mass fuel + mass air = mass combustion gas
MULTIUNIT SYSTEMS
A process usually consists of several units. The mass balance
calculation for a multi-unit system is basically the same as
that for a single unit system. An overall mass balance and
component balances are set around the whole system. Then,
Mass Balance I: No Chemical Reactions Involved
115
balances around individual units are made as well. Likewise,
balances around points where mixing of streams occur and
points of splitting of streams are set up. One output from one
unit may be the input for another and so on. Usually the
solution proceeds from one unit to another, with a solved
quantity used for the next unit. Pure algebraic solution will
result in a system of equations in many unknowns that
require computer solution. For our case, a combination of
algebraic and arithmetic solution is adequate. Fig. 3-21 shows
a schematic diagram of a multi-unit system.
1
DRAFT
3
2
4
Figure 3-21 An example of a multi unit system. The broken
lines indicate boundaries for mass balances around each unit
and the entire system.
We discuss some examples and present the techniques in
solving them.
Example 3-10: Multiple-effect Evaporation
A triple-effect evaporator concentrates a 10% caustic soda to
50% NaOH. Assuming equal amounts of evaporation in each
effect, calculate the concentration of the solution leaving each
effect.
Solution:
Required: xp1 and xp2
Basis: 100 kg of feed containing 10% NaOH (Note: In problems
with unspecified amount of materials being processed, the
most convenient basis is 100 units of that material.)
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National Engineering Center CEEC Seminar June 2 -4, 2004
Take mass balances around the solution side of the whole
system.
S
E1
F
P1
Effect 1
xF = 0.1
E3
E2
P2
Effect 2
xP1
C1
xP2
Effect 3
C2
P3
xP3 = 0.5
C3
Figure 3-22
Overall mass balance:
F = 100 = E 1 + E 2 + E 3 + P 3
3-50
NaOH balance:
DRAFT
Fx f = P 3 x p3
3-51
(100)(0.1) = P 3 (0.5)
P3 =
(100)(0.1)
= 20 kg
(0.5)
From Eq. 3-50
E 1 + E 2 + E 3 = 100 − 20 = 80 kg
3-52
Since the evaporations are all equal,
E 1 = E 2 = E 3 = 80 = 26.67 kg
3
3-53
Consider a mass balance around effect 1.
Overall mass balance around effect 1 (solution side):
F = E1 + P1
3-54
Since we know F and E1,
P 1 = 100 − 26.67 = 73.33
Since the NaOH is not volatile, the NaOH in the feed is the
same as the NaOH in P1 and is equal to 10 kg. Therefore ,
x p1 =
10 = 0.136
73.33
or, by an NaOH balance around effect 1,
Mass Balance I: No Chemical Reactions Involved
117
Fx f = P 1 x p1
Fx
x p1 = P f
1
x p1 =
3-55
(100)(0.1)
= 0.136
73.33
Mass balance around effect 2 (solution side)
Overall mass balance:
P1 = E2 + P2
3-56
P 2 = 73.33 − 26.67 = 46.67 kg
NaOH balance:
P 1 x p1 = P 2 x p2
P 1 x p1
P2
(73.33)(0.136)
= 0.214
x p2 =
(46.67)
DRAFT
x p2 =
3-57
Example 3-11:
NaC1 crystals are to be produced from 1,000 kg of a solution
containing 10% NaCl, 1% KOH, and 89% H2O. The solution is
first sent to an evaporator where 790 kg of the water from the
solution is removed. The hot concentrated solution is then
sent to a crystallizer where the solution is cooled to 20°C and
NaC1 crystals precipitate. (No H2O evaporates.) Each kg of
crystals carries with it 0.1 kg of adhering mother liquor. The
wet crystals are then treated in a drier where 95% of the water
is removed. Calculate:
(a) the composition of the concentrated solution from the
evaporator.
(b) the composition of the final crystal product.
(c) the percentage recovery of NaC1 from the original solution.
Solve using the arithmetic method.
36 g NaCl
, that
100 g H 2 O
is, 100 g of water can hold only a maximum of 36 g of NaC1 in
solution.
Data: At 20°C, the solubility of NaC1 in H20 is
KOH is non-volatile and very soluble in H2O.
Assume that the presence of KOH does not affect the solubility
of NaCl in H2O.
Solution:
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National Engineering Center CEEC Seminar June 2 -4, 2004
This is a multi-unit system and it is easier to apply the
arithmetic method. Mass balances should be considered
around each unit.
790 kg H2O
1,000 kg
solution
Evaporator
10% NaCl
1% KOH
89% H2O
H2O removed = 95% of the
water in the
wet crystals
Crystals plus
adhering
Drier
mother liquor
Crystallizer
Final
0.1 kg mother liquor
crystal
kg crystals
product
20oC
36 g NaCl
100 g H2O
Figure 3-23
Required:
a) composition of the solution from the evaporator
b) final crystal product composition
c) % recovery of NaC1.
DRAFT
Basis: 1000 kg feed to the evaporator
With the above basis, it is convenient to make a mass
balance around the evaporator first. The amount and composition of the concentrated solution to the crystallizer can be
solved for and used for a balance in the next unit.
a) Balance around the evaporator
H2O
1,000 kg
solution
10% NaCl
1% KOH
89% H2O
Evaporator Concentrated
solution
Figure 3-24
Feed composition:
NaC1 = 0.1(1000) = 100 kg
KOH = (0.01)(1000) = 10 kg
H2O = (0.89)(1,000) = 890 kg
Water balance:
H 2 O in the concentrated solution = H 2 O in the feed - H 2 O evap.
= 890 - 790 = 100 kg
The concentrated solution consists of:
NaCl
KOH
Mass, kg
100
10
%
47.62
4.76
Mass Balance I: No Chemical Reactions Involved
119
H2 O
Total
100
210
47.62
100
b) The information obtained from the balances around the
evaporator is now used in the balance around the crystallizer.
Mass balance around the crystallizer:
It is easier to calculate first for the total crystals produced
and then find the amount of adhering mother liquor.
100 kg NaCl
10 kg KOH
100 kg H2O
Crystals plus
adhering
mother liquor
Crystallizer
mother liquor
36 g NaCl
100 g H2O
Figure 3-25
kg crystals
Crystals produced = 100 kg H2 O % ( 100 − 36 )
100 100 kg H2 O
= 64 kg
Adhering mother liquor =
0.1 kg mother liquor
% 64 kg crystals
kg crystals
DRAFT
= 6.4 kg
Composition of the mother liquor
H2O
KOH
NaCl
Total
Mass, kg
100
10
36
146
%
68.49
6.85
24.66
100
Mass balance around the drier:
H2O
64 kg NaCl
Drier
6.4 kg mother liquor
Dried
crystals
Figure 3-26
The adhering mother liquor consists of:
0.6849(6.4) = 4.38 kg H2O
0.0685(6.4) = 0.438 kg KOH
0.2466(6.4) = 1.58 kg NaCl
H20 evaporated = 0.95(4.38) = 4.l6 kg
H2O remaining with crystals = 0.05(4.38) = 0.22 kg
Total NaCl in crystal product = 64 + 1.58 = 65.58 kg
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National Engineering Center CEEC Seminar June 2 -4, 2004
Composition of the final crystal product:
NaCl
KOH
H2O
Total
Mass, kg
65.58
0.44
0.22
66.24
%
99.01
0.66
0.33
100
c) % recovery of NaCl = 65.58 % 100 = 65.58%
100
You can try solving this problem using purely algebraic
method. However, the resulting system of simultaneous
equations is difficult to evaluate.
RECYCLE STREAMS
Recycling operations refer to processes in which part of the
Recycle stream
DRAFT
Feed
Product
Fig. 3-27
product is returned and mixed with the feed entering a
process equipment. Fig. 3-27 shows a recycle stream.
In any process, some of the objectives are to enhance the
yield and to obtain products with minimum impurities. In the
manufacture of ammonia, recycle streams are indispensable.
Conditions do not allow 100% conversion to NH3. The
unreacted nitrogen and hydrogen are recycled and mixed with
the fresh feed. In the petroleum industry, the reactors yield
mixtures consisting of complex components, the result of
incomplete conversion. Distillation is used to separate these
mixtures. After the recovery of the desired products, the other
components are recycled back to the appropriate reaction
units. In the production of sulfuric acid, the fluid used for
absorbing SO3 is commonly part of the product that is
recycled. Make-up water or dilute sulfuric acid solution is
added to this stream to enhance the absorbing capacity.
Other unit operations employ recycle operations for energy
conservation and for better control of product quality. In distillation, in which high purity products are desired, the enrichment of the product is facilitated by recycling or "refluxing"
part of the product back to the distillation column. In drying
operations, we can conserve heat energy by recycling part of
the exhaust air and mixing it with the incoming fresh medium.
Since less quantity of incoming fresh air feed has to be conditioned (by dehumidification and heating), less amount of heat
energy is necessary. In absorption, part of the liquid from the
tower bottom is recirculated. This results in an increased
liquid rate in the tower and contributes to a more efficient
wetting of the tower packing material.
Mass Balance I: No Chemical Reactions Involved
121
In Fig. 3-28, stream A is the fresh feed while stream D is
the net product obtained from the system. Stream C, on the
other hand, is the total product obtained from the process.
This splits to form streams D and E. Stream E is the recycle
stream, which is mixed with the fresh feed to form the gross
feed, stream B. In some cases, stream E may be recycled to
other units located ahead of the unit where it comes from.
E
1
A
2
3
B
C
4
D
Figure 3-28 Mass balances around a process
involving recycle stream.
DRAFT
Mass balance calculations involving recycle operations
should pose little difficulty if balances are set up properly. As
usual, the arithmetic or the algebraic method is suitable
depending on the presence of a tie component. Mass balances
can be taken around four positions: (See Fig. 3-28.)
1. Taking a balance around the whole system, (boundary
represented by 1) the mass in stream A equals the mass
in stream D. Note that the recycle loop is within the
system so that stream E is not involved.
2. Taking a balance around the process within the recycle
loop (represent by line 2), the mass in stream B equals
the mass in stream C.
3. Taking a balance around the mixing point of the recycle
and the fresh feed, (line 3), the mass in stream E plus the
mass in stream A equals the mass in stream B.
4. Taking a balance around the point of splitting of the total
product (line 4), the mass of stream C equals the mass of
the recycle stream E plus the mass of the net product,
stream D.
The above four mass balances are not independent,
however. Combinations of any three give the fourth balance.
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National Engineering Center CEEC Seminar June 2 -4, 2004
Do not make a mistake of setting up a mass balance around
the boundary shown in Fig. 3-29.
Boundary
DRAFT
Figure 3-29
Setting up the balance around the whole system, but cutting
through the recycle line is a poor choice if the recycle line is
unknown. Note that the effect of the recycle line cancels out.
In setting up an overall mass balance around the whole
system (outside the recycle loop), the recycle stream is not
involved in the mass balance equation.
Two important terms in recycle operation are the recycle-tofresh-feed ratio and the total-feed-to-fresh-feed ratio.
Comparing the balance around the whole system and the
balance within the recycle loop, the net effect of the recycle is
to lower the capacity of an equipment. For instance, the
maximum capacity of an equipment is 1000 kg per hour, a
once-through flow (without recycle), results in a 1000 kg per
hour feed being processed. If, however, a recycle stream is
provided, say 500 kg per hour, then only 500 kg product
stream can be obtained. The equipment can handle only 1000
kg: 500 kg recycle and 500 kg fresh feed. Capacity is sacrificed
for other more important factors. It can be a choice between a
product with less impurity and a product with high level of
impurity. The use of recycle becomes factor in optimization of
the process.
Either the arithmetic or algebraic method can be used for
recycle calculations. With the use of the mass balances
around other boundaries, we can easily solve the unknown
quantities.
Remember that with steady state operation, the flow rates
are constant. The recycle line is just a mass stream that is
circling within a loop, combining with a feed at a certain point
(near the entrance of an equipment or a system), and splitting
up with the net product at another point (after the exit of an
equipment or a system). See Figure 3-30
Recycle
loop
Net feed
Equipment
or system
Net product
Figure 3-30
We must keep in mind that the above mass balances are
only true for steady-state operations. For batch, semi-batch,
and other unsteady state operations, the method of
Mass Balance I: No Chemical Reactions Involved
123
calculation is different. In these cases, we would encounter
accumulation of mass in the system. Non-chemical engineers
become confused with the concept.
Example 3-12
In a pilot process, a sticky material containing 80% water is to
be dried. To facilitate the operation, a part of the dried product
containing 5% water is recycled and mixed with the feed. If the
material entering the drier contains 30% water, calculate
(a) the kg water removed per 2000 kg fresh feed
(b) the recycle-to-feed ratio.
Solve using the arithmetic and algebraic methods.
Solution:
Basis: 2000 kg of fresh feed
Part 1: Using the arithmetic method.
Water evaporated
Recycle
DRAFT
Fresh feed
80% water
2000 kg
Mixe
r
30%
water
Dried product
Drier
5% water
Figure 3-31
To obtain the amount of water evaporated, consider a
balance around the entire system (outside the recycle loop).
Using ratios,
kg H 2 O evaporated = 2,000 kg fresh feed %
0.2 kg bone-dry solids
kg fresh feed
kg H 2 O evaporated
% ( 0.80 − 0.05 )
1 − 0.80 1 − 0.05 kg bone-dry solids
= 1579 kg.
1579 kg water
30% water
Drier
5% water
Figure 3-32
= 1579 kg H 2 O evaporated %
%
124
Consider
a
balance
around the drier inside
the recycle loop.
Amount of the mixture
entering the drier
1
kg H 2 O evaporated
0.30
0.05
(
−
)
1 − 0.30 1 − 0.05 kg bone-dry solids
kg mixed feed
= 6000 kg
(1 − 0.30) kg bone-dry solids in the mixed feed
National Engineering Center CEEC Seminar June 2 -4, 2004
R
2000 kg
Mixe
r
2000+R
E
R+P
Drier
P
Figure 3-33
Recycle
2000 kg
Mixer
6000 kg
DRAFT
Figure 3-34
Consider a mass balance around the mixer
Recycle = 6000 - 2000 = 4000 kg
Therefore,
the recycle-to-feed ratio = 4000 = 2.0
2000
Part 2: Using the algebraic method.
Let
R = recycle
P = product
E = water evaporate
Overall mass balance (outside the recycle line)
2000 = E + P
Bone-dry solid balance:
(1 - 0.8) (2000) = (1 - 0.05)P
P=
(0.2)(2000)
= 421 kg
0.95
Solving for E
E = 2000 - 421 = 1579 kg
Balance around the drier (inside the recycle loop)
Mass Balance I: No Chemical Reactions Involved
125
Bone-dry solid balance
(1 - 0.3) (2000 + R) = (1 - 0.05)(R + 421)
0.7(2000 + R) = (0.95)(R + 421)
Solving for R
R = 4000 kg
and
recycle 4000
=
=2
2000
feed
BYPASS
The bypass arrangement is closely related to recycle. A main
stream is split into two. One stream goes to the process equipment while the other is mixed with the outlet stream from the
equipment. Fig. 5-8 shows this.
DRAFT
Bypass stream
Net Feed
Feed
Process
Equipment
Product
Figure 3-35
The splitter consists of control valves. Bypassing a stream is
useful in attaining precise control of concentration. In general,
it is easier to make a large change in concentration or property
in a small mass rather than make a small change in a large
mass. For instance, if we desire to lower the water content of
an air-water mixture, which could be done by cooling the
mixture to condense part of the water vapor, it is easier to use
a bypass stream. A small portion of the air is cooled to the
appropriate temperature and is mixed with the bypassed
stream after water removal. Aside from attaining accurate
control, energy is also conserved.
A dehydrating agent can also be used for H2O removal. A
bypass stream can be used as a control to offset the gradually
diminishing power of the dehydrating agent.
Another example of the use of a bypass is in lowering the
salt content of a saline solution by evaporation. It will not be
necessary to evaporate all the water from the entire solution.
Only a portion of the saline water is evaporated. The rest
bypasses the evaporator and is mixed with the salt-free water
from the evaporator, resulting in the desired lower salt
content.
Mass balance calculations involving bypass streams are
similar to that for recycle operations. Either the arithmetic or
the algebraic method is suitable. Mass balances can be taken
126
National Engineering Center CEEC Seminar June 2 -4, 2004
around four positions: (See Fig. 3-36.)
1
Process
Equipment
3
2
4
Figure 3-36 Mass balances in bypass operation (1) around the
whole system, (2) around the process within the
bypass loop, (3) around the point of splitting and
(4) around the point of mixing.
Comparing the balance around the whole system and that
within the bypass loop, we can see that a bypass stream
results in the use of a smaller-sized equipment in contrast to
that of a recycle stream where the equipment capacity has to
be larger.
DRAFT
Example 3-13
We desire to lower the n-hexane content of a nitrogen-hexane
mixture at 45oC (the partial pressure of n-hexane = 310 mm
Hg) and a total pressure of 757 mm Hg to a mixture containing 15% mole n-hexane. This could be achieved by chilling the
mixture to 10°C to condense out some of the n-hexane. The
gas leaves saturated at 10oC (the partial pressure of n-hexane
= 64 mm Hg) and at a pressure of 750 mm Hg. The bypass gas
then mixes with the gas from the chiller to form the 15%
mixture. 300 m3/min of the original gas mixture is to be
treated.
Bypass stream
15% n-
Net Feed
Nitrogen gas
T = 45oC
PT = 757 mm Hg
PnHex = 310 mm Hg
Chiller
hexane
T = 10oC
PT = 750 mm Hg
PnHex = 64 mm Hg
Condensed n-hexane
Figure 3-37
a.How much n-hexane is condensed in the chiller?
b.What is the volume of the gas that bypasses the chiller?
Assume ideal gas behavior.
Solution:
Basis: One minute operation.
The arithmetic method will be used since the n-hexane-free
nitrogen can be considered as a tie component.
Mass Balance I: No Chemical Reactions Involved
127
By a n-hexane balance, the n-hexane condensed is equal to
the n-hexane in the N2 entering the system (point 1) minus the
n-hexane in the product leaving the system (point 4). See Fig.
3-38.
1
2
3
4
Figure 3-38
Nitrogen entering =
%
300 m3 % 757 % 0 + 273 % 1 kmol
760 45 + 273 22.4 m3
min
(757 − 310) kmol N 2
= 6.76 kmol/min
757 kmol gas mixture
n-hexane entering =
DRAFT
6.76
kmol N2 310 kmol n-hexane
%
= 4.69 kmol/min
min
(757 − 310) kmol N2
Nitrogen balance
Nitrogen entering = nitrogen leaving
Therefore the n-hexane in the nitrogen leaving
= 6.76 kmol N 2 % 0.15 kmol n-hexane = 1.19 kmol/min
0.85 kmol N 2
The amount of n-hexane that condensed
= 4.69 - 1.19 = 3.50 kmol /min
Considering the chiller alone:
Phex= 310 mm Hg
PT = 757 mm Hg
Chiller
Phex= 64 mm Hg
PT = 750 mm Hg
Condensed n-hexane
= 3.50 kmol
Figure 3-39
Using the ratio method:
310
64
mole n-hexane condensed
=
−
mole n-hexane free nitrogen gas 757 − 310 750 − 64
The moles of nitrogen entering the chiller
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National Engineering Center CEEC Seminar June 2 -4, 2004
= 3.5 kmol n-hexane condensed
min
%
1
310
64
kmol
n-hexane condensed
(
−
)
757 − 310 750 − 64
kmol nitrogen gas
= 5.83 kmol/min
Volume entering the chiller
3
757
% 760 % 45 + 273
= 5.83 kmol N2 % 22.4 m %
0 + 273
757 − 310 757
kmol
= 258.63 m3/min
Volume that bypasses the chiller
= 300 - 258.63 = 41.37 m3/min
DRAFT
PURGE
Purge or bleed is a flow arrangement used in recycling operations where a fraction of the recycle stream becomes an
output stream. In some processes with recycle stream, a
separator removes only the desired product from the stream
going out of a process unit. The inerts and impurities, which
enter the system in the fresh feed, gradually accumulate in the
recycle stream. We can avoid the accumulation by bleeding
out or purging a portion of the recycle line. The arrangement
is shown in Fig.5-13.
Recycle stream
Process
Equipment
Feed
stream
Gross
product
Purge stream
Separator
Product
Figure 3-40
If the desired product does not contain any inert material or
impurity, then by a component balance around the whole
system (outside the recycle loop) the inerts and impurities
entering in the feed stream is equal to the inerts and impurities in the purge stream.
An example of a process where a purge stream is applied is
in the manufacture of ammonia from N2 and H2. The nitrogen
comes from the air and contains inert gases such as argon.
The mixture goes to a reactor in which the reaction goes only
to about 70% completion. The product gases lead to a cooler
where most of the ammonia condenses out and is separated. If
the residual gas containing N2, H2, NH3, and inerts is recycled
Mass Balance I: No Chemical Reactions Involved
129
and mixed with the feed, the inerts will gradually accumulate
in the system. By providing a purge stream, the concentration
of the recycle line is kept constant.
Another example is in the use of recirculating cooling water
in refrigeration or air-conditioning systems. The water used
usually contains dissolved mineral impurities. As the water
recirculates, some is lost by evaporation and make-up water
has to be added. Since the impurities are nonvolatile, they will
accumulate and cause formation of scale deposits within the
system. To prevent the accumulation, a portion of the water is
bled-off or purged.
Calculations involving purge are similar to that for recycle
problems. They should not pose any problem as long as proper
mass balances are set up.
Example 3-14: Purge
DRAFT
A cooling tower cools water by contacting it with air intimately.
Some water evaporates. It gets the latent heat required from
the bulk of the liquid water. This provides the cooling effect.
This evaporated water is replenished by addition of make-up
water. The water used normally contains CaCO3. To avoid
scale formation in the process pipes, the CaCO3 should be
kept below 130 ppm. This is done by providing a purge
stream.
In a certain cooling tower, the make-up water contains 30
ppm CaCO3. The evaporation is 1,200 kg/h. What should be
the minimum flow of the purge stream in order to prevent
deposition of CaCO3? Assume no entrainment of water in the
air occurs.
Process
Make-up water M
30 ppm CaCO 3
Evaporation
= 1200 kg/h
Cooling
tower
P
Purge
130 ppm CaCO 3
Figure 3-41
Solution:
We will use the algebraic method.
130
National Engineering Center CEEC Seminar June 2 -4, 2004
Basis: 1 h
Let M = kg make-up water
P = kg purge
Overall mass balance
M = P + 1200
CaCO3 balance
(30 % 10 −6 )M = (130 % 10 −6 )P + (0)E
Solving the system of equations,
30(P + 1200) = 130P
100P = 36, 000
DRAFT
P = 360 kg/h
PROBLEMS
3-1
We desire to make up 100 kg of a solution containing 40% acid in
water. Two solutions are available, a 15% acid solution and a 65%
solution. How many kg of each solution is required?
3-2
The used mixed acids coming from three different sources has the
following compositions (by mass %):
a. 30% HCl, 10% HNO3, and 60% H2O
b. 50% HCl, 45% HNO3, and 5% H2O
c. 5% HCl, 80% HNO3, and 15% H2O
If 120 kg of (a), 300 kg of (b) and 150 kg of (c) are mixed together,
what is the composition of the resulting mixture?
3-3
When pure furfural (solvent) is added to a mixture containing 0.2
mass fraction diphenylhexane and 0.8 mass fraction docosane, two
separate layers can be obtained. If the composition of the two
layers are:
Upper layer
Lower layer
furfural
diphenylhexane
docosane
furfural
diphenylhexane
docosane
mass fraction
0.88
0.101
0.019
0.004
0.104
0.892
How much solvent was added to 500 kg of solution of docosane
and diphenylhexane? What are the amounts of the layers?
Mass Balance I: No Chemical Reactions Involved
131
3-4
Three solutions each composed of KNO3, NaNO3 and H2O with the
following compositions:
% KNO3
%NaNO3
M1
6
25
M2
29
57.5
M3
35
5
DRAFT
are mixed together and then heated until some of the water evaporates. The final solution contains 39% KNO3 and 25% NaNO3. The
original masses of M1 and M2 are equal. Find the masses of each of
the original solutions and the amount of the water that is evaporated.
3-5
100 kg. of pure salt, NaCl, is added to 500 kg of NaOH solution
containing 60% NaOH and 40% H2O. The mixture is heated until
100 kg of H2O has evaporated. What is the composition of the
resulting mixture?
3-6
Two mixtures consist of NaCl, NaOH, and water. The composition
of the first is 10% NaOH, 40% NaCl and 50% water. The second
consists of 50% NaOH, 15% NaCl, and 35% water. Both solutions
are subjected to heat and lose water. The first solution weighs
1500 kg and loses 500 kg of water. When the two solutions are
mixed, the resulting mixture analyzes 35% NaOH, 40% NaCl, and
25% H2O. What is the original mass of the second mixture? How
much water evaporates from it?
3-7
A producer gas consisting of 8.4% CO2, 0.8% O2, 21.2% CO, 7.9%
H2, and 61.7% N2 at a temperature of 30°C and 770 mm Hg total
pressure, with Pwater = 25 mm Hg, flowing at a rate of 500 m3 per
minute is mixed with a natural gas consisting of 81.11% CH4,
6.44% C2H6, 2.1% C3H8, 0.74% C4H10, 0.42% CO2, and 9.19% N2 at
a total pressure of 2 atm and a temperature of 25°C and flowing at
a rate of 100 m3 per minute. Its water vapor content has a partial
pressure of 30 mm Hg. What is the composition of the resulting
mixture on a dry basis? What is the water vapor content in grams
per kmol of dry gas?
3-8
Moist air at 40°C and at a total pressure of 765 mm Hg is flowing
through a pipeline. The partial pressure of water is 20 mm Hg. In
order to measure the rate of flow, pure carbon dioxide gas is bled
into the gas at rate of 20 kg/min. At a point downstream where
complete mixing has occurred, it is found that the air contains 5%
CO2 by volume. Find the volumetric flow rate of the moist air in
m3/min. The % CO2 is on a dry basis.
3-9
A tunnel drier is designed to dry wet sand continuously. 1,000 kg
of wet sand is to be dried per hour from 30% to 9% moisture. How
much water is evaporated per hour?
3-10 Wood chips are to be dried in a rotary drier from 35 to 12%
moisture. If 10 kg of wet wood is charged per minute, how much
product can be obtained over an 8-hr period? Find the ratio kg H2O
evaporated/kg wet wood charged
3-11 A drier is being used to reduce the moisture content of bagasse to
be used for fiberboard manufacture. It contains 50% moisture.
5.75 kg of wet bagasse is charged into the drier. After 3 hours, it
was found that the weight has gone down to 3.25 kg. What is the
moisture content of the final product?
132
National Engineering Center CEEC Seminar June 2 -4, 2004
3-12 2,000 kg/h of wet crystals of naphthalene containing 10 % methanol on a dry basis is to be dried on a belt drier. How much methanol would be picked up by the air if the naphthalene should
contain only 0.1% methanol on a dry basis?
3-13 A granular material is to be dried using a batch tray drier. The
trays measure 0.5 % m % 0.5 m % 3 cm thick. The material
contains 25% moisture. The drier can remove only 200 kg of water
per hour. If the product is to contain not more than 5% moisture,
how much fresh material should be charged to the drier? How
many trays are needed? The density of the bone-dry material is
1600 kg/m3. Assume there is no shrinkage during drying.
3-14 It is desired to concentrate a 6% KN03 solution (in water) to 20%
KN03 solution. 7000 kg of product liquor is to be needed per hour.
How much solution should be charged? How much water is
evaporated?
3-15 A solution containing 15% dissolved solids is to be concentrated to
60% dissolved solids. If the evaporator will evaporate 20,000 kg of
water per hour, what must be the feed rate? How much product is
obtained per hour?
DRAFT
3-16 A evaporator is used to produce 10,000 kg/h of 50% sodium
hydroxide. The feed is 20% sodium hydroxide. How much water is
evaporated? What is the flow rate of the feed?
3-17 A mixture contains 60% styrene and 40% ethylbenzene (by mass)
flowing at 1,500 kg/h is to be separated by distillation. The distillate contains 90% styrene and the bottoms, 10% styrene. Calculate
the flow rate of the distillate and the bottoms.
3-18 1,000 kmol/h of a mixture containing 45% ammonia and 55%
water is to be distilled. The distillate contains 91% ammonia which
is 83% of the ammonia in the feed charged to the distillation
column. How much is the bottoms and the distillate?
3-19 An evaporator is concentrating solutions coming from two different
sources. The solution from the first source containing 10% NaCl
and 10% NaOH flows at the rate of 50 kg per minute. The other
solution containing 8% NaC1 and 12% NaOH flows at the rate of
70 kg per minute. The two streams are fed directly to the evaporator. If 50% of the total water is to be evaporated, calculate the
composition and the flow rate of the product.
3-20 A mixture containing 70% methanol and 30% water is to be
distilled. If the distillate product is to contain 99.9% methanol and
the bottoms product 0.004% methanol, how much distillate and
bottoms product are obtained per 100 kg of feed distilled?
3-21 100 kmol of a benzene-toluene mixture in equimolal amounts is
distilled. 96% of the benzene is obtained in the distillate product
while 90% of the toluene is in the bottom. Calculate the composition of the product streams.
3-22 A process plant needs two ethyl-alcohol-water solutions, one
containing 89% ethyl alcohol and the other 60% ethyl alcohol. A
dilute mixture containing 35% ethyl alcohol is available. The 60%
solution is withdrawn at the rate of 1000 kg/h If the bottoms is to
contain no more than 3% ethyl alcohol and that only 5% of the
Mass Balance I: No Chemical Reactions Involved
133
alcohol in the feed should be lost in this stream, calculate the flow
rates of all the unknown streams.
3-23 A mixture consisting of n-butane, isopentane, and n-pentane is to
be distilled. The following tabulation gives the composition of the
streams in mole fraction:
Component
n-butane
isopentane
n-pentane
feed
0.23
0.32
0.45
distillate
x
y
0.024
bottoms
0.129
0.3
z
Per 100 moles of feed, find the amount of the distillate and the
bottom streams. Complete the tabulation of the composition.
3-24 Ethane is to be removed in an absorber from an ethane-hydrogen
mixture by using a mixture of hexane containing 2% ethane. The
feed contains 40% ethane and 60% hydrogen. If 99% of the ethane
is removed, what is the composition of the effluent at the top and
at the bottom? The liquid to gas molar ratio is 2.
DRAFT
3-25 Aqueous triethanolamine is normally used to absorb carbon
dioxide from gaseous mixtures. A mixture containing 30% carbon
dioxide is to be stripped with a 1.5 molar solution of triethanolamine. In order to decrease the concentration of the carbon dioxide
in the mixture by 90%, how much kg triethanolamine solution
should be used per kmol of carbon dioxide entering?
3-26 In order to recover the benzene from a benzene-air mixture, the
mixture is scrubbed with kerosene in a packed tower. 8000 m3/h
of a benzene-air mixture containing 2 mole % benzene (T = 25°C, P
= 765 mm Hg) is treated with 10,000 kg/h of kerosene. If 98% of
the benzene is recovered, find the composition of the gas stream
and liquid stream leaving the tower. What is the volumetric flow
rate of the outgoing gas stream (T = 25°C, P = 759 mm Hg)?
3-27 In a plastic manufacturing concern, air is used to dry plastic
sheets in which acetone (CH3COCH3) is used as the solvent. The
acetone in the air is recovered by absorption with pure water in a
packed tower. 1000 kmol/h of an air-acetone mixture containing 4
mole % acetone is to be scrubbed to a mixture containing 0.06
mole % acetone. The tower is to operate such that the ratio,
kg absorbing H2O
= 1.5. What is the flow rate and composition of
kg acetone-free air
the outgoing liquid stream? What is the volumetric flow rate of the
outgoing gas stream if its pressure is 758 mm Hg and its temperature at 30°C? Neglect the water vapor in the outgoing gas stream.
3-28 A certain plant requires SO2 solution in water which can be
prepared by absorbing SO2 gas in pure water. The source of the
SO2 is a sulfur burner which issues gas consisting of 12% SO2,
10% O2 and 78% N2. 90 kmol/h of this gas is available. The
resulting solution is 1% SO2 by weight. The outgoing gas has a
partial pressure of SO2 equal to 2.6 mm Hg and a partial pressure
of water equal to 20 mm Hg. The total pressure is 760 mm Hg and
the temperature is 35°C. Find the liquid water requirement in
kg/h, the volumetric flow rate of the outgoing gas, and the percent
recovery of the SO2 gas.
3-29 To remove propane from a mineral oil, steam is used. The oil
contains 2% propane by mass. 0.06 kg steam is used per 100 kg of
134
National Engineering Center CEEC Seminar June 2 -4, 2004
oil-propane mixture. If 98% of the propane is stripped, what is the
composition of the mixture issuing out of the absorber? The
mineral oil is nonvolatile.
3-30 A mixture of a non-volatile oil (MW = 290) and butane (C4H10) is fed
to a packed tower. It contains 4% mole butane. Superheated
steam at 140°C fed at 1 kmol per 30 kmol butane-free oil is used.
The outgoing liquid stream contains 0.04% butane. Calculate the
composition of the leaving gas stream and the mass flow rate of the
outgoing liquid stream.
3-31 Oleic acid dissolved in corn oil (25% acid) is to be recovered using
pure propane. 100 kg/min of the oleic acid-corn oil solution is to
be treated. The extract (product stream) contains 86% acid and
negligible propane while the raffinate contains 4% acid. How much
propane should be used? What is the composition of the raffinate?
DRAFT
3-32 4,000 kg/h copra containing 55% oil (mass) is to be extracted with
16,000 kg/h of n-hexane solvent. The solvent contains 98%
n-hexane and 2% oil. The underflow contains 2.5 kg solution/kg
inerts. This solution consists of 0.2 mass fraction of oil with the
rest being n-hexane. What is the percentage recovery of the oil
from the copra?
3-33 A liquid contains 60% styrene and 40% ethylbenzene. Styrene is to
be extracted with diethylene glycol. The extraction efficiency
desired is 91%. The extract contains 10% styrene. If the feed rate is
2.4 tons per day, what is the flow rate of fresh solvent? Forty
metric tons of sugar beets analyzing 47% water, 40.5% pulp, and
12.5% sugar is to be extracted with pure water. The resulting
sugar solution is 17% sugar. 96.5% of the sugar in the beets is
recovered. If each ton of pulp retains 2.5 tons of water, find the
amount of water used and the resulting sugar solution.
3-34 Eight hundred kg/h of halibut livers containing 23% oil is
extracted with 570 kg/h of pure ether. The extracted livers is
analyzed and is found to contain 1.12% oil, 32.96% ether, and
65.92% oil-free livers. Find the composition and the weight of the
extract, and the percentage recovery of the oil.
3-35 One thousand kg of an oilseed containing 20% oil is to be extracted
with 800 kg of n-hexane containing 2% oil (n-hexane solvent is
being reused after separating the oil). 91% of the oil in the oilseed
is to be recovered. If the extract contains 55% oil, calculate the
composition of the raffinate stream. (% oil, % oil-free solids, and %
n-hexane).
3-36 Roasted copper ore consisting of 86% gangue (water-insoluble),
10% copper sulfate, and 4% water is extracted with pure water.
The extract phase contains 8% by mass of copper sulfate, 97% of
which is recovered from the ore. Per ton of gangue in the raffinate
phase, 2.2 tons of water is retained plus the unrecovered sulfate.
Per 10 tons of the ore to be processed, calculate the masses of the
three other streams.
3-37 Activated carbon is used to recover acetone from an acetone-air
mixture. The air-acetone mixture contains 15% acetone and flows
into the absorber at 180,000 liters/min at 80°C and 100 kPa. The
rate of flow of the activated carbon is at 160 kg/min. The acetone
is reduced to 2% by volume. What is the flow rate of the outgoing
Mass Balance I: No Chemical Reactions Involved
135
activated carbon with the adsorbed acetone? What is the volumetric flow rate of the outgoing gas at STP?
3-38 Silica gel is to be used to separate a mixture of butane and
butylene. The feed containing 55% butane and 45% butylene is fed
at the bottom of the column. The silica gel is introduced at the top
and travels from top to bottom. The adsorbed gases (the product) is
stripped from the silica gel. It contains 92% butane and 8%
butylene (by mol). What is the composition of the gas exiting the
adsorption column?
3-39 An activated charcoal is used to remove color from a sugar syrup.
Calculate the mass of the char required to remove 99.9% of the
colored material from 100 kg of the sugar syrup. The syrup
contains 0.003 mass fraction of color. The spent char contains
0.002 mass fraction of color.
DRAFT
3-40 A calcium carbonate slurry is to be filtered in a plate-and-frame
filter press with 15 frames measuring 90 cm × 120 cm × 2.5 cm.
The solids in the slurry is 5 mass percent. The filter cake is 32%
porous. Assuming that the filter cake is incompressible, how much
volume of the slurry is required to fill the filter press? The specific
gravity of the solids in the slurry is 2.63. The density of the liquid
is 1.0. Assume that the specific gravity of the slurry is almost equal
to 1.
3-41 A slurry to be filtered is composed of 4% (by volume) solids. The
density of the liquid component is 1.04 g/cm3 and that of the
solid, 2.8 g/cm3 If the 1,000 kg of the slurry is to be filtered, calculate the volume of the filter cake. (porosity = 30%) and the
minimum amount of filtrate that could be obtained. Hint: The
minimum amount of filtrate corresponds to the condition when the
void volume of the filter cake are filled up with the liquid.
3-42 A centrifuge is used to separate a solution containing 45% oil and
55% water by mass at a rate of 100 kg per hour. The specific
gravity of the oil is 0.9 while that for water, 0.998. The oil fraction
contains 5% volume water. If the oil recovery is 95%, what is the
flow rate of the oil fraction?
3-43 Coconut water can be concentrated to 20% solids by freeze concentration wherein water freezes and leaves most of the solid constituent in the liquid portion. Coconut water normally contains 1.5%
solids. How kg product (20% solids) can be obtained from 1000
liters of fresh coconut water if every kilogram of the water frozen
carries with it 0.01 kilogram of liquid solution? Assume a density
of 1 g/cm3 for all liquids.
3-44 Sea water can be desalinated by using reverse osmosis. A plant
treats 4,000 m3/day of feed water containing 27,000 ppm of salinity. The product salinity is 300 ppm. What is the product recovery
(desalinated water) based on the volume of the feed? The salinity of
the brine output is 52,000 ppm.
3-45 Ultrafiltration is a membrane process used to separate solution
components on the basis of molecular size and shape. With the
solution subjected under pressure, the smaller molecules pass
through the membrane while the larger molecules are retained. We
desire to separate protein molecules with a molecular weight of
150,000 daltons from salts in solution. A fermentation broth
contains 0.01% (mass) protein and 3% (mass) inorganic salts. If
136
National Engineering Center CEEC Seminar June 2 -4, 2004
one cubic meter of the broth is reduced to 10 liters, what is the
concentration of the protein in the product in kmol per liter?
Assume that the density is essentially the same as that of water.
3-46 Foam fractionation can be used to reduce the alkyl benzene sulfonate content to 1 ppm in sewage. If 20,000 cu meters is to be
treated and the volume of the collapsed foam is 0.03 of the feed,
estimate the alkyl benzene sulfonate concentration in the feed in
ppm. The specific gravity of all liquids is essentially 1.
3-47 A double-effect evaporator is to concentrate 1,000,000 kg/day of a
liquor containing 5% solids to 40% solids. Assuming equal evaporations are obtained from such effect calculate the composition of
the solution from the first effect (if forward feeding is used) and the
flow rate of the product in kg/h. How much evaporation is
obtained?
3-48 It is desired to produce 7% NaNO3 solution continuously.
Bypass stream
DRAFT
NaNO3
Water
500kg/hr
Saturator
47.9%
NaNO3
7%
NaNO3
Figure 5-1P
The water line (NaNO3-free) is split into two. 500 kg per hour is
sent to a tank where NaNO3 is added. The mixture is stirred well to
form a saturated solution of NaNO3 (47.9%). The other line
bypasses the tank and is mixed with the 47.9% solution. What is
the flow rate of the bypass stream and the final product?
3-49 A process needs an air supply which should contain 0.12 mol
H2O/ mol dry air exactly. 1500 m3/min of air at 25°C and 101.3
kPa is to be treated. Part of this air goes to a spray chamber where
the air picks up water and goes out with 0.3 mol H2O/mol dry air.
The other part of the air feed bypasses the water spray and is
mixed with the humidified air to produce a mixture containing 0.12
mol H2O per mol dry air. What is the water consumption? What is
the ratio of the flowmeter readings of the bypass stream and the
stream to the water spray?
3-50 An air conditioning system supplies 1000 m3/min of air containing
0.01 mol H2O/mol dry air. It is at 20°C and 1 atm. To conserve
energy, part of the exhaust air containing 0.08 mol H2O/mol dry
air is recycled and mixed with the fresh air from the air conditioner
to produce a gross air feed to the room containing 0.035 mol
H20/mol dry air. How many kg of water is picked up by the air per
minute? What is the volumetric flow rate of the recycle stream?
(27°C and 99 kPa)
3-51 In Problem 3-49, if the recycle–to-feed ratio is doubled keeping the
water picked up constant, what variables will change? What are
the values?
Mass Balance I: No Chemical Reactions Involved
137
3-52 Wax from candelilla plant can be produced by solvent-extraction.
(Chemical Engineering, May 19, 1980, p. 85). The process involves
Hexane recycle
Residue with
solvent
Distillation column
Candellila
stems
300 kg/day
Recovered
Hexane
Extractor
Makeup hexane
Condenser
Wax + hexane
Wax
Residue
Figure 5-5P
DRAFT
loading candelilla stems into a vertical column and circulating
hexane down through the column to dissolve the wax. The product
is then recovered by distillation, with the hexane condensed and
recycled. In the pilot plant used, 300 kg/day of candelilla stems is
processed with a 96% recovery. The extract phase contains 40%
wax and 60% hexane. The residue from the extractor contains 10%
hexane of which 60% is recovered and recycled. The total loss of
hexane in the process is 5%. The wax, similar to carnauba, makes
up about 8% (by wt.) of the plant. How much is the kg wax recovered? What is the flow rate of the hexane entering the extractor?
3-53 Silvinite ore (43% KCl and 57% NaCl) is a source of potassium
chloride, which is recovered in the following manner:
The ore is crushed, screened, and sent to a dissolver kept at 110°C.
At this temperature the mother liquor contains 16.6% NaCl and
22.8% KCl. The rest of the ore is discarded as waste. The solution
then goes to a cooler-crystallizer where it is cooled to 30°C. KCl
crystals form. The crystals are filtered off and washed with water.
The mother liquor from the crystallizer and the wash water are
recycled to the dissolver to treat the fresh ore.
The KCl output from the washer has a composition of 88% KCl, 2%
NaCl, and 10% water. The washings contain 20% NaCl and 5%
KCl. The overall recovery of KCl is 80%. The waste ore discharged
contains 30% water. For every 1,000 kg of ore processed, 2,500 kg
of solution results. Calculate the amount and composition of the
recycle stream. How much wash water is used?
138
National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
—Chapter 4—
Mass Balance II:
Chemical Reactions
Involved
The knowledge of chemistry is important to the chemical
engineer. The development of a future chemical engineer starts
with the integration of chemistry to his knowledge. The
student should absorb chemistry as though it will be his
future profession. This is what will set him apart from other
engineers, who usually do not find use for knowledge in
chemistry.
This chapter is important because, for the first time, the
student will encounter a different approach in dealing with
chemical reactions. We discuss chemical reactions under
industrial conditions in contrast to ideal conditions assumed
in most chemistry courses. Treat the discussion as though it
were an extension of your chemistry course. As you are aware
of what is going on, conflict would not arise.
When chemical reactions are not involved in a system, we
can easily track the components since they remain unchanged
chemically. When chemical reactions occur, we have to
account for the changes in some chemical species. We have to
incorporate the balanced chemical equations in mass balance
calculations.
STOICHIOMETRY
Stoichiometry is derived from the Greek words "stoicheion,"
meaning elementary constituent, and "metrein", meaning to
measure. Stoichiometric calculation refers to the calculation of
the weights of substances involved in chemical reactions.
Most chemical reactions take place when different
substances come in contact with one another under appropriate conditions. The mass of the inputs equals the mass
outputs but some of the resulting chemical species are now
Mass Balance II: Chemical Reactions Involved
139
different. We have to use the information provided by a
balanced chemical equation, which is discussed next.
THE BALANCED CHEMICAL EQUATION
A chemical equation expresses what happens during a chemical reaction. For example, methane reacts with oxygen to form
carbon dioxide and water. Instead of describing or writing
what happens during the chemical change, we can impart the
same idea by using chemical symbols written in the form of a
chemical reaction:
CH4 + 2O2 d CO2 + 2H2 O
Eq. 4-1
DRAFT
Equation 4-1 is a compact representation that shows that one
molecule of methane reacts with two molecules of oxygen to
form one molecule of carbon dioxide and two molecules of
water.
To some students, chemistry is daunting because the
impression is that they must memorize the reactants and
products in chemical reactions. The fear is baseless since the
chemical changes represented by chemical equations are
determined from experiments (and most of these are reported
in journals). These are compiled in references and even in
encyclopedias (e.g., The Encyclopedia of Chemical Reactions.)
It will not hurt if students memorize the common ones. The
ease of dealing with chemical reactions comes with practice.
As mentioned above, chemical engineering students do not
have problems with the concept. From his chemistry subjects,
he knows that chemical equations should be balanced, that is,
the number and type of elements on the left side of the
equation is exactly the same as that on the right side.
Equation 4-2 represents a balanced chemical equation
a A + b B + c C + ... t m M + l L + n N + ...
Eq. 4-2
where a, b, c, ... m, l, n ... are the number of moles of reactants
A, B, C, ... and products M, L, N, ..., respectively. By the law of
conservation of mass, the sum of the masses of compounds on
the left side of Eq. 4-2 equals the sum of the masses of the
compounds on the right. The compounds that react form other
compounds but the elements remain the same. The number
and types of atoms of each element on the left are identical to
the number and types of atoms on the right, although
combined differently.
A balanced equation gives the exact proportion of the
reactants and products of a chemical reaction. a moles of A
react with b moles of B to produce m moles of M, n moles of N,
and so on. The relationship is fixed. Consider the equation
pyrites + oxygen d ferric oxide + sulfur dioxide
The balanced chemical equation is
140
National Engineering Center CEEC Seminar June 2 -4, 2004
4 FeS 2 + 11O 2 d 2Fe 2 O 3 + 8 SO 2
Eq. 4-3
Eq. 4-3 shows that 4 moles of FeS2 react with 11 moles of
oxygen to produce 2 moles of Fe2O3 and 8 moles of SO2. It also
tells us that each side of the equation has four atoms of iron,
22 atoms of oxygen, and eight atoms of sulfur.
Although a balanced equation provides the number and
types of atoms involved in a reaction, we still have to perform
calculations for a mass balance. We have to use the molecular
weights. For instance, referring to Eq. 4-3, how many kg SO2
and kg Fe2O3 will 150 kg FeS2 produce? How much O2 is
needed? We have to apply the balanced chemical equation.
Take a basis of 150 kg of FeS2 and convert to kmol.
(The molecular weight of FeS2, [MW FeS2] = 55.45 + 2(32) =
119.45).
kmol FeS 2 =
150
= 150 = 1.256
MW FeS 2 119.45
From the balanced chemical equation, 4 moles of FeS2 will
produce 8 moles SO2. Using ratio and proportion,
DRAFT
SO 2 produced = 1.256 % 8 = 2.51 kmol
4
= 2.51 % [MW SO 2 ] = 2.51 % 64 = 160.6 kg
Using a similar procedure, we can find the Fe2O3 produced
and O2 needed.
O 2 needed = 1.256 % 11 % [MW O 2 ] = 110.5 kg
4
Fe 2 O 3 needed = 1.26 % 2 % [MW Fe 2 O 3 ] = 100.1 kg
4
CHEMICAL OR GRAVIMETRIC FACTORS
We used ratio and proportion to solve the problem above. We
can systematize this procedure by using chemical or gravimetric factors. In Eq. 4-4, the chemical compounds are A, B, C, D,
with the respective a, b, c, and d moles of each.
a A+b Bd c C + d D
Eq. 4-4
The stoichiometric masses of each compound based on the
equation are a[MW A], b[MW B], c[MW C], and d[MW D], where
MW is the molecular weight. We can set up mass ratios of any
two of the four compounds, such as:
Mass A = a % [MW A]
Mass C c % [MW C]
Mass D = d % [MW D]
Mass B b % [MW B]
Mass C = c % [MW C]
Mass B b % [MW B]
Mass Balance II: Chemical Reactions Involved
141
and so on.
We refer to each ratio a gravimetric or chemical factor. To
illustrate its use in mass balance calculations, let us consider
the reaction given previously. (Eq. 4-3)
4FeS 2 + 11O 2 d 2Fe 2 O 3 + 8SO 2
Again, given 150 kg of FeS2 find the oxygen needed and the
Fe2O3 and SO2 produced. The calculation starts with 150 kg
FeS2 and uses appropriate gravimetric factors.
O2 needed
= 150 kg FeS 2 %
mass O 2
mass FeS 2
= 150 kg FeS 2 %
11 [MW O 2 ] kg
4 [MW FeS 2 ] kg
= 150 kg FeS 2 %
11 (32) kg
4 (119.45) kg
= 110.5 kg O 2
DRAFT
Fe2O3 produced = 150 kg FeS 2 %
mass Fe 2 O3
mass FeS2
= 150 kg FeS 2 %
2 [MW Fe 2 O 3 ] kg
4 [MW FeS 2 ] kg
= 150 kg FeS 2 %
2 (158.9) kg
4 (119.45) kg
= 99.77 kg
SO2 produced
= 150 kg FeS 2 %
mass SO2
mass FeS 2
= 150 kg FeS 2 %
8 [MW SO 2 ] kg
4 [MW FeS 2 ] kg
= 150 kg FeS 2 %
8 (64) kg
4 (119.45) kg
= 160.74 kg
In the above calculations, we started from a given amount of
FeS2. Then we use gravimetric factors that have MW FeS2 as
the denominator. The molecular weights of the desired quantities are the respective numerators.
The gravimetric factor is convenient to use in instances
where we have to calculate a mass balance without writing the
balanced chemical equation. It is also useful in finding the
equivalent amount of a substance in terms of another
substance. An example is finding the K2O equivalent of, say 50
142
National Engineering Center CEEC Seminar June 2 -4, 2004
kg of KCI. By using the gravimetric factor,
calculate the required quantity as follows:
kg K 2 O = 50 kg KCl %
= 50 kg KCl %
[MW K 2 O]
, we
2 % [MW KCl]
[MW K 2 O]
2[MW KCl]
94.19 kg K2 O
(2)(74.56) kg KCl
= 31.58 kg
Note the presence of the factor 2 in the denominator. Here, the
balance is on the number of atoms of K. Since K2O contains
two atoms of potassium, the denominator should be multiplied
by 2. The factor is based on the equation:
DRAFT
K2O + 2HCI → 2KCI + H2O
However, sometimes, finding the multiplier is not always
obvious. To be safe, write the balanced chemical equations
involved if possible and base the gravimetric factors on these
equations. Use the method when you are confident about the
stoichiometry. With experience, you can mentally see a chemical equation.
Mass balances involving chemical reactions are common in
unit processes such as combustion, oxidation, neutralization,
etc. We also encounter them in some unit operations where
hydration of compounds is involved, such as in
crystal-lization. In some cases, we use stoichiometric relationship even if chemical reactions are not involved. Example 4-1
illustrates this.
Example 4-1
We desire to manufacture a fertilizer with a formula 2-12-6
(These numbers represent total nitrogen expressed as % N,
available phosphate expressed as % P2O5, and soluble potash
expressed as % K2O, respectively.) from ammonium nitrate,
calcium phosphate, potassium chloride of 90% purity, and
inert fillers. How much of each material should we mix to
obtain 1000 kg of the final product?
Solution:
NH4NO3
Ca3(PO4)2
KCl 90% pure
Mixer
Mixed fertilizer
2% N
12% P2O5
6% K2O
Inert materials
Figure 4-1
Although the problem is plain mixing, the use of gravimetric
factors or chemical factors is essential. For fertilizer, the
Mass Balance II: Chemical Reactions Involved
143
essential components N, P, and K are not expressed in terms
of the elements themselves, but in terms of other compounds
that may or may not be present (such as P2O5 and K2O). The %
P2O5 and % K2O represent equivalent amounts of P and K,
respectively.
Basis: 1000 kg of product
Note that N, P, and K are tie components. N comes from
NH4NO3, P2O5 from Ca3(PO4)2, and K2O from KCl. Therefore, we
can use the arithmetic method.
kg N in product = (0.02)(1000) = 20 kg
kg P2O5 in product = (0.12)(1000) = 120 kg
kg K2O in product = (0.06)(1000) = 60 kg
molecular weights (MW):
MW of NH4NO3 = 80.06
MW of P2O5 = 141.94
MW of Ca3(PO4)2 = 310.18
DRAFT
MW of K2O = 94.20
MW of KCI = 74.55
MW of N = 14.01
kg NH 4 NO 3 needed = 20 kg %
= 20 %
[MW NH 4 NO 3 ]
2[MW N]
80.06 = 57.14 kg
2(14.01)
kg Ca 3 (PO 4 ) 2 needed = 120 kg %
[MW Ca 3 (PO 4 ) 2 ]
[MW P 2 O 5 ]
= 120 % 310.18 = 262.2 kg
141.94
kg KCl needed = 60 kg %
= 60 %
2 [MW KCl]
1
% 0.9
[MW K2 O]
2(74.55)
% 1 = 105.5 kg
0.9
94.20
By an overall mass balance,
Inert needed = 1, 000 − (57.14 + 262.23 + 105.52)
= 575.1 kg
144
National Engineering Center CEEC Seminar June 2 -4, 2004
CRYSTALLIZATION
Most often in crystallization, the materials that precipitate out
are hydrated crystals. We have to use the gravimetric factors
in the mass balance calculations. The solution that remains
after crystallization is the "mother liquor." This contains the
impurities. As the mother liquor wets the crystals, we cannot
totally free the product from impurities. To minimize the
impurities, we can wash the crystals with pure water or
another solvent before drying them. In solving crystallization
problems, we need data on solubility of substances in water or
other solvents.
Example 4-2
A hot solution containing 35% MgSO4 and 65% H2O is cooled
to 60oF. During cooling, 4 percent of the total water in the
system evaporates. How many kilograms of MgSO4·7H2O
crystals, are obtained per metric ton of the original mixture?
Data:
At 60°F, a water-MgSO4 solution contains a maximum of
g MgSO 4
0.245 g solution
. (This is the solubility of MgSO4 at 60oF.)1
The amount of MgSO4·7H2O crystals formed.
DRAFT
Required:
Water evaporated
MgSO4 solution
Crystallizer
Mother liquor
35% MgSO4
65% H2O
.
MgSO4 7H2O crystals
Figure 4-2
One ton original mixture
Basis:
We can observe that no tie or key component is involved.
Therefore, we should use the algebraic method.
Let
C=
kg MgSO4⋅7H2O crystals obtained
L=
kg mother liquor
E=
kg water evaporated
Molecular weights:
MgSO4:
1
120.37
Perry, Chemical Engineer's Handbook, 4th edition, McGrawHill Book Co., New York, page 17-10.
Mass Balance II: Chemical Reactions Involved
145
MgSO4·7H2O:246.48
H2O:
18.02
Overall mass balance:
1000 = L + C + E
Eq. 4-5
MgSO4 balance:
(0.35)(1, 000) = (L)(0.245) + C (
MW MgSO 4
)
MW MgSO 4 $ 7H 2 O
Eq. 4-6
350 = 0.245L + C ( 120.4 )
246.5
Water evaporated balance:
(0.65)(1, 000)(0.04) = E
Eq. 4-7
E = 26 kg
DRAFT
Substituting the value of E in Eq. 4-5 and solving for L,
1000 - 26 = L + C
L = 974 - C
Substituting Eq. 4-8 into Eq. 4-7
350 = 0.245(974 − C) + 120.37 C
246.5
C = 457.7 kg
Another complication in the calculation arises if a certain
amount of the mother liquor adheres to the crystals.
CHEMICAL REACTIONS IN INDUSTRIAL PROCESS
A mass balance calculation that involves only a single chemical reaction that goes to completion is simple. In actual cases,
however, the situation is far from simple. We encounter
complex reactions such as the following.
y
Consecutive (or series) reaction,
AdBdD
y
Parallel (side or competing) reactions,
AdB
AdD
AdS
BdS
A+B d D
146
A+B d E
National Engineering Center CEEC Seminar June 2 -4, 2004
y
Reversible reactions. The products react to form the
reactants. The reaction apparently does not go to
completion.
A+BdC+D
or
C+DdA+B
y
A+B î C+D
Mixed reactions. This is a combination of consecutive,
parallel, and/or reversible reactions
As a result, the products obtained contain side-products
from undesired reactions and unreacted or excess reactants.
Furthermore, in actual practice the reactants are seldom pure.
Consider the earlier reaction given.
DRAFT
4FeS 2 + 11O 2 d 2 Fe 2 O 3 + 8 SO 2
Eq. 4-8
Finding the stoichiometric amounts of the compounds
involved is simple. Only the gravimetric factors are necessary.
However, in actual practice many complications arise. For
instance the reaction may not go to completion. In this case
the products will contain unreacted O2 and FeS2. To increase
the yield, we have to use excess O2. Moreover, the reactants
are not pure. Pyrites contain about 20% impurities that may
include sulfides of mercury and copper. To burn pyrite, we use
air (21% O2 and 79% N2) instead of pure oxygen because it is
expensive. Therefore, N2 is found in the gaseous products.
Aside from the above reaction, the following reaction also
occurs:
4FeS 2 + 15O 2 d 2 Fe 2 O 3 + 8 SO 3
Eq. 4-9
Eq. 4-9 accounts for around 10% of the two reactions.
Because of the above factors, the products will contain
several substances regarded as impurities, such as side
products and unreacted excess reactants. To complicate
matters, the temperature, pressure, and concentration affect
the nature of the reaction (in terms of completion and yield).
EXCESS AND LIMITING REACTANTS
When the reactants are in stoichiometric ratios, the amount of
each reactant is the "theoretical" amount of that particular
reactant.
To hasten or promote the reaction, we have to supply one
reactant in excess (amount more than the theoretical needed).
This reactant is the "excess" and the other, the "limiting"
reactant. The reactant kept in excess is usually the cheaper
one. For instance, in combustion, excess air is used to burn
the fuel, which is the limiting reactant.
CONVERSION AND YIELD
Mass Balance II: Chemical Reactions Involved
147
In evaluating the efficiency of a process based on chemical
reactions, we use conversion and yield. “Conversion” is the
fraction of the limiting reactant transformed or converted to
products. The "yield" refers to the ratio of the actual amount of
desired product to the amount of product had the reaction
gone to completion. For multiple products, a conversion can
refer to one product and other conversions for the other
product(s). The conversion to a specific product has to be
specified. One hundred percent yield refers to the fact that the
expected amount of desired products has been obtained.
Undesired side reactions, incomplete reactions, and chemical
equilibrium are often the factors that reduce yield.
COMPARISON OF IDEAL AND INDUSTRIAL REACTIONS
Table 4-1 shows a summary of the comparison between an
ideal or theoretical reaction and actual industrial reaction. The
degree of difficulty in solving problems depends on the actual
condition of the reaction. In mass balance calculations involving industrial reactions, we should consider all the factors
mentioned in the said table.
DRAFT
Table 4-1
Ideal (Theoretical Reaction)
Industrial Reaction
Complete reaction
Seldom complete
Stoichiometric amounts of
Some reactants are in excess
reactants are used
Pure materials are used or are
Raw materials are impure or
assumed to be used.
mixed with other substances
Assumed to occur at only one
Occurs over a wide range of
condition
temperature, pressure, and
concentration.
No competing side reactions are
Several reactions may occur
involved
simultaneously.
Stoichiometric amounts of
The products will contain side
products are obtained.
products and unreacted or excess
reactants.
Example 4-3
Sodium nitrite is manufactured based on the following
equation:
Na2CO3 + 2NO + 12 O2 d 2NaNO2 + CO2
The Na2CO3 used is first dissolved in water to form a 40%
solution. The NO and O2 used are 10% in excess of the
theoretical requirement and the reaction is 90% complete. Per
100 kg of Na2CO3, calculate:
a. the amount of NaNO2 obtained
b. the composition of the resulting solution
c. the composition and volume of the issuing gas (30°C, 1
atm)
Solution:
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National Engineering Center CEEC Seminar June 2 -4, 2004
Basis: 100 kg Na2CO3
Molecular weights:
MW Na2CO3 = 105.99
MW NO = 30.01
MW O2 = 32.00
MW NaNO2 = 69.00
MW CO2 = 44.01
2 MW NaNO2
% 0.9
a. NaNO2 obtained =100 kg Na2 CO3 %
MW Na2 CO3
2(69.00)
= 100 %
% 0.9 = 117.2 kg
105.99
b. Composition of the resulting solution
Na 2 CO 3 unreacted = (1 − 0.9)(100) = 10 kg
H 2 O in solution = 100 kg Na 2 CO 3 %
Composition:
kg
117.17
10
150
DRAFT
NaNO2
Na2CO3
H2 O
60 kg H 2 O
= 150 kg
40 kg Na 2 CO 3
%
42.27
3.6
54.12
c. Composition of gaseous stream
NO out = 100 kg Na 2 CO 3 %
= 100 %
2 MW NO % (0.1 + 0.1)
MW Na 2 CO 3
2(30.01)
% 0.2 = 11.33 kg
105.99
1
2 MW O2
(0.1 + 0.1)
O2 out = 100 kg Na2CO3 % MW
Na CO %
2
= 100 %
1
2 (32)
105.99
3
% 0.2 = 3.02 kg
CO 2 produced = 100 kg Na 2 CO 3 %
MW CO 2
% 0.9
MW Na 2 CO 3
= 100 % 44.01 % 0.9 = 37.37 kg
105.99
Gas
NO
O2
CO2
total
kg
11.33
3.02
37.37
51.72
kmol
0.38
0.09
0.85
1.32
mol %
28.79
6.82
64.39
100
gas volume (at 30oC, 1 atm)
3
= 1.32 kmol % 22.4 m % 30 + 273
0 + 273
kmol
= 32.82 m3
Mass Balance II: Chemical Reactions Involved
149
Example 4-4
Commercial hydrochloric
produced by the reaction:
acid
and
sodium
sulfate
are
2 NaCl (s) +H 2 O (g) +SO 3 (g) d Na 2 SO 4 (s) + 2HCl (g)
The reaction is carried out at 482oC at which temperature
NaCI and Na2SO4 exist as solids, and H2O, SO3, and HCl are
gases. After separation of the Na2SO4, water will absorb the
HCl. Per 1000 kg of sodium chloride charged, how much
Na2SO4 and 37% by weight hydrochloric acid solution are
obtained if the reaction goes to 90% completion? If the SO3 is
supplied 20% in excess of the theoretical amount and the
water 100% in excess, how much 98% H2SO4 (by weight) and
H2O should be fed into the reactor?
Solution:
Products
DRAFT
Reactor
NaCl
SO3
90% conversion
H2O
Figure 4-3
Reaction:
2NaCl + H2O + SO3 → Na2SO4 + 2HCl
Basis: 1000 kg NaCl
MW NaCl: 58.44
MW H2O: 18.02
MW SO3: 80.06
MW Na2SO4: 142.04
MW HCl: 36.46
MW H2SO4: 98.08
kg Na2SO4 obtained
= 1, 000 kg NaCl%
MW Na 2 SO4
% 0.9
2 MW NaCl
= 1, 000 % 142.04 % 0.9 = 1, 094 kg
2(58.44)
kg 37% HCl solution
1kg HCl solution
= 1, 000kg NaCl % MW HCl % 0.9 %
0.37 kg pure HCl
MW NaCl
= 1, 000 %
(36.46)(0.9)
= 1, 518 kg
(58.44)(0.37)
kg 98% H2SO4 needed
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National Engineering Center CEEC Seminar June 2 -4, 2004
= 1, 000 kg NaCl %
%
1.2 kg total SO3 required
MW SO3
%
2 MW NaCl
kg theoretical SO3
kg 98% solution
MW H 2 SO 4
%
MW SO 3
0.98 kg pure H 2 SO 4
80.06 % 1.2 % 98.08 % 1
(2)(58.44)
80.06 0.98
= 1,027.5 kg
= 1, 000 %
kg theoretical and excess H2O needed
= 1, 000 kg NaCl%
= 1, 000 %
MW H2 O
% (1 + 1)
2 MW NaCl
18.02 % 2 = 308.4 kg
(2)(58.44)
H2O in 98% H2SO4
= water in solution + chemically combined water
DRAFT
= (1027.5)(0.02) + (1027.5)(0.98) % 18.02
98.08
= 205.6 kg
H2O that should be supplied
= 308.35 - 205.56 = 102.79 kg
The calculations in Examples 4-1 to 4-4 were in mass units
and employed the gravimetric factors. In some problems, the
mole units are more convenient. The next example makes use
of both mass and mole units. Note that the basis used is
consistent throughout.
Example 4-5
Phosphorus can be prepared by reacting phosphate rock,
sand, and coke in an electric furnace. A phosphate rock
consisting of 70% Ca3(PO4)2 reacts to form SiO2. The inerts is
7%. SiO2, C, and Ca3(PO4)2 react to form a gaseous product
consisting of P, CO2, and CO, and a slag consisting of CaSiO3,
inerts, excess sand, unreacted carbon, and ash from coke.
Sand is supplied 6% in excess of the amount that should be
added to the SiO2 already present in the rock for complete
reaction to CaSiO3. Coke is added 20% in excess of the
theoretical amount for complete conversion to CO2. 95% of the
carbon forms CO2 and the remainder CO. The coke used
contains 90% carbon and 10% ash. Per 1000 kg of the rock
fed, what are the masses and compositions of the outgoing
stream?
Solution:
Mass Balance II: Chemical Reactions Involved
151
Required:
Masses and composition of the different streams per 1000
Phosphate rock
70% Ca3(PO4)2
23% SiO2
7% inerts
Gas product
P
CO2
CO
Electric Furnace
Sand, 6% in excess
Coke, 20% in excess
90% C 10% ash
Slag
CaSiO3
Inerts
Excess sand
Unreacted carbon
Ash from coke
Figure 4-4
kg of rock fed.
The major chemical reaction involved on which the theoretical and excess amounts are based is
Ca3 (PO4 ) 2 + 3SiO2 + 52 C d 3CaSiO3 + 2P + 52 CO2
Eq.4-10
DRAFT
which is the desired reaction.
The other minor reaction involves the formation of carbon
monoxide:
Ca 3 (PO 4 ) 2 + 3SiO 2 + 5C d 3CaSiO 3 + 2P + 5CO
Molecular weights:
MW Ca3(PO4)2: 310.18
60.08
MW SiO2:
MW C: 12.01
MW CaSiO3: 116.16
Basis:
Eq. 4-11
MW P:
30.97
MW CO2:
44.01
MW CO:
28.01
1000 kg of phosphate rock
Constituents of the rock:
kg Ca3(PO4)2 = 700 kg
kg SiO2 = 230 kg
kg inerts = 70 kg.
In this problem, the balanced chemical reaction has been
set up and it is convenient to use mole units. The following are
based on the major stoichiometric equation:
Ca3(PO4)2 in rock =
700 = 2.26 kmol
310.18
CaSiO3 formed = 3(2.26) = 6.78 kmol
P formed = 2(2.26) = 4.52 kmol
SiO2 needed to react with all of the Ca3(PO4)2
= 3(2.26) = 6.78 kmol
SiO2 available in rock = 230 = 3.83 kmol
60.08
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National Engineering Center CEEC Seminar June 2 -4, 2004
Theoretical sand to be added = 6.78 - 3.83 = 2.95 kmol
Excess sand needed = 0.06(2.95) = 0.177 kmol
Total sand needed = 2.95 + 0.177 = 3.13 kmol
Theoretical carbon needed =
5
2 (2.26) = 5.65 kmol
Excess carbon needed = 0.2(5.65) = 1.13 kmol
Total carbon needed = 6.78 kmol
kg carbon needed = (6.78)(12) = 81.4 kg
Coke needed = 81.4 = 90.35 kg
0.9
Ash in coke = 0.10 (90.44) = 9.04 kg
Calculation of the CO2 and CO evolved:
Since the ratio of C converted to CO2 to C converted to CO is
given as 95/5,
DRAFT
Let
x = the carbon converted to CO2 (kmol)
x (5) = the carbon converted to CO (kmol)
95
Using the stoichiometric Eq. 4-10
kmol Ca3(PO4)2 equivalent to x kmol of carbon
(x) 2
5 =
converted to CO2.
Using the stoichiometric Eq. 4-11
x (5)( 1 ) = kmol of Ca (PO ) equivalent to x (5)
3
4 2
5
95
95
kmol of carbon converted to CO.
Then, using a Ca3(PO4)2 balance,
(x)( 25 ) + x (5)( 15 ) = 2.26
95
Solving for x,
x = 5.51 kmol CO 2 formed
x (5) = 0.29 kmol CO formed
95
Carbon reacted = 5.51 + 0.29 = 5.80 kmol
Carbon unreacted = 6.78 - 5.80 = 0.98 kmol
kg sand needed = (3.13)(60.08) = 188.05 kg
Coke needed = 90.36 kg.
Masses and composition of the slag:
Substances
CaSiO3
Inerts
Excess sand
Unreacted C
kmol
6.78
-0.177
0.98
mass, kg
787.56
70.00
10.63
11.77
mass %
88.58
7.88
1.20
1.32
Mass Balance II: Chemical Reactions Involved
153
Ash from coke
Total
-7.94
9.04
889
1.02
100.00
Amounts and composition of the
gaseous products:
Gas
Phosphorous
CO2
CO
Total
kmol
4.52
5.51
0.29
10.32
mol %
43.80
53.39
2.81
100.00
COMBUSTION
DRAFT
Combustion is a commonly employed industrial process for
heat generation. It is the chemical reaction between a fuel and
oxygen, which is usually from air. Fuels can be gaseous,
liquid, or solid. They may occur naturally or they may be
products or by-products of industrial processes.
Some examples of gaseous fuels are natural gas, liquefied
petroleum gas, and producer gas. Natural gas is a mixture of
methane, ethane, butane, propane, carbon monoxide, hydrogen and helium. Producer gas consists mainly of CO and H2
prepared from reaction of steam and air with coal.
Most liquid fuels are light or heavy oils that are products of
petroleum refining, such as gasoline, kerosene, fuel oil, and
naphtha, among others. A fuel from renewable resources is
ethyl alcohol produced from fermentation of sugars. Some
solid fuels are coal, wood, and charcoal.
Combustion depends on the amount of carbon and the
available hydrogen of a fuel. The reaction of these two components with oxygen produces heat, the desired effect of the
chemical reaction. (See the section on heat of combustion
discussed later in this chapter.) The usual reactions involved
are:
C + O 2 dCO 2 + heat
2C + O 2 d 2CO + heat
2H 2 + O 2 d 2H 2 O + heat
The first reaction represents the complete combustion of
carbon. The second is an incomplete reaction and should be
prevented if possible. Carbon monoxide is still combustible
and therefore, its formation means loss of energy. (CO can also
be used as a fuel.)
The reaction is
CO + 12 O2 d CO2 + heat
Hydrogen always burns completely to water.
When air is used as the source of oxygen, nitrogen is
involved in the mass balance. The reaction of N2 with O2
results in different oxides of nitrogen, which are usually
considered negligible in the calculations. These are pollutants
and government regulations limit their presence in gaseous
emissions to the atmosphere.
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National Engineering Center CEEC Seminar June 2 -4, 2004
To ensure complete reactions, we supply excess air. Excess
oxygen is the amount of oxygen that is supplied over that is
theoretically needed for complete combustion.
% excess oxygen =
% excess air =
mols excess oxygen
% 100
mols theoretical oxygen
mols excess air % 100
mols theoretical air
Obviously, % excess air = % excess oxygen.
DRAFT
Of interest in mass balance calculations are the products of
combustion, also called the combustion gases, stack gases, or
flue gases. Obviously, the common products are CO2, CO, O2,
N2, and H2O. We present the analysis in volume percent and
is on a dry basis. This is also known as the “Orsat analysis.”
If the fuel is a pure compound, then, the stoichiometric
reaction with oxygen is sufficient for mass balance calculations. For most calculations, we should segregate the carbon
from the hydrogen.
The following examples give techniques in combustion calculation starting with burning pure carbon and progressively
becoming complicated.
Example 4-6
Pure carbon is burned with theoretical amount of air. Calculate the composition of the combustion gases if
a. the combustion is complete.
b. if 95% of the carbon burns to CO2 and 5% to CO.
Solution:
Basis:
100 kmol pure carbon
a. The reaction is
C + O 2 d CO 2
kmol O2 required = 100 kmol
kmol CO2 produced = 100 kmol
Nitrogen from air = 100 % 79 = 376 kmol
21
kmol combustion gas = 100 + 376 = 476 kmol
Composition of the combustion gas:
% CO2 = 100 % 100 = 21%
476
% N2 = 376 % 100 = 79%
476
b. The reactions involved are:
C + O 2 d CO 2
Mass Balance II: Chemical Reactions Involved
155
C + 12 O2 d CO
carbon converted to CO2 = 95 kmol
oxygen reacted for CO2 formation = 95 kmol
carbon converted to CO = 5 kmol
oxygen reacted for CO formation = 2.5 kmol
theoretical oxygen required = 100 kmol
oxygen from theoretical air unreacted due to CO
formation = 2.5 kmol
nitrogen from air = 100 % 79 = 376 kmol
21
Composition of the combustion gas:
Comp.
CO2
CO
O2
N2
kmol
95
5
2.5
376
%
19.85
1.04
0.52
78.58
Example 4-7
DRAFT
100 kg of charcoal per hour is burned with 30% excess air.
The charcoal consists of 95% carbon and 5% ash. Calculate
the composition of the flue gas if the combustion is complete.
Solution:
Basis:
100 kg charcoal
carbon from charcoal = 95 kg
kmol carbon burned = 95 = 7.92 kmol
12
theoretical oxygen required = 7.92 kmol
excess oxygen = 0.3 % 7.92 = 2.38 kmol
Total oxygen used = 7.92 + 2.38 = 10.3 kmol
nitrogen from air = 10.3 % 79 = 38.7 kmol
21
Composition of the combustion gas:
CO2
O2
N2
kmol
7.92
2.38
38.7
%
16.16
4.86
78.98
Example 4-8
A liquid fuel consists of 90% carbon and 10% hydrogen by
mass. It burns in a furnace using air 20% in excess of the
theoretical amount necessary for complete combustion. (20%
excess air is also the same as 20% excess oxygen.) All of the
carbon is burned to CO2 and all the hydrogen to water. The
combustion gases exit at 300oC and at 1 atmosphere.
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National Engineering Center CEEC Seminar June 2 -4, 2004
a. What is the analysis of the combustion gases on a dry
basis?
b. What is the partial pressure of the water in the gas?
c. What is the volumetric flow rate of the combustion gases in
m3/s if 100 kg of fuel are burned per minute?
The air is substantially dry and contains 21% O2 and 79% N2.
Solution:
It is better to make use of the arithmetic solution so that the
process involved will be more clearly understood.
Basis:
100 kg fuel
Carbon in the fuel = (0.9) (100) = 90 kg
= 90 = 7.50 kmol
12
Hydrogen in the fuel = (0.1)(100) = 10 kg
DRAFT
Air
supplied 20%
in excess
Combustion
gases
Furnace
Liquid fuel
90% C
10% H
CO2
N2
O2
H2O
Figure 4-5
= 10 = 4.95 kmol
2.02
The reactions involved are:
C + O 2 d CO 2
H2 + 12 O2 d H2O
Theoretical O2 needed by C = 7.50 kmol
Theoretical O2 needed by H2 = 4.95/2 = 2.475 kmol
Theoretical O2 needed by the fuel = 7.5 + 2.475
= 9.98 kmol
Excess O2 = 0.2 (9.98) = 2.00 kmol
Total O2 supplied = 9.98 + 2 = 11.98 kmol
Nitrogen from the air = 11.98 (79/21) = 45.07 kmol
(a) The analysis of the combustion gases on a dry basis:
Mass Balance II: Chemical Reactions Involved
157
(b) The partial pressure of water in the gas:
The amount of H2O is equal to the kmol of H2 in the fuel since
all the H2O formed = 4.95
gas
CO2
O2
N2
total
kmol
7.5
2
45.07
54.57
mol %
13.74
3.67
82.59
100
total moles of wet combustion gases = 54.57 + 4.95 = 59.52
Since the total pressure is 1 atmosphere,
4.95 (101.325) = 8.43 kPa
the partial pressure of H2O = 59.52
(c) The volumetric flow rate of the combustion gases in m3/s
(300°C, 1 atm):
DRAFT
100 kg of fuel produces 59.52 kmol of combustion gases
that include the water vapor.
Therefore,
3
flow rate = 59.52 kmol % 22.4 m % 300 + 273 % 1 min
60 s
0 + 273
kmol
= 46.64 m3/s
Example 4-9
Ethane is completely burned with 30% excess air. Calculate
the composition of the combustion gases.
Solution:
The chemical reaction is:
C 2 H 6 + 72 O 2 d 2 CO 2 + 3H 2 O
Basis:
1 kmol ethane
theoretical oxygen required = 3.5 kmol
excess oxygen = 3.5 (0.3) = 1.05 kmol
total oxygen required =
3.5 + 1.05 = 4.55 kmol
nitrogen from air = 4.55(79/21) = 17.12 kmol
oxygen in combustion gases = 1.05 kmol
Composition of the combustion gases:
kmol
CO2
O2
N2
2
1.05
17.12
%
9.92
5.2
84.88
Example 4-10
A fuel gas consists of the following components: 10% CH4,
15% C2H6, 15% C3H8, 20% C4H10, and 40% C5H12 (by volume)
at two atm and 30°C. This is burned with 25% excess air
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National Engineering Center CEEC Seminar June 2 -4, 2004
(assumed dry at 30°C and at 102.06 kPa). Due to some
problems in combustion, only 90% of the total carbon burns
to CO2. The remaining 10% is converted to CO. The combustion gas is at 400°C and at 100.66 kPa. Calculate the:
(a)
combustion gas analysis on a dry basis
(b)
the partial pressure of water in the stack gases
(c)
cubic meters combustion gases per cubic meter
of fuel
(d)
cubic meters air used per cubic meter of fuel
Solution:
Although a chemical reaction can be written for each fuel
component, an easier method is to find the total carbon and
hydrogen in all of the components of the fuel. Proceed with the
calculation based on these amounts.
Fuel gas
90% C
10% H
2 atm 30 oC
Combustion
gases
Burner
CO 2
CO
N2
O2
H 2O
Dry air
DRAFT
25% excess
30 oC 102.06 kPa
Figure 4-6
Basis: 100 kmol of fuel
Let us make a tabulation the C and H2 in the fuel.
Components
CH4
C2H6
C3H8
C4H10
C5H12
Total
kmol
k-atom C
kmol H2
10
15
15
20
40
100
10
30
45
80
200
365
20
45
60
100
240
465
Theoretical O2 required for the C = 365 kmol
Theoretical O2 required for the H2 = 465/2 = 232.50 kmol
Theoretical O2 required = 365 + 232.5 = 597.50 kmol
Excess O2 required = 0.25(597.50) = 149.38 kmol
Total oxygen required = 597.50 + 149.38 = 746.88 kmol
N2 from air = 746.88 (79/21) = 2809.69 kmol
CO2 produced = 0.9(365) = 328.50 kmol
CO produced = 0.1(365) = 36.50 kmol
H2O produced = 465 kmol
O2 in combustion gas = excess O2 + O2 that should have
reacted with CO for the complete combustion of C to CO2
Mass Balance II: Chemical Reactions Involved
159
= 149.38 + 36.50/2 = 167.63 kmol.
Note that the excess oxygen is based on the complete
conversion of C to CO2 and H2 to H2O. If any incomplete
reaction is encountered, the unreacted oxygen goes out with
combustion gases. In this case, the unreacted oxygen is equal
to half of the amount of CO (based on CO + ½ O2 → CO2).
(a)
The combustion gas analysis:
Gas
CO2
CO
O2
N2
Total
(b)
kmol
328.5
36.5
167.63
2,809.69
3,342.32
mol %
9.83
1.09
5.02
84.06
100
The partial pressure of water in the combustion gas:
kmol H2O = 465
kmol dry combustion gases = 3342.32
total pressure = 100.66 kPa
DRAFT
465
(100.66) = 12.29 kPa
Partial pres. of H2O = 3342
+ 465
(c) Cubic meters of combustion gas per cubic meter of fuel:
3807.32 kmol of wet combustion gases is produced per 100
kmol of fuel.
Therefore, using the ideal gas law indirectly,
m 3 combustion gases
=
m 3 fuel
3
3807 kmol % 22.4 m % 400 + 273 % 101.325
0 + 273
100.66
kmol
3
30
+
273
22.4
m
%
%1
100 kmol %
0 + 273
2
1 kmol
= 170.2
(d) Cubic meters of air used per cubic meter of fuel:
746.88 + 2809.69 = 3556.57 kmol of air is used per 100 kmol
of fuel. Therefore,
m 3 air used =
m 3 fuel
3
3556kmol % 22.4 m % 30 + 273 % 101.3226
0 + 273
102.06
kmol
3
30
+
273
22.4
m
%
%1
100 kmol %
0 + 273
2
1 kmol
= 70.62
Example 4-11
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National Engineering Center CEEC Seminar June 2 -4, 2004
Burning a liquid fuel consisting only of carbon and hydrogen
results in a gas analyzing 11.72% CO2, 1.3% CO, 4.32% O2,
and 82.66% N2. From this data, calculate the analysis of the
liquid fuel. Assume theoretical amount of air is used.
Solution:
This problem is the reverse of Examples 4-6 to 4-10. This
calculation can be used to check the consistency of experimental results.
The most convenient basis is 100 mol of dry flue gas. As
usual, the flue gas contains H20 but this is not reported.
DRAFT
Components
CO2
CO
O2
N2
Total
mol
11.72
1.30
4.32
82.66
100.00
mol O2
11.72
0.65
4.32
16.69
From the gas analysis, the oxygen in the dry flue gas is
16.69 mols. This oxygen comes only from the air. Since the air
is 21% oxygen and 79% nitrogen, 82.66 mols of nitrogen
corresponds to 82.66(21)/(79) = 21.97 mol O2. The difference,
21.97 - 16.69 = 5.28 mol O2 corresponds to the water formed
by the reaction of the hydrogen in the fuel. From the reaction
H 2 + 12 O 2 d H 2 O
the H2 in the fuel = 2 (5.28) = 10.56 mol. The fuel analysis,
therefore, is
Atom
C
H
mol
13.02
21.12
Mass
156.24
21.33
Mass %
87.99
12.01
Example 4-12
Pyrites (FeS2 ore) is being used as a source of sulfur for the
production of sulfuric acid. The ore contains 88% FeS2 and
12% inert, non-volatile constituents. Air, 40% in excess of the
theoretical amount required for complete reaction according to
the chemical equation:
4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2
is being used to burn the FeS2. If all the FeS2 burns but with
only 90% conversion of the sulfur to SO2 and the remaining
10% is converted to SO3 (4 FeS2 + 15 O2 → 2 Fe2O3 + 8SO3),
what is the analysis of the combustion gases?
Mass Balance II: Chemical Reactions Involved
161
Solution:
Pyrites
88% FeS2
12% Inerts
SO2
SO3
O2
N2
Burner
Air
40% in excess
Inerts
Figure 4-7
A convenient basis is 100 kg of pyrites.
Molecular weights:
FeS2:
O2:
Fe2O3:
120.0
32.0
159.7
SO3:
SO2:
80.06
64.06
kg FeS2 = (0.88)(100) = 88 kg
kmol FeS2 = 88 = 0.733 kmol
120
DRAFT
From the reaction,
4FeS 2 + 11O 2 d 2Fe 2 O 3 + 8SO 2
Theoretical O2 needed = (0.733)( 11 ) = 2.02 kmol
4
Excess oxygen
= (0.4)(2.02) = 0.81 kmol
Total oxygen
= 2.02 + 0.81 = 2.83 kmol
N2 from the air
= (2.83)( 79 ) = 10.65
21
With 90% conversion of FeS2 to SO2
= (0.9)(0.733)( 8 ) = 1.32 kmol
4
O2 used for the SO2 = (0.9)(0.733)( 11 ) = 1.814 kmol
4
SO2 produced
From the reaction,
4FeS 2 + 15O 2 d 2Fe 2 O 3 + 8SO 3
SO3 formed
O2 needed
= (0.733)(0.1)( 8 ) = 0.147kmol
4
= (0.733)(0.1)( 15 ) = 0.275 kmol
4
Oxygen balance:
O2 in combustion gas
= total O2 available - O2 for SO2
= 2.82 - 1.815 - 0.275
= 0.73 kmol
162
- O2 for SO3
National Engineering Center CEEC Seminar June 2 -4, 2004
Composition of the combustion gas
gas
SO3
SO2
O2
N2
total
kmol
0.15
1.32
0.73
10.65
12.85
mol %
1.17
10.27
5.68
82.88
100
DRAFT
Example 4-13: Purge Stream
Coco gas (gas obtained from the pyrolysis of coconut coir dust)
can be transformed to form synthesis gas for the manufacture
of methanol. A typical synthesis gas analysis is 32.98% CO,
65.96% H2, 0.34% CH4 and 0.72% air. The synthesis gas is
sent to a converter where 7% of the reactants is transformed
to methanol according to the reaction
CO + 2H2 → CH30H
The gases from the converter are sent to a condenser and
separator where all of the methanol is removed. The gases
containing CO and H2 and the inert materials are recycled and
mixed with the fresh feed. To avoid accumulation of the
components not involved in the reaction, a portion of the
recycle gas is purged. What should be the purge to recycle
ratio to keep the components not involved in the reaction (air
and CH4) at 9.4%? What is the composition of the purge gas?
What is the % yield of CH3OH?
The mole ratio of H2 to CO is kept at 2 at all points in the
system.
Solution:
Recycle
Purge
Fresh feed
32.98% CO
65.96% H 2
0.34% CH4
0.72% air
Converter
9.4 %
air & CH4
Condenser
-Separator
Gross
product
Product
Figure 4-8
We can solve this problem in different ways. In problems
involving recycle and purge, no fixed rule for the procedure
can be given. Analysis is needed in finding the best method.
For this particular problem, it will be best to take a basis of
100 mol of gas entering the converter.
Basis: 100 mol of gas entering the converter
The gas consists of
Mass Balance II: Chemical Reactions Involved
163
9.4 mol air + CH4
90.6 mol of CO and H2
Since
mol H2
= 2, then
mol CO
mol CO = 90.6 = 30.20
3
mol H2 = 2 % 90.6 = 60.4
3
Considering a balance around the converter and following the
reaction:
CO + 2H 2 d CH 3 OH
9.4 mol air + CH 4
30.20 mol CO
60.40 mol H 2
Converter
Figure 4-9
9.4 mol air + CH 4
CH3OH
CO
H2
and with 7% conversion, the
CH3OH produced = 0.07 (30.20) = 2.11 mol
DRAFT
CO remaining = 30.20 - 2.11 =28.09 mol
H2 remaining = 60.40 - 2(2.11) = 56.18 mol
Considering a balance around the condenser-separator,
2.11 mol CH 3OH
28.09 mol CO
56.18 mol H 2
9.4 mol air +CH 4
SeparatorCondenser
2.11 mol CH 3OH
Recycle + Purge
28.09 mol CO
56.18 mol H 2
9.4 mol air + CH 4
Figure 4-10
the recycle and the purge consist of
28.09 mol CO
56.18 mol H2
9.4
mol air + CH4
-----93.67 mol total
and the analysis in % is
29.99% CO
59.98% H2
10.03% air + CH4
Considering a balance at the point of mixing of the feed and
the recycle
164
National Engineering Center CEEC Seminar June 2 -4, 2004
Recycle:
29.99% CO
59.98% H 2
10.03% air + CH 4
Feed
32.98% CO
65.96% H 2
1.06% air +CH 4
To Converter
30.20 mol CO
60.40 mol H 2
9.4 mol air + CH 4
Figure 4-11
(Note that
composition.)
Let
the
purge
and
recycle
have
the
same
F = moles feed
R = moles recycle
Overall mole balance:
F + R = 100
DRAFT
Air + CH4 balance:
0.0106F + 0.1003R = 9.4
0.0106(100 − R) + 0.1003R = 9.4
0.0897R = 9.4 − 1.06
R = 92.98 mol
The purge therefore is 93.67 - 92.98 = 0.69 mol
mol purge
= 0.069 = 0.0075
mol recycle 92.98
F = 100 - 92.97
= 7.03 mol
CO in feed = 0.3298 (7.03) = 2.32 mol
% yield of CH 3 OH =
mol CH 3 OH produced
%100
mol CO in feed
= 2.11 % 100
2.32
= 91.95%
Mass Balance II: Chemical Reactions Involved
165
DRAFT
PROBLEMS
4-1 Find the number of kmol contained in 100 kg of each of the
following compounds:
a) acrylic acid, C3H4O2
b) azoxybenzene, C12H10N2O
c) cetyl alcohol, C16H34O
d) cyclohexanol, C6H12O
e) carbon oxyselenide COSe
f) cobalt chloride, CoCl2
g) iron pentacarbonyl, Fe(CO)5
h) hexachlorodisloxane, (SiCl3)2O
i) uranium hexaflouride, UF6
4-2 Balance the following equations
a) MnO2 + KBr + H2SO4 → MnSO4 + K2SO4 + Br2 + H2O
b) KMnO4 + NaHSO3 + H2SO4
→ K2SO4 + MnSO4 + NaHSO4 + H2O
c) KMnO4 + HCl → KCl + MnCl2 + Cl2 + H2O
d) Cl2 + KI → KC1 + I2
e) H2SO4 + HI → H2S + I2 + H2O
f) Fe + AgNO3 → Fe(NO3)2 + Ag
4-3 For the following reactions, calculate the amount of the
compounds asked for: (Note: The equations are not balanced.)
a) Benzene + Oxygen → Carbon dioxide + water
per 120 kg of benzene burned, how many kilograms CO2 and
kilograms H2O are produced? How much oxygen is needed?
b) NaCl + H2O + SO3 → Na2SO4 + HC1
Per 300 kg NaC1, how much Na2SO4 and HC1 are formed?
c) Ca3(PO4)2 + SiO2 + C → CaSiO3 + P + CO2
Per 100 tons Ca3(PO4)2 how much of each of the products are
formed?
d) C2H6 + O2 → CO2 + H2O
How many kg CO2 and H2O can be produced from 13 kg of
ethane?
e) C12H22O11 (sucrose) + H2O → C6H12O6
C6H12O6 → C2H5OH (ethyl alcohol) + CO2
How much ethyl alcohol and carbon dioxide can be produced
from 500 kg of sucrose?
4-4
a) What is the mass percent of nitrogen found in pure barium
nitrate, Ba(NO3)2?
b) Find the mass percent of each element in MgCO3.
c) Find the mass percent of each element in CaSO4.
d) What is the mass percent of carbon and hydrogen in
carminic acid, C22H20O13?
4-5 A certain compound contains 26.585% potassium, 35.890%
chromium, and 38.1035% oxygen. What is its formula?
4-6 Find the amount of each of the following components, necessary to make 1,000 kg of a fertilizer containing 2% N2, 12%
P2O5, and 6% K2O:
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National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
NaNO3, 98% pure
Ca3(PO4)2
KCl, 97% pure
Inert fillers
4-7 Repeat Prob. 4-6 for the following:
Calcium nitrate, pure
Phosphoric acid
KNO3
Inert fillers.
4-8 To make 5,000 kg of mixed fertilizer containing 6% N2, 12%
P2O5 and 12% K2O, find the amount of the following ingredients necessary:
(NH4)3PO4
Ca3(PO4)2
KNO3
Inerts fillers
4-9 10,000 kg of a 5-10-5 fertilizer is available. It is desired to
convert it to a 3-12-12 fertilizer by the addition of appropriate
amounts of each of the following components:
NaNO3
CaHPO4
KCl
Inerts filler
4-10 A saturated solution of Na2SO4 at 30°C is cooled in a crystallizer to a saturated solution at 15°C. Assuming no water is
evaporated, how many kg of Na2SO4⋅10H2O is expected to be
produced from a metric ton of the solution at 30°C
Solubility data: at 30°C, 40 parts Na2S04 per 100 parts H2O
at 15°C, 13.5 parts Na2SO4 per 100 parts H2O
4-11 Recalculate Prob. 4-10 if 10% of the original water in the
solution is lost in cooling.
4-12 CaC12⋅6H2O crystals are to be obtained from a hot solution
containing 34% CaCl2, 8% inert soluble and non-volatile
material, and 58% water. The solution is cooled to 5°C and
the products withdrawn. Most of the adhering liquor are
removed by centrifuging them out until the adhering liquor is
0.1 kg per kg of CaCl2⋅6H2O crystals. The wet crystals are sent
to a drier where all of the free moisture is removed. What is
the purity of the final product? The solubility of CaCl2 at 5°C
is 62.25 parts CaC12 per 100 parts H2O.
4-13 For the following reactions, find the composition of the final
product. If the second reactant used is 10% in excess of the
theoretical amount or complete reaction. (The first is the
limiting reactant) The reaction is 100% complete and the
reactants are pure.
a. 2KCl + H2SO4 → K2SO4 + 2HCl
b. 3Na2CO3 + 3Br2 → 5NaBr + NaBrO3 + 3CO2
c. 2CuI + K2CO3 → 2KI + CO2 + Cu2O
d. CH4 + 2O2 → CO2 + 2H2O
4-14 For the following, the reactions go only to 95% completion.
Pure reactants and theoretical amounts are used. For every
500 kg of the impure products, how much of each of the
reactants are used?
a. KCl +NaNO3 → KNO3 + NaCl
b. MgO + CaCl2 + CO2 → MgCl2 +CaCO3
Mass Balance II: Chemical Reactions Involved
167
c. K2SO4⋅2MgSO4 + 2K2Cl2 → 3K2SO4 + 2MgCl2
d. C6H6 + Cl2 → C6H5Cl + HCl
e. C6H5NH2 + 2CH3OH → C6H5N(CH3)2 + 2H2O
4-15 Sodium sulfate can be manufactured from sulfuric acid and
sodium chloride by the Mannheim process according to the
reactions
NaCl + H2SO4 → NaHSO4 + HCl
NaHSO4 + NaCl → Na2SO4 + HCl
NaCl, 99.9% pure and 90% H2SO4 are used in a certain batch.
The first reaction goes to completion, while the second is only
95% complete. Theoretical amounts of chemicals are used.
For every 500 kg of NaCl used, find the amount and composition of the final cake. Assume that in heating, all of the HCl
are given off. How much 35% HCl (in H2O) solution can be
recovered if the HCl gas is absorbed by water?
4-16 Ammonium sulfate can be produced as indicated in the
following reaction:
DRAFT
(NH4)2CO3 (aq.)+CaSO4⋅2H2O → CaCO3 + 2H2O +(NH4)2SO4 (aq.)
A 40% (NH4)2CO3 solution in water is reacted with 5% excess
of CaSO4⋅2H2O to produce CaCO3 which is insoluble in H2O
and (NH4)2SO4 which is soluble in H2O. The reaction is only
95% complete. The CaCO3 and CaSO4⋅2H2O are filtered off
from the solution and the filtered solids retain 0.1 kg
solution/kg solids. What is the amount and composition of
the new liquid solution containing the product? If all the H2O
is removed, calculate the amount and purity of the final
(NH4)2SO4 product.
4-17 Calcium carbide is manufactured from limestone and coal
CaCO3(s) → CaO(s) → CO2 (g)
CaO(s) + 3C → CaC2(s) + CO(g)
Limestone containing 91 % CaCO3 an 9% inerts is available.
The first reaction is 95% complete while the second is 90%
complete. 4.8 Mg limestone and 5.85 Mg coal containing 60%
by weight are available. Find the amount and composition of
the final impure CaC2. The coal contains 7% ash that is left in
the CaC2 product.
4-18 100 kg of pure carbon is burned. For each of the following
cases, calculate the composition of the combustion gases:
a. Theoretical amount of pure O2 used; complete combustion
b. Theoretical amount of air used; complete combustion
c. 30% excess air is used: 90% of the C burns to CO2 and 10%
to CO
d. 20% excess air is used: 90% of the C burns to CO2 while the
10% goes out as unburned solid.
4-19 The following pure compounds are burned with 25% excess
of air required over the theoretical. Find the composition of
the flue gases if combustion is complete.
a. methane, CH4
b. benzene, C6H6
c. ethanol, CH3CH2OH
d. naphthalene, C10H10
e. carbon monoxide, CO
f. nitrocellulose, (C5H4O2(CH2ONO2)(ONO2)2)x
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National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
4-20 A liquefied petroleum gas analyzes 0.5% ethane, 0.1% ethylene, 16.4% propane, 2.1% propylene, 74% butane and 6.9%
butene. If the mixture is burned with 15% excess air and 95%
of the carbon burns to CO2 and 5% to CO and all the hydrogen to H2O, what is the Orsat Analysis of the flue gas?
4-21 A fuel gas supplied at 25°C and 1000 mm Hg analyzes 20%
H2, 5% CO, 35% CH4, 10% C2H6, 20% C3H8, and 10% N2. It is
burned completely with 25% excess air (755 mm Hg, 30°C).
For every 100 m3 of the fuel, calculate the volume of the flue
gas (800°C, 4 cm H2O draft) produced and the volume of the
air used.
4-22 A liquid fuel analyzing 90% C and 10% H by weight is burned
with 20% excess air. Only 90% of the carbon burns to CO2
and the rest to CO. Per 100 kg of the fuel used, find the flow
velocity of the combustion gases flowing through the stack 20
cm in diameter if it is at 200°C and 10 cm H2O draft.
4-23 The flue gases from a furnace using a liquid fuel consisting
only of carbon and hydrogen analyze 9.85% CO2, 0.52% CO,
and 4.78% O2 and 84.85% N2 on a dry basis. Find the analysis of the fuel used and the percent excess air used.
4-24 Ethyl alcohol, CH3CH2OH can be used as a motor fuel. It has
the advantage of not emitting air pollutants although its
heating value is lower. Ordinary alcohol is 92.423% by
weight alcohol, the rest being water. If alcohol is used as a
fuel using 15% excess air, what is the flue gas analysis
obtained on a dry basis if complete combustion is obtained? If
its total pressure is 755 mm Hg, what is the partial pressure
of water?
4-25A natural gas contains 0.5% H2S. To prevent the formation of
SO2 when the gas is used as fuel, the H2S is separated by
absorption and then transformed to elemental sulfur according to the following reactions:
H2S + 3/2 O2 → SO2 + H2O
SO2 + 2 H2S → 3S + 2H2O
How many Mg (megagram) of elemental sulfur can be recovered from 1 million m3 of natural gas (30°C, 2 atm) if the
absorption is 98% effective and the two chemical reactions go
to completion?
4-26 Pyrites ore is used for the production of sulfuric acid. The
pyrites is 80% FeS2 and 20% inert materials. The FeS2 is
burned with 30% excess air based on the reaction
4FeS2 + 11O2 → 2Fe2O3 + 8SO2
Only 90% 0f the FeS2 follows this reaction. The rest goes as
follows:
4FeS2 + 15O2 → 2Fe2O3 + 8SO3.
What is the composition of the gases resulting in the
combustion?
Mass Balance II: Chemical Reactions Involved
169
4-27 In Problem 4-26, the resulting gases are sent to the converter
where most of the SO2 is changed to SO3 according to the
reaction
SO2 + ½ O2 → SO3
The source of the oxygen is the O2 in the combustion gases. If
the reaction is 90% complete, what is the resulting composition of the gases from the converter?
4-28 The converter gases in Prob. 7-27 are sent to an absorber
where all of the SO3 is absorbed by the H2O to form H2SO4
solution. The plant produces 20 Mg of 60% H2SO4 in water
per day. How much pyrites is required to produce this
amount? What is the composition of the final gas issuing out
of the absorber? (Neglect the H2O in this gas.)
4-29 Producer gas is a fuel manufactured by passing steam and
limited amount of air through a hot bet of solid fuel. A typical
analysis obtained from coconut coir dust (solid fuel) is the
following (I.E. Cruz, Producer Gas from Indigenous Solid
Fuels, U.P. Engineering Research Journal, Vol. V, No. 1,
1976, p. 4):
DRAFT
6% CO2, 1% O2, 14% CO, 19% H2, and 1% CH4; residual, N2
This fuel is used for combustion.
Find the flue gas analysis if 20% excess air is used. Assume
combustion is complete.
Note: Here, the CO is considered a fuel. In computing the
theoretical air, the amount of free O2 is subtracted from
that required by the combustible materials (CO, H2, CH4).
4-30 From a boiler efficiency test data, the following flue gas
analyses were obtained:
a
b
c
d
e
f
% CO2
% O2
% CO
% N2
12.4
12.3
12.2
11.9
11.2
11.8
4.6
5.1
5.0
5.0
5.3
5.2
0.1
0.0
0.0
0.0
0.0
0.1
82.9
82.5
82.8
83.1
83.5
82.9
Calculate the % excess air and the fuel analysis (%C and %H
by wt, liquid fuel) for each set. Explain the reason for the
variation in the answers for the fuel analysis, assuming the
analysis are all correct. What is the most probable correct
value for the fuel analysis.
4-31 To absorb SO2 from a gas mixture containing 12% SO3, 7%
SO2, 6% O2, and 75% N2, a solution containing 30% SO3 and
70% sulfuric acid issues at the bottom of the tower. Part of
this stream is recycled back and mixed with water to form
98% sulfuric acid and 2% water. (SO3 + H2O = H2SO4). For
every 100 kmol/h of SO3 absorbed, calculate the recycle rate
170
National Engineering Center CEEC Seminar June 2 -4, 2004
and the amount of make-up water necessary. What is the rate
and composition of the outlet gas if all of the SO3 is absorbed?
4-32KClO4 can be produced by heating up KClO3. The reaction
that takes place is
4KClO3 → KCl + 3KClO4
450 kg of impure KClO3 containing 5% KCl is desired to be
converted to KClO4. If the reaction goes only to 95% completion, what is the composition of the final product? How much
KClO4 is obtained?
4-33 The basic chemical reaction in the making of soap is
expressed by the following:
DRAFT
+
3NaOH
+ (C17H35COO)3 C3H5 → 3C17H35COONa
C3H5(OH)3
caustic soda glyceryl stearate
sodium stearate (soap
glycerin
A 20% NaOH solution (in H2O) is used in excess of the
amount theoretically required. Per metric ton glyceryl
stearate, how much NaOH (in kg) is required? If the reaction
is 95% complete, how much sodium stearate is produced?
4-34 500 m3 per min of a fuel gas at 30°C and 800 mm Hg consisting of 15% CH4, 40% C2H6, 15% C3H8, 10% C6H6, and 20% N2
is burned with 20% excess air. All the C burns to CO2 and H2
to H2O. The air enters at a temperature of 60°C and a total
pressure of 760 mm Hg wing PH2O = 12 mm Hg. Calculate:
a. the flue gas analysis (dry basis)
b. the partial pressure of water in the flue gas
c. m3 humid air used/m3 fuel at the given conditions
PT of flue gas = 755 mm Hg.
4-35 Sodium bisulfate is produced by passing sulfur dioxide
through solutions containing small amounts of sodium bisulfite and in suspension a considerable amount of soda ash.
The reaction is:
2NaHSO3 + 2Na2CO3 + 2H2O + 4SO2 → 6NaHSO3 + 2CO2
A solution containing 5% NaHSO3 by weight is available. In
one batch, 2 metric tons of Na2CO3 is suspended in 40 metric
tons of the solution. Sulfur dioxide is passed to the system
20% in excess of the amount required for complete reaction.
The reaction goes only to 90% completion. How much NaHSO3
is formed
Make a schematic diagram with appropriate labels.
4-36 Ethylene dibromide is produced by the liquid-phase bromination of ethylene. Ethylene and bromine are fed in equimolar
amounts into a vessel where they are reacted at 35 to 85°C
and atmospheric pressure. The reaction involved is
CH2 = CH2 + Br2 → BrCH2CH2Br
Mass Balance II: Chemical Reactions Involved
171
For every 1,000 kg of ethylene dibromide product, 860 kg of
bromine is used. Two-thirds of the ethylene dibromide liquid
from the stripper is recycled to be used as the solvent in the
reactor. Another reaction that occurs is
Br2 + BrCH2CH2Br → BrCH2CHBr2 + HBr
What is the yield on the ethylene? The yield on the bromine?
What is the composition of the liquid that goes out of the
main reactor?
Nitrogen
Hydrogen
Water
Air
Reactor
Acetone
Scrubber
Isopropyl
Alcohol
Distillation column
Dist. column
DRAFT
Isopropyl alcohol
and water
Isopropyl alcohol and water
Water
Figure 4-12
4-37 Acetone is produced by the catalytic dehydrogenation of
isopropyl alcohol. Air and vaporized isopropyl alcohol are
reacted in contact with heated brass or copper catalyst at
500°C and 300 kPa. The hot reaction gases containing
acetone, isopropyl alcohol, and hydrogen pass through a
water-cooled condenser and thence into a water scrubber
where the noncondensable gases nitrogen and hydrogen are
removed. The condenser product and solution from the
scrubber are fed to the fractionating column. Concentrated
acetone is distilled off at the overhead while an isopropyl
alcohol-water mixture is removed from the bottom of the
column. The isopropyl alcohol mixture is sent to the column
where it is separated into a 91% alcohol constant boiling
solution (an azeotrope) and pure water at the bottom. The
91% isopropyl alcohol is recycled without further
purification. The reactions involved are
172
National Engineering Center CEEC Seminar June 2 -4, 2004
(CH3)2CHOH + ½ O2 (air) → CH3COCH3 + H2O
(CH3)2CHOH → CH3COCH3 + H2
The air fed is ½ the theoretical amount to react with the fresh
isopropyl alcohol. The conversion of isopropyl alcohol per
pass in the reactor is 60%.
Assuming 100% overall conversion to acetone (actual yields
are from 85-90%), calculate the amount of nitrogen and
hydrogen that issues out of the reactor for every 1,000 kg of
acetone produced. What is the amount of 91% isopropyl
alcohol recycled? What is the composition of the liquid from
the scrubber if 500 kg of water is used in the scrubber?
Assume 100% conversion.
4-38 Propylene glycol is produced by pressure hydration of propylene oxide in a tower reactor. After dehydration and vacuum
distillation, the propylene glycol is recovered as overhead
from the distillation column. Dipropylene glycol is obtained
as a by-product.
DRAFT
The reaction
CH3CH-CH2 + H2O
\ /
O
→ CH3CHOHCH2OH
For every 10,000 kg of propylene glycol produced, calculate
(a) the amount of make-up chemicals required
(b) the amount of dipropylene glycol byproduct
(c) the amount and composition of the recycle
Of the total propylene oxide entering the reactor, 98% is
converted to propylene glycol and 2% to dipropylene glycol
when the mole ratio of water to propylene = 1:2.
4-39 Ethylene glycol monoethyl ether is produced by reacting
ethylene oxide and ethyl alcohol in the presence of acidic or
basic catalysts. Diethylene glycol monoethyl ether and triethylene glycol monoethyl ether are likewise formed. Ethylene
oxide is mixed with an excess of preheated alcohol, containing recycled alcohol and fresh makeup alcohol, and the
reactants are fed to the reactor. Excess alcohol is removed as
a distillation column top product and is recycled to the
reactor. The mixture of crude glycol ethers are separated in a
series of distillation columns (under vacuum). The reaction is
O
/ \
CH2 CH2 + CH3CH2OH → CH3CH2OCH2CH2OH
+
CH3CH2OCH2CH2OCH2CH2OH
CH3CH2OCH2CH2OCH2CH2OCH2CH2OH)
Mass Balance II: Chemical Reactions Involved
+
173
Recycle
Propylene glycol
Dehydrator
Tower Reactor
Propylene
oxide
Vacuum Column
Water
Dipropylene
glycol
Figure 4-13P
DRAFT
The reaction has an 86% yield of ethylene glycol monoethyl
ether; 10% of the ethylene is converted to diethylene and 4%
to triethylene. If the recycle alcohol is twice that of the
makeup alcohol, calculate the composition of the reactor
effluent.
4-40 Ethylene oxide is made by the catalytic oxidation of ethylene
with air in the presence of a silver catalyst. Some carbon
dioxide is also formed by the complete oxidation of some of
the ethylene. Ethylene and air are mixed in a volume ratio of
1:8 and passed over a catalyst. To suppress the formation of
carbon dioxide, vent gases from the absorber is recycled to
the reactor in such quantity as to keep the ethylene concentration in the feed at 4%.
Recycle
Air
Reactor
Ethylene
Purge
AbsorberStripper
Ethylene
oxide
Figure 4-13
The effluent gases from the reactor are washed with water
under pressure in an absorber. The ethylene oxide is
absorbed and sent to a vacuum stripping column, where it is
liberated from solution and passed overhead to a fractionating column for final purification. All the water is removed
from the recycle line. The reaction is
O
/ \
CH2=CH2 + ½ O2 → CH2-CH2 + (CO2 + H2O)
174
National Engineering Center CEEC Seminar June 2 -4, 2004
If the yield is 65%, the rest being oxidized to CO2, calculate
the flow rate and composition of the recycle stream. What is
the purge rate? Use a basis of 1,000 kg of ethylene charged.
4-41 A 1:3 ratio of nitrogen and hydrogen react catalytically at
high temperature and pressure to produce ammonia. The
nitrogen source is air and the source of hydrogen is fuel oil,
natural gas, etc. The N2-H2 mixture is compressed and sent
to the reactor. At the inlet of the compressor, the fresh feed is
joined by a recycle stream of unconverted nitrogen and
hydrogen. The gases leaving the reactor are cooled and some
ammonia liquefies. Part of the gas is purged to prevent
accumulation of inert gases. The remaining gas is recycled.
The overall yield of ammonia is 85%. Only 80% of the
ammonia from the reactor is liquefied; the rest goes with the
recycled gas. The purge gas contains 11.5% inerts while the
fresh N2--H2 feed mixture contains 2.3%. On a basis of 1,000
kg of ammonia product, calculate the amounts and compositions of all streams. Calculate the conversion per pass in the
reactor.
Purge
stream
Recycle stream
H
DRAFT
F
Nitrogen A
Hydrogen
Inerts
B
Compressor
C
Reactor
D
G
Separator
E
Ammonia
Figure 4-14
Mass Balance II: Chemical Reactions Involved
175
DRAFT
176
National Engineering Center CEEC Seminar June 2 -4, 2004
--Chapter 5--
Heat Balance
THE ENERGY BALANCE EQUATION
The energy balance equation is similar to the mass balance
equation
DRAFT
Energy inputs -
Energy outputs = Accumulation
5-1
Energy exists in many forms: work, heat, kinetic energy,
potential energy, internal energy, chemical energy, light,
electricity, sound, magnetic energy, nuclear energy, etc. It can
be transformed from one form to another. The transformation
as well as the production of energy is seldom 100% efficient.
For example, in transforming the chemical energy of a fuel to
mechanical work using an engine cannot exceed 40%; 60% or
more becomes heat energy. Heat is the product of inefficiency.
While this implies that heat is a lower grade of energy, heat is
an important form of energy in processes. In this chapter, we
shall discuss only three forms of energy - heat, internal
energy, and enthalpy. We shall discuss the other forms
relevant to chemical engineering in Chapter Eight. We should
note that a mass balance is necessary when performing an
energy balance. Heat balance is simple enough to be included
in this chapter.
Our discussion is introductory. The concepts being introduced are part of thermodynamics, the principles of which are
important in chemical engineering. All other scientists and
engineers need thermodynamics. We are fitting thermodynamics in the chemical engineering knowledge seamlessly.
HEAT
Closely related to temperature changes is the transient form of
energy called heat. We can observe that whenever a difference
in temperature exists, heat transfers from a body at a higher
temperature to a body at a lower temperature. This manifestation of energy transfer is known as heat transfer. A body with
a high temperature has its molecules with more energy than
those of a body at a lower temperature. Contact of these two
bodies allows transfer of the molecular energy manifested as
heat. In mathematical equations, heat given off by a body (an
Heat Balance
177
exothermic process) is denoted by a negative quantity while
heat taken on by a body (an endothermic process) is positive.
(This is a common convention but the reverse is also used).
The unit of heat is the joule and the symbol is Q.
THE HEAT BALANCE EQUATION
The heat balance equation is
Heat inputs −
Heat outputs = Accumulation
5-2
At steady state, accumulation = zero and
Heat inputs =
Heat outputs
5-3
INTERNAL ENERGY AND ENTHALPY
DRAFT
Heat is the result of the vibrational, rotational, translational
movement of molecules. This movement of the molecules is
what we normally feel as heat and normally indicated by the
temperature of a body. We refer to this energy as internal
energy of the molecules. It is postulated that at absolute zero
(T = 0 kelvin), the internal energy of the molecules is zero. We
designate U as internal energy. Related to this quantity is
enthalpy or heat content, H.
H = U + PV
5-4
where P = pressure and V = volume
The unit of U and H is kJ/kmol in SI. The customary units are
cal/mol and Btu/lbmol.
The internal energy is a function of temperature and volume.
U = f(T, V)
5-5
dU = ØU dV + ØU dT
ØV T
ØT V
5-6
Taking the differential
At constant volume,
dU = 0 + ØU dT
ØT V
5-7
We define ØU as Cv , heat capacity at constant volume, with
ØT V
units, kJ/(kmol$K) in the SI units. The customary units are
cal/mol$ o C and Btu/lbmol$ o F. If the mol and lbmol are
replaced by g and lbmass, respectively, we refer to heat capacity as specific heat.
dU = C v dt
Integrating from U1 to U2 and from T1 to T2
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National Engineering Center CEEC Seminar June 2 -4, 2004
5-8
and
¶UU dU = U2 − U1 = U = ¶TT Cvdt for 1 kmol
2
2
1
1
U = n ¶ T1 Cv dt
T2
5-9
for n kmols
We use internal energy when the process occurs at constant
volume. Normally this is the volume of a closed rigid container.
The enthalpy is a function of temperature and pressure
H = f(T, P)
5-10
dH = ØH dP + ØH dT
ØP T
ØT P
5-11
Taking the differential,
At constant pressure,
dH = 0 + ØH dT
ØT P
5-12
ØH as C , heat capacity at constant pressure
P
ØT P
Its unit is the same as that for Cv.
DRAFT
We define
dH = Cp dt
5-13
Integrating from H1 to H2 and from T1 to T2
and
¶HH dH = H2 − H1 = H = ¶TT Cpdt for 1 kmol
2
2
1
1
H = n ¶ T1 Cp dt
T2
5-14
for n kmols
We use enthalpy when the process proceeds at constant
pressure. Industrial processes are usually carried out at
constant pressure rather than at constant volume. (Can you
imagine the difference?) Therefore the use of enthalpy is more
common. We will employ this in most of the heat balances that
we will set up.
DATUM OR REFERENCE TEMPERATURE
The true absolute values of internal energy and enthalpy are
impossible to evaluate. Therefore, relative values are used in
calculations. We do this by assigning a value of zero to internal energy and enthalpy at a reference temperature. 0°C is the
natural choice for most applications although the ambient
temperature of 20°C or 25°C is also common. In some cases,
the choice is by convention. In tables of values of enthalpy or
internal energy, 0°C is a common reference. However, Appendix 9 which gives the properties of saturated steam the reference temperature is 0.01°C at which the internal energy of
liquid water is zero.
Heat Balance
179
Since we are only interested in the changes in internal
energy or enthalpy, we can use any reference temperature as
long as we are consistent. In instances where the properties
have different reference temperatures, we should adjust them
to a common datum when they are used in solving problems.
HEAT QUANTITIES IN HEAT BALANCE WITHOUT REACTION:
SENSIBLE HEAT AND LATENT HEAT
In this chapter, we will consider two types of heat: sensible
heat and latent heat. The other types of heat will be discussed
in later chapters.
Sensible heat is the change in the internal energy or
enthalpy of a substance if it is heated or cooled from T1 to T2
without undergoing a change in state. This definition applies
to gases, liquids, and solids. Some examples in which sensible
heats are involved: Heating liquid water from 50°C to 90°C;
Heating solid metal at 300°C to 400°C; Cooling air at 120°C to
70°C.
In the following discussion, we assume that the process
occurs at constant pressure.
For a solid, sensible heat = ¶ T1 nC p,solid dT
T2
For a liquid, sensible heat = ¶ T1 nCp,liquid dT
DRAFT
T2
For a gas, sensible heat = ¶ T1 nCp,gas dT
where
T1 is the initial temperature
T2 is the final temperature
n is the mol of substance
Cp, gas is the heat capacity of a gas
Cp, liquid is the heat capacity of liquid
Cp, solid is the heat capacity of a solid
T2
For processes that take place at constant volume, we have to
replace Cp by Cv.
Latent heat is the heat required to change a quantity (mass
or mol) of a substance from one state to another. This occurs
at constant temperature and the energy involved is responsible for the rearrangement of the molecules in changing the
state.
Latent heat of fusion is the heat required to melt a solid to
form its liquid state. The reverse is latent heat of solidification,
which is the heat evolved when a liquid freezes. The two latent
heats for a particular substance are equal in magnitude but
opposite in sign. An example is the freezing of liquid water at
0°C.
Latent heat of vaporization is the heat needed to vaporize a
liquid. For example this is the heat involved in boiling liquid
water at 100°C. Latent heat of condensation is the opposite.
This is the heat given off when a vapor condenses.
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National Engineering Center CEEC Seminar June 2 -4, 2004
Latent heat of sublimation is the heat involved when a solid
changes directly to gas. This is the heat needed in the sublimation of dry ice (solid CO2) at 1 atm.
The above quantities are examples of thermophysical
properties needed in setting up heat balance without chemical
reaction. We will discuss how to obtain and use these data.
HEAT CAPACITY OF SOLIDS AND LIQUIDS
We can obtain the heat capacity of liquids and solids from
experiments. The data are available in the literature and in
databases. They vary with temperature and are satisfactorily
represented by equations
or
CP = a + bT
5-15
CP = a + bT - c/T2
5-16
DRAFT
where
a, b, and c, are constant
Appendix 2 gives the heat capacity of selected inorganic solids
in the form or over given temperature ranges. Equations correlated from experimental data are of the form CP = a + bT.
Appendix 3 gives the heat capacity of selected liquids.
Several empirical methods allow us to predict the heat
capacity of solids and liquids. Their usefulness is quite
limited.
HEAT CAPACITY OF GASES
From calculations employing statistical mechanics and the
kinetic theory of gases, the heat capacity of monatomic gases
such as krypton, argon, etc., is
Cv = 52 R
5-17
Cp = 72 R
5-18
C p − Cv = R
5-19
where
R = ideal gas constant = 8.31 kJ/(kmol·K)
For other gases, the heat capacity is a function of temperature. This temperature dependence can be experimentally
determined and expressed as empirical relations. Some of
these may be polynomials such as:
Cp = a + bT + cT 2 +dT 3 + eT 4 ...
5-20
where a, b, c, and d are constants that depend on the type
of molecules. T is the temperature in K. A tabulation of the
values of the constant a, b, c, d, and e are given in Appendix
4.
Heat Balance
181
Using the relation CP - Cv = R, we can express Cv as
Cv = a + bT + cT2 + dT3 + eT4 - R
5-21
Example 5-1
What is the heat capacity of carbon dioxide at 100°C?
Use the data on heat capacity in Appendix 4.
Solution:
From Appendix 4, the values of the constants are
a = 27.437
b = 0.0423150
c = -1.9555 × 10-5
d = 3.9968 × 10-9
e = -2.9872 × 10-13
Therefore,
Cp = 27.437 + (42.3150 % 10−3 )(373) - (1.9555% 10−5 )(373) 2
+(3.9968 % 10 −9 )(373) 3 - (2.9872 % 10 −13 )(373) 4
DRAFT
= 40.701 kJ/(kmol·K)
Example 5-2
Find the change in enthalpy involved in heating 100 kg of
carbon dioxide from 25°C to 700°C.
Solution:
H = n ¶ T 1 C p dT
T2
The CP of carbon dioxide is found in the last example.
H = 100 ¶ 298 (27.437 + 42.3150 % 10 −3 T − 1.9555 % 10 −5 T 2
44
973
+3.9968 % 10−9 T3 - 2.98732 % 10−13 T4)dT
= 100 27.437T + 42.3150 % 10
2
44
−3
T 2 − 1.9555 % 10
3
= 100 [27.437(973 − 298) + 42.3150 % 10
2
44
+ 3.9968 % 10
4
−9
−3
−5
T 3 + 3.9968 % 10
4
−9
(973 2 − 298 2 ) − 1.9555 % 10
3
(973 4 − 298 4 ) − 2.9872 % 10
5
−13
T 4 − 2.9872 % 10
5
−5
(973 3 − 298 3 )
(973 5 − 298 5])
= 71,990 kJ
HEAT CAPACITY OF MIXTURES OF GASES
The heat capacity of a mixture of gases depends on the mole
fraction of each component. The following formula is valid
when no interaction between different kinds of molecules
exists so that no heat is involved in mixing.
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National Engineering Center CEEC Seminar June 2 -4, 2004
−13
T5
973
298
Cp mix = y1 Cp 1 + y2 Cp 2 + y3 Cp 3 + ... + yn Cp n =
where
n
i=1
y i C p,i
5-22
yi: mole fraction of component i
Cp,i: heat capacity of component i
Example 5-3
Find the heat capacity of a gas mixture consisting of 12%
CO2, 3% CO, 6% O2, 74% N2, and 5% H2O at 200°C. What is
the change in enthalpy in cooling 10 kmol of this mixture from
1000°C to 200°C at constant pressure?
Solution:
C pmix = 0.12C pCO 2 + 0.03C pCO + 0.06C pO 2 + 0.74C pN 2 + 0.05C pH 2 O
Getting the values of Cp from Appendix 4,
Cpm =
0.12(27.437 + 0.042315T − 1.9555 % 10−5T2 + 3.9968 % 10−9T3 − 2.9872 % 10−13T4) +
DRAFT
0.03(29.556 − 0.006581T + 2.013 % 10−8T2 + 1.2227 % 10−8T3 + 2.2617 % 10−12T4) +
0.06(29.526 − 8.8999 % 10−3T + 3.8083 % 10−5T2 − 3.2629 % 10−8T3 + 8.8607 % 10−12T4) +
0.74(29.342 − 3.5395 % 10−3T + 1.0076 % 10−5T2 − 4.3116 % 10−9T3 + 2.5935 % 10−13T4) +
0.05(33.933 − 8.4186 % 10−3T + 2.9906 % 10−5T2 − 1.7825 % 10−8T3 + 3.6934 % 10−12T4)
= 29.36041 +1.306 % 10−3T + 8.89052 % 10−6T2 − 5.19315 % 10−09T3 + 9.40236 % 10−13T4
H = n ¶ T 1 C pm dT
T2
= 10 ¶ T 1 (29.36041 +0.001306T + 8.89052 % 10 −6 T 2 − 5.19315 % 10 −09 T 3 + 9.40236 % 10 −13 T 4 )dT
T2
2
3
4
5 473
= 10 29.36041T + 0.001306 T2 + 8.89052 % 10−6 T3 − 5.19315 % 10−09 T4 + 9.40236 % 10−13 T5
1273
= −2.787 % 105 J
MEAN HEAT CAPACITIES
Evaluating enthalpy or internal energy from empirical
equations, graphs, or tables is quite tedious as the examples
show. We can make life easier by using mean heat capacity,
CPm. We express the change in enthalpy from T1 to T2 as
T2
H = ¶ T1 CP dT = Cpm (T 2 − T 1 )(for one kmol)
5-23
Solving for the mean molal heat capacity,
¶ TT C p dT
C pm =
¶ TT dT
2
1
5-24
2
1
Heat Balance
183
Fig. 5-1 shows Eq. 5-24 graphically. The area under the curve
between T1 and T2. represents ∆H. It is also equal to the area
of the rectangle with a height CPm and a base T2 -T1. For practical purposes,
Cp vs T
Cpm
Cp
T1
T
T2
Figure 5-1
T1 is assigned 25°C, the reference or datum temperature.
Therefore, the mean heat capacity becomes only a function of
the upper temperature boundary.
T
¶ 298
C p dT
DRAFT
2
C pm =
T 2 − 298
5-26
Using Eq. 5-26 we can obtain a plot of CPm vs. T. Fig. 5-2
shows such plots for different gases. Naturally, using the
value of Cpm from these plots will give us the change in
enthalpy between 25°C and a higher temperature. However,
since we are interested only in changes in enthalpy, we can
express the ∆H in heating a gas from Ta to Tb as
H = Hb − Ha = C pm b (T b − 298) − C pm a (T a − 298)
5-27
where Cpm b = mean heat capacity for the interval 298 K and Tb
Cpm a = mean heat capacity for the interval 298 K and Ta
As Fig. 5-3 shows, the datum temperature is 25°C, where
implicitly, the enthalpy is assigned a value of zero. Hb is the
enthalpy that results in heating the substance from 25°C to
Tb, while Ha is the enthalpy that results in heating the
substance from 25°C to Ta.
We should always take note of the datum temperature used
in different tables and graphs.
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National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
25oC
25oC
Figure 5-2
Average heat capacity of some gases with datum
temperature of 25oC.
Heat Balance
185
Tb
Hb
∆H = Hb - Ha
∆H = Hb - 0
Ha
Ta
∆H = Ha - 0
H=0
25oC
DRAFT
Figure 5-3
Example 5-4
Find the change in enthalpy involved in heating 100 kg of
carbon dioxide from 25°C to 700°C using the mean capacity
values given in Fig. 5-2.
Solution:
From Fig. 5-2 , for carbon dioxide,
Cpm = 47.2 kJ/(kmol⋅K) (bet. 25°C and 700°C)
H = nC pm (T 2 − 298)
= 100 (47.2)(973 − 298)
44
= 72,400 kJ
Example 5-5:
Sensible Heat
A fuel gas burns to give a stack gas analyzing 12% CO2, 3%
CO, 6% O2, 74% N2, and 5% H2O at 600°C. If this mixture is
cooled down to 100°C in a waste heat extractor, what % of the
sensible heat of the gas (datum 25°C) is recovered? Use mean
heat capacities.
Solution:
In calculating enthalpy for mixture using CPm, it is best to use
tabulations.
The sensible heat lost by the gas is H = H600oC − H100oC, and
the % sensible heat recovered = HHo % 100.
600 C
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National Engineering Center CEEC Seminar June 2 -4, 2004
Basis: 100 kmol of gas mixture (H25°C = 0)
Components
n,
kmol
CO2
CO
O2
N2
H2 O
12
3
6
74
5
Cpm bet. 25
and 600oC,
kJ/(kmol.K)
46.2
30.6
31.9
30.3
36.3
H at 600oC =
Sensible heat recovered =
=
H
at 600oC =
nCpm(575), kJ
319,000
53,000
110,000
1,290,000
104,000
1,876,000
Cpm bet. 25
and 100oC,
kJ/(kmol⋅K)
38.9
29.2
29.7
29.2
33.8
H at 100oC =
H
at 100oC =
nCpm(75), kJ
35,000
6,600
13,400
162,000
12,000
229,700
1,876,000 - 230,000
1,646,000 kJ
1, 646, 000
% sensible heat recovered = 1, 876, 000 % 100 = 87.8%
DRAFT
PHASE CHANGES AND LATENT HEATS
The energy of the molecules at the lowest temperature, the
absolute zero, is postulated to be zero and gradually increases
as the temperature increase. The molecules acquire energy so
that they can move freely in rotational, vibrational, and translational modes. There are specific temperatures (keeping
pressure constant) where the state of a substance changes. It
can change from one crystalline solid form to another crystalline form. It can change from the crystalline solid state into
the liquid state, then the liquid state into the gaseous state.
The molecules in the liquid state move more freely than the
molecules in the solid state and thus possess more energy.
Therefore these phase changes involve the absorption or evolution of energy, e.g., changing a solid into a liquid requires heat
absorption. This energy involved in phase transitions is called
latent heat.
The energy involved in melting a solid or freezing a liquid is
the latent heat of fusion, ∆Hf, kJ/kmol. In melting, energy is
absorbed (heat is positive), while in freezing, given off (heat is
negative.) The energy involved in vaporizing a liquid is the
latent heat of vaporization, ∆Hv, kJ/kmol (+). In condensing a
vapor, the energy involved is the latent heat of condensation,
numerically equal to the latent heat of vaporization but of
opposite sign (-). The change from a solid state to the gaseous
state without going through a liquid state is also possible and
is referred to as sublimation, ∆Hsub, kJ/kmol. Experimental
values of the latent heats of several substances are available
in the literature. Experimental data on latent heats are available. Furthermore, several techniques can be used to estimate
them. For example the latent heat of vaporization can be
estimated using Trouton's Rule:
H v
BP l 88
5-28
where Hv is the latent heat of vaporization in kJ/kmol at BP,
the normal boiling point in K. The constant 88 varies at high
temperatures making the rule sometimes unreliable.
Heat Balance
187
A more accurate method of predicting the latent heat of
vaporization is the Riedel correlation
9.09(ln P c − 1)
H vn
=
BP
0.930 − BP/T c
where
BP
Hvn
Pc
Tc
5-29
= normal boiling point, K
= latent heat of vaporization at BP, kJ/kmol
= critical pressure, atm
= critical temperature, K
Example 5-6
Find the change in enthalpy involved when 10 kg of benzene
at 10°C is heated to form benzene vapor at 110°C. The
pressure is kept constant at 101.325 kPa. The melting point of
benzene is at 5.3°C.
Solution:
From Appendix 1, for benzene
DRAFT
Normal boiling point:
Critical temperature
Critical pressure
80.1°C.
288.5°C
47.7 atm
The change in enthalpy is
H = H1 + Hv ben +H2
H1 = ¶ 283 nCp liqdT
353.1
From Appendix 3 ,
Cp at 10°C = 1.63 kJ/(kg⋅K)
Cp at 80.1°C = 1.95 kJ/(kg⋅K)
Since the variation of Cp with temperature is linear,
C pm bet. 10 & 80.1o C =
C p at 10oC + C p, at 80.1oC
2
= (1.63 + 1.95)/2 = 1.79 kJ/(kg⋅K)
H1 = 10(1.79)(80.1 − 10) = 1255 kJ
Latent heat of vaporization of benzene:
9.09(ln P c − 1)
H vn
=
BP
0.930 − BP/T c
H vn = (80.1 + 273)
(9.09)(ln 47.7 − 1)
= 30,535 kJ/kmol
0.930 − 80.1 + 273
288.5 + 273
For 10 kg benzene
Hv = (10/78)(30535) = 3915 kJ
For heating the benzene vapor from 80.1°C to 110°C
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National Engineering Center CEEC Seminar June 2 -4, 2004
H = n ¶353.1 Cp dT
383
From Appendix 4
C p = − 31.368 + 0.4746T − 3.1137 % 10 −4 T 2 + 8.5237 % 10 −8 T 3 − 5.0524 % 10 −12 T 4
10 −31.368T + 0.4746 T 2 − 3.1137 % 10 −4 T 3 + 8.5237 % 10 −8 T 4 − 5.0524 % 10 −12 T 5
5
3
4
78
2
= 403.5 kJ
Therefore
DRAFT
H = 1255 + 3915 + 403.5= 5574 kJ
HEAT BALANCE WITH CHEMICAL REACTION; HEAT OF
REACTION
In this section we shall consider heat produced or consumed
by chemical reactions. In general, heat evolution or heat
absorption accompanies chemical reactions. This energy
results from the rearrangement of the atoms of the reacting
molecules to form the products. The chemical bonds that keep
the atoms of the reactants together are different from those of
the products. The difference in the energy content of the
reactants and products represents the “heat of a chemical
reaction” or simply “heat of reaction”. As most chemical
reactions occur at constant pressure, we use enthalpy change
(∆H) in the calculations. ∆HR represents the heat of reaction in
kJ/mol based on one of the reactants or products in the
chemical reaction.
If the products possess more energy (or stronger chemical
bonds) than the reactants, then energy must be supplied. This
reaction is an “endothermic” reaction (positive quantity by
convention). If the products have less energy, then energy is
given off. This is an “exothermic” reaction (negative quantity).
The other heat quantities we consider in these chapter are
heat of formation and heat of combustion. We discuss the
relationship of the three types of heat mentioned.
STANDARD HEAT OF REACTION
Heats of reactions vary with temperature and pressure. We
usually report values of heat of reaction at a temperature of
25°C and a pressure of 1 atm. This heat of reaction is the
“standard heat of reaction,” H oR .
Heat Balance
189
383
353.1
The chemical equation should indicate the state at which
the reactant or products are in, whether gas, liquid, solid, or
other forms. In the reaction below,
a A(g) + b B(l) d c C(s) + d D(l)
A is a gas; B, a liquid; C, a solid; and D, a liquid. The symbol
“aq” designates that a reactant or product is diluted in an
infinite amount of liquid water,
The following are some examples of standard heats of
reaction: (The basis is 1 kmol × the coefficient of a particular
compound in the reaction.)
Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(s)
HoR = 219.5 MJ
2P(s) + 3Cl2(g) → 2PCl3(I)
HoR = 635.6 MJ
3Be(s) + 2NH3(g) → Be3N2(s) + 3H2(g)
HoR = 495.7 MJ
HESS LAW
DRAFT
We cannot determine the heats of reaction for all chemical
reactions. Some may be impossible or difficult to determine.
Instead we can use Hess' Law to determine the values
indirectly from a set of chemical reactions with known heats of
reaction. We manipulate the set of chemical reactions as
though they are a system of algebraic equations: add,
subtract, multiply or divide by a constant. We do the same
operations to the respective heat of reaction. Let us illustrate
the method for the following chemical equation:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Given the following chemical reactions and the respective
heats of reaction
[1] CaCO3(s) → CaO(s) + CO2(g)
HoR = +177.9 MJ
[2] Ca(OH)2(s) + 2HCl(aq) → CaCl2(aq) + 2H2O(l) HoR= +158.3 MJ
[3] Ca(OH)2(s) → CaO(s) + H2O(1)
HoR = +65.3 MJ
Add (1) and (2),
[4]
CaCO3(s) + Ca(OH)2(s) + 2HCl(aq) →
CaO(s) + CO2(g) + CaCl2(aq) + 2H2O(l)
The corresponding heat of reaction 177.9 + 158.3 = 336.2 MJ
We then subtract [3] from [4] and do the same to the heat of
reaction
[5] CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
HoR = 336.2 - 65.3 = +270.9 MJ
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National Engineering Center CEEC Seminar June 2 -4, 2004
Equation [5] is our desired reaction and the corresponding
heat of reaction is 270.9 MJ.
LATENT HEAT IN HEAT OF REACTION
Some substances can exist in more than one possible natural
states. Water, benzene, hexane, etc., can exist both as a gas
and a liquid at ordinary conditions. The formation of water
from oxygen and hydrogen can be represented by either of the
following:
H2(g) + 12 O2(g) d H2O(g)
HoR = −241.99 MJ
H2(g) + 12 O2(g) d H2O(l)
HoR = −286.03 MJ
Subtracting the latter from the former (applying Hess Law),
H oR = 44.04 MJ
H 2 O(l) d H 2 O(g)
DRAFT
Understandably,
vaporization.
the
difference
is
the
latent
heat
of
STANDARD HEAT OF FORMATION
A way of simplifying numerous data on heat of reaction is to
have a compilation of the “standard heat of formation.” This
quantity designated as H oF , is the ∆H involved in the formation of one mol of a compound from its elements, with both
the product and the reactants in their natural state at 25°C
and 1 atm. We must indicate the natural state of the elements
in the formation reaction, such (g) for gases, (l) for liquids, (s)
for solids. For example, the following reaction gives us the
formation equation for methane:
C(s) + 2H2 (g) → CH4 (g)
T= 25°C; P = 1 atm
The above equations tells us that one mol of solid carbon
reacts with two mols of hydrogen gas to form one mol of
methane. At 25°C and 1 atm, carbon is solid and both hydrogen and methane are gases. The heat involved here is the
standard heat of formation of methane.
Appendices 5 and 6 are compilations of the standard of heat
of formation of selected compounds. By applying Hess Law, we
can calculate the standard heat of reaction for specific chemical reactions as long as the H 0F for each compound is
available.
For the reaction,
aA + bB + ... → qQ + rR + ...
H0R = (q H 0FQ + rH0F R + ...) - (aH0F A + bH 0F B + ....)
Heat Balance
191
STANDARD HEAT OF COMBUSTION
Another useful compilation of data is the standard heat of
combustion, H oC . This is the heat involved in the complete
chemical reaction between an organic compound (usually a
hydrocarbon containing carbon, hydrogen, oxygen, sulfur, and
chlorine) and stoichiometric molecular oxygen to form carbon
dioxide gas, liquid water, nitrogen gas, aqueous hydrogen
chloride, and sulfur dioxide gas at 25°C and 1 atm. For
example
C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l)
has a standard heat of combustion =
-1367.83 MJ/kmol
Appendix 5 is a tabulation of the standard heat of combustion of various organic compounds. Aside from its value in
determining the heat given off in a combustion reaction, we
use them to calculate standard heat of reaction by applying
Hess Law. For the reaction
DRAFT
aA + bB + ... → qQ + rR + ...
where A, B, ... Q, R, ... are compounds that under go
combustion reaction.
H oR = (q H oCQ + rHoC R + ...) - (aHoC A + bHoC B + ....)
Example 5-7
Calculate the standard heat of reaction of
C4H10 (g) (n-butane) → C4H8 (g) (n-butene) + H2(g)
from standard heats of combustion.
Solution:
The standard heat of reaction is:
HoR = HoC butene + HoC H2 − HoC butane
From Appendix 5,
H oC butane = -2879.01 MJ/kmol
H oC butene = -2720.40 MJ/kmol
From Appendix 6,
HFo H2O(l) = -286.03 MJ/kmol = H oC H 2 (g)
Note that the standard heat of formation of liquid water is
identical to the standard heat of combustion of hydrogen.
Since the chemical reaction is the same:
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National Engineering Center CEEC Seminar June 2 -4, 2004
H2(g) + ½ O2(g) → H2O(1)
HoR = (-2720.40 - 286.03) - (-2879.01) = -127.42 MJ
DRAFT
HEATING VALUE OF FUELS
The heating value of a fuel denotes the amount of heat that
can be provided by a given fuel. It is synonymous to the
standard heat of combustion, which normally is used for pure
compounds. Common fuels such as coal, oil, wood, gasoline.
etc., are mixtures of different compounds and may contain
non-combustible substances.
Heating values are measured using a bomb calorimeter. If
the water product is condensed, then the heating value is
called the higher heating value (HHV). If the water remains as
vapor, the heating value is the lower heating value (LHV).
Calorimetric determinations are conducted with a water
condenser and the heating value corresponds to the HHV.
Unless otherwise specified, a heating value is understood to be
the higher heating value. The approximate heating values of
some typical fuels are:
Anthracite coal:
31,400 kJ/kg
Bituminous coal: 33,700 kJ/kg
Lignite:
16,700 kJ/kg
Fuel oil:
44,200 kJ/kg
Natural gas:
41,000 kJ/m3
Producer gas:
4,800 kJ/m3
Butane:
119,000 kJ/m3
The heating value of a mixture of some combustible
compounds can be obtained from standard heats of
combustion.
Example 5-7
Find the heating value of the following fuel gas mixture at
25°C and 101.325 kPa in MJ/m3
30% CH4, 10% C2H6, 20% H2, and 40% N2
Use standard heat of combustion data.
Solution:
Basis:
100 kmol of mixture
From Appendix 5,
H oC (CH4) = -890.95 MJ/kmol
H oC (C2H6) = -1560.92 MJ/kmol
From Appendix 6,
H oF (H 2 O) = -286.03 MJ/kmol = HoC (H2)
Heat Balance
193
The total heat available
= 30(-890.95) + 10(-1560.92) + 20(-286.03)
= -48,058 MJ
3
The volume = 22.4 m % 100 kmol % 25 + 273 = 2445 m 3
0 + 273
kmol
HHV = 48,058/2,445
= 19.66 MJ/m3
PROBLEMS
DRAFT
5-1
1400 grams of copper is heated from 30°C to 80°C? How much
heat did it absorb?
5-2 One kg of iron is initially at 35°C. It absorbs 500 joules of heat?
What will be its final temperature?
5-3 Calculate the number of joules of heat lost to the atmosphere per
kmol of combustion gases leaving a chimney if the gases are at
270°C and the surrounding atmosphere is 30°C. The composition
of the gases is 14% CO2, 4% O2,, and 82% N2.
5-4 Sodium carbonate must be heated to
°C before it begins to melt.
How many calories are necessary to change one Mg from a solid at
the melting point to a liquid?
5-5 Liquid ammonia is used as a refrigerant. When liquid ammonia is
allowed to vaporize at -28.9°C, its latent heat of vaporization at
that temperature is 82.8 J/mol. Under these conditions, calculate
the kJ of heat absorbed by the vaporizing 750 grams of ammonia.
5-6 The ice tray in your refrigerator holds 400 grams of water.
Compute the number of joules that must be removed from that
amount of water at 25°C to cool it to 0°C, freeze it to 0°C and cool
it further to -10°C.
5-7 A vessel contains 300 g of water at a temperature of 25°C. 200 g of
water at 60°C is poured in. After stirring the contents, what is the
resulting temperature?
5-8 70 grams of pure ethanol is cooled from 30°C to 4°C. How much
heat is removed?
5-9 How much heat is needed to transform ice at -20° to steam at
130°C?
5-10 Find the number of calories released as 300 grams of mercury
cools from 86 to 31°C.
5-11 What is the heat capacity of N2 at 500°C?
5-12 Find the heat needed to raise the temperature of a 10 kmol of a gas
mixture consisting of 30% CH4, 20% C2H6, 20% H2 and 30% N2
from 35°C to 150°C.
5-13 Find the heat that can be abstracted from 100 kmol of a flue gas
consisting of 20% CO2, 5% O2, and 75% N2 in cooling it from 600°C
to 110°C. The partial pressure of water is 50 kPa while the total
pressure is 103 kPa.
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National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
5-14 The water evaporated from an evaporator is used to heat the feed
from 100 to 144°F. How much condensate is obtained in the
process? The latent heat of condensation is
kJ/kg.
5-15 The following reactions are possible during the pyrolysis of biomass
in the total absence of oxygen:
a. C + H2O → CO + H2
b. C + 2H2O → CO2 + 2H2
c. C + 2H2 → CH4
d. C + CO2 → 2CO
e. CO + H2O → CO2 + H2
f. 2CO + 2H2 → CH4 + CO2
g. CO + 3H2 → CH4 + H2O
h. CO2 + 4H2 → CH4 + 2H2O
Calculate the heat of reaction of each at 25°C and 1 atm from data
on heat of formation.
5-16 The following are the possible reactions during the pyrolysis of
biomass in the partial presence of oxygen:
a.
C + ½O2 → CO
b.
CO + ½O2 → CO2
c.
CH4 + 2O2 → CO2 + H2O
d.
C + 2H2O → CO2 + 2H2
Calculate the heat of reaction at 25°C and 1 atm from data on heat
of combustion.
5-17 Find the heat of formation of each compound from the given heat
of combustion:
a.
n-hexane (l)
H oC = -4166 kJ/mol
b.
Benzene (l)
H oC = -3270 kJ/mol
c.
Cyclopentane (l)
H oC = -3293 kJ/mol
d.
n-propyl benzene (l) H oC = -5222 kJ/mol
e.
Isobutane (g)
H oC = -2871 kJ/mol
5-18 Find the heat of combustion of each compound from the data on
heat of formation:
a.
Acetylene (g)
H oF = -226.9 kJ/mol
b.
Cumene (l)
H oF = -41.23 kJ/mol
c.
Ethane (g)
H oF = -87.72 kJ/mol
d.
Ethyl alcohol (l)
H oF = -277.8 kJ/mol
e.
n-heptane (l)
H oF = - 224.5 kJ/mol
5-19 In the manufacture of Na2CO3 using the Solvay process, the following are the approximate reaction steps and the respective heats of
reaction at 25°C.
CaCO3 → CaO + CO2
H oR = 43.4 kcal
CaO + H2O → Ca(OH)2
H oR = -15.9 kcal
o
H R = -15.2 kcal
NH3 + H2O + CO2 → NH4HCO3
H oR = -52.7 kcal
NH4HCO3 + NaCl → NH4Cl + NaHCO3
o
H R = 30.7 kcal
2NaHCO3 → Na2CO3 + CO2 + H2O
2NH4Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2O H oR = 10.7 kcal
Using Hess' Law, manipulate the equations to produce the net
equation for the Solvay process and the corresponding heat of
reaction:
CaCO3 +2NaCl → Na2CO3 + CaCl2
Heat Balance
195
5-20 We can decompose water to hydrogen and oxygen by electrolysis.
How much energy is needed to decompose 100 ml of water? The
equation is 2H2O (l) → 2H2 (g) + O2 (g); ∆H = 571 kJ
5-21 The chlorophyll in the plants is responsible for photosynthesis, the
reaction that allows plants to utilize light energy from the sun to
manufacture sugar from carbon dioxide and water. The reaction is
6 CO2 (g)+ 6 H2O(l) → C6H12O6 (s) + 6 O2 (g) ∆H = 2813 kJ
How much energy is needed to form 1 kg of glucose?
5-22 LPG (liquefied petroleum gas) comprises several hydrocarbons that
include butane. How much energy can be provided by 10 kg of
butane?
C4H10 (g) - 13/2 O2 (g) → 4CO2 (g) + 10H2O ∆H = -2884
DRAFT
5-23 Magnesium metal is produced by the electrolysis of magnesium
chloride. Calculate the amount of energy needed to produce 100 kg
of magnesium metal.
5-24 Biogas produced from animal wastes consist of 65% methane and
35% carbon dioxide. Calculate the heating value of 1 m3 of biogas
at 30°C and 1 atm. (kJ/m3)
5-25 How much energy is required to oxidize 20 m3 (25°C and 755 mm
Hg) of sulfur dioxide to sulfur trioxide?
5-26 Compare the heating potential of gasoline (assume consisting of
pure octane) and anhydrous ethanol. How many kg ethanol should
be burned in order to match the heat given off by burning 1 kg of
gasoline?
5-27 Gasohol consist of 80% gasoline and 20% anhydrous ethanol.
Calculate the lower heating value of 1 kg of the mixture.
5-28 Calculate the heat released in burning 5 kg of acetylene.
5-29 Find the higher heating value of a mixture of 20% methane, 30%
ethane, 40% propane, and 10% butane.
5-30 Compare the lower heating values of 1 m3 methane and 1 m3 of a
mixture containing 50% methane and 50% ethane.
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National Engineering Center CEEC Seminar June 2 -4, 2004
--Chapter 6--
Graphical Solution
In this chapter, we present graphical techniques of solution,
which are essential in chemical engineering. Rather than
using either the arithmetic or algebraic method, some
problems are solved and understood better by manipulating
graphs hands-on.
We can solve simple addition and subtraction of mixtures of
substances graphically. We limit our discussion to mixtures
consisting of up to three components. Graphical methods are
applicable to graphs that have concentration as a variable in
one of its coordinate axes. Consider a mixture consisting of
three components A, B, and C, with masses Ma, Mb, and Mc,
respectively. The mass fractions of each component are
M
x a = M + Ma + M
DRAFT
a
b
6-1
c
xb =
Mb
Ma + Mb + Mc
6-2
xc =
Mc
Ma + Mb + Mc
6-3
C
0.0 1.0
0.1
0.9
0.2
B
of
on
cti
fra
ss
ma
0.5
0.5
0.6
0.4
0.7
D
C
of
xb ,
0.6
on
cti
fra
ss
ma
0.7
0.4
0.8
x c,
0.8
0.3
0.3
0.2
0.9
0.1
1.0
0.0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
B
A
xa, mass fraction of A
Figure 6-1
Considering xa, xb, and xc as variables, a 3-coordinate system
with scales ranging from 0 to 1.0 is suitable for this system.
Fig. 6-1 shows a triangular coordinate system. Diagonal lines
Graphical Solution
197
parallel to line CA represent mass fractions xb. Horizontal lines
parallel to line BA represent mass fraction xc. (What diagonal
lines represent xa?)
Any point on the triangle represents the equation xa + xb + xc
= 1. As an example, if xa = 0.4, xb = 0.4, and xc = 0.2 this
composition is represented by the intersection of lines xa = 0.4,
xb = 0.4, and xc = 0.2 at D.
To determine a point on the triangle, only two of the
variables have to be specified since the third can be obtained
from the equation xa+ xb+ xc = 1. xa = 1 means 100% A and is
represented by point A on the graph. xb = 1 is represented by
point B and xc = 1 by point C.
The equilateral triangular graph is used in chemistry. This
triangle, however, requires a special type of graphing paper
that is quite difficult to obtain or prepare. For engineering
purposes, the right triangle, like that of Fig. 6-2, is practical.
We can use ordinary cross-section paper. The 90° vertex is the
origin. With a range from 0 to 1.0, xa is assigned on the
abscissa, while xc, on the ordinate. The points xa = 1 on the
horizontal axis and xc = 1 on the vertical axis are connected by
a straight line AC.
1.0
C
DRAFT
0.9
xc, mass fraction C
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
A
B 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
xa, mass fraction A
Figure 6-2
Lines parallel to AC are drawn, passing through xa = 0.9,
0.8, 0.7, ..., 0.1 on the horizontal axis. Mass fraction xa is
represented by reading values along the abscissa, while xc is
similarly obtained by reading values along the ordinate. xb can
be calculated by using the relation xb = 1 - xa - xc or by reading
along the diagonal lines. Line AC represents xb = 0, the next
parallel line is xb= 0.1, the next xb = 0.2, and so on. Point B
represents xb = 1; point A, xa = 1; and point C, xc =1.
Any point inside the triangle is a mixture of 3 components
and any point on the side, a mixture of 2 components. The
vertices represent pure components.
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National Engineering Center CEEC Seminar June 2 -4, 2004
ADDITION
If two mixtures with masses M1 and M2 and respective
compositions xa1, xb1 and xc1, and xa2, xb2 and xc2 are mixed, a
mixture with mass M3 and composition xa3, xb3, and xc3 results.
M1 and M2 are represented by corresponding points on Fig.
6-3.
xc
M2
xc2
xc1
M1
DRAFT
xa1
xa2
xa
Figure 6-3
By a component balance,
x a1 M 1 + x a2 M 2 = x a3 M 3
6-4
x c1 M 1 + x c2 M 2 = x c 3 M 3
Overall mass balance
M1 + M2 = M3
6-5
6-6
Substituting Eq. 6-6 in Eq. 6-4
x a1 M 1 + x a2 M 2 = x a3 (M 1+ M 2 )
6-7
M 1 x a2 − x a3
M 2 = x a3 − x a1
6-8
x c1 M 1 + x c2 M 2 = x c3 (M 1 + M 2 )
6-9
Rearranging
From Eq. 6-5
Rearranging
M 1 x c2 − x c3
6-10
M 2 = x c3 − x c1
In Fig. 6-4 the sum of M1 and M2 lies anywhere within the
triangle and within the range of the compositions of M1 and M2
Graphical Solution
199
and denoting the angles they make with the horizontal as α
and β respectively, the following are obtained
x c3 − x c1
tan = x a3
− x a1
6-11
x c2 − x c3
tan = x a2
− x a3
Combining Eq. 6-8 and Eq. 6-10
6-12
x a2 − x a3 x c2 − x c3
x a3 − x a1 = x c3 − x c1
6-13
x c2 − x c3
x c3 − x c1
x a3 − x a1 = x a2 − x a3
6-14
Rearranging
This gives
DRAFT
xc
M2
xc2
xc3
xc1
M3
M1
β
α
xa1
xa3
Figure 6-4
xa2
xa
tan = tan 6-15
=
6-16
which is only possible if M3 lies on the line joining M1 and M2.
The addition point is dependent on the relative amounts of M1
and M3, as will be shown.
From Fig. 6-5,
M1 x a2 − x a3 DE
M2 = x a3 − x a1 = AD
6-17
where DE and AD are line segments.
Using similar triangles
DE = BC
AD AB
Therefore,
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National Engineering Center CEEC Seminar June 2 -4, 2004
6-18
M1 BC
M2 = AB
6-19
Similarly, the following relations can be established
M1
M1
BC
=
=
= BC
M 3 M 1 + M 2 AB + BC AC
6-20
M2
M2
AB
=
=
= AB
M 3 M 1 + M 2 AB + BC AC
6-21
xc
M2
C
DRAFT
M3
B
M1
A
D
E
xa1
xa3
xa2
xa
Figure 6-5
The above relationship for graphical addition, known as the
"inverse lever arm rule", states that the addition point of two
mixtures M1 and M2 on the ordinary scale lies on a line joining
M1 and M2 so located such that the ratio of the distance
between M1 and M3 to the distance between M2 and M3 is equal
to the mass ratio M2/M1.
M1
A
M2
B
C
M3
Figure 6-6
Consider AC as the lever and B as the fulcrum in Fig. 6-6.
BC is the arm opposite M1. AB is the arm opposite M2. Following the rule, M1/M2 = BC/AB. The lengths of BC, AB, and AC
represent the relative amounts of M1, M2, and M3, respectively.
Thus, if M2 is 100 kg and M1 is 50 kg, then M2/M1 = 100/50 =
Graphical Solution
201
2. Hence, AB/BC = 2. If the length of AC is 3 units, then AB =
2 units and BC = 1 unit.
Example 6-1
Four mixtures of equal masses, M1, M2, M3, and M4 with
components A, B, and C are mixed. If the compositions of the
mixtures are given below, find the composition of the resulting
mixture graphically.
M2
M1
xa = 0.1
xc = 0.4
xb = 0.5
M3
xa = 0.3
xc = 0.55
xb = 0.15
xa = 0.55
xc = 0.2
xb = 0.25
M4
xa = 0.3
xc = 0.1
xb = 0.6
Solution:
xa is set along the abscissa while xc along the ordinate. M1,
M2 M3, and M4 are plotted on the graph. (Fig. 6-7) The addition
is (M1 + M2) + (M3 + M4) .
1.0
DRAFT
0.9
0.8
0.7
0.6
xc
0.4
0.3
M2
Σ1
0.5
M1
ΣT
M3
0.2
0.1
0
M4
Σ2
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
xa
Figure 6-7
The parentheses represent groupings, i.e., adding M1 and M2
first and then M3 and M4. Let Σ1 represent (M1 + M2) and Σ2, (M3
+ M4).
A straight line is drawn between M1 and M2. Since M1= M2
the addition point Σ1 must be on the midpoint of M1 and M2.
Likewise, a straight line is drawn between M3 and M4. The
midpoint is taken and it represents Σ2.
The total mixture is now represented by Σ1 + Σ2. To add Σ1
and Σ2, connect the two points by a straight line. Since Σ1 and
Σ2 have equal masses, the addition point ΣT is on the midpoint
of that line. The resulting mixture, therefore, has a composition xa = 0.32 and xc = 0.31, as read from the graph.
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National Engineering Center CEEC Seminar June 2 -4, 2004
Other solutions are possible. (See Fig. 6-8). For instance,
the addition can be grouped as
ΣT = (M1 + M4) + (M2 + M3)
ΣT = Σ' +Σ"
1.0
0.9
0.8
0.7
0.6
M2
xc 0.5
0.4
0.3
M1
Σ'
0.2
DRAFT
0.1
0
Σ"
ΣT
M3
M4
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
xa
Figure 6-8
The addition is similarly performed on the graph. The
answer is identical to the first solution. Depending on the
arrangement of terms, we can vary the solution and choose
the most convenient one.
GRAPHICAL SUBTRACTION
Graphical subtraction is the inverse of graphical addition and
can be similarly proved. However, by just observing the procedure for addition, we can see that the same diagram in graphical addition can represent the one on subtraction. If the bigger
quantity is M3 from which M1 is subtracted, M2 results. (Refer
to Fig. 6-9.)
M3 -
M1 = M2
M1
M3
M2
A
B
C
Figure 6-9
The subtraction point of
two masses lies on the
extension of the straight
line joining the two on the
side of the larger mass.
From the figure, segment
AB represents M2; BC, M1;
and AC, M3 or
M 1 BC
=
M 3 AC
Graphical Solution
203
M1 BC
M2 = AB
6-22
M 2 AB
=
M 3 AC
6-23
Example 6-2
Three masses M1, M2, and M3 are added together. Each has the
following composition:
M2
xa = 0.3
xc = 0.55
M1
xa = 0.1
xc = 0.4
M3
xa = 0.5
xc = 0.1
By a conventional method, part of the resulting mixture is
separated. If the amount removed is M4 with composition xa =
0.1 and xc = 0.1, what is the composition of the final mixture if
the masses of M1, M2, M3, and M4 are all equal? Solve
graphically.
Solution:
DRAFT
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
M2
A
C
M1
B
0.2
0.1
0
M4
M3
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Figure 6-10
All the points representing the mixtures are plotted on the
graph as shown in Fig. 6-10 . First, M1 is added to M2, then M3
is added to (M1 + M2) and finally M4 is subtracted from (M1 + M2
+ M3).
To add M1 and M2, connect M1 to M2 by a straight line and
the addition point is the midpoint of the line M 1 M 2 since M1=
M2. Call this point A.
To add M3 to (M1+M2), connect A to M3. AM 3 is 4.86 units
long. Denoting B as the addition point of (M1 + M2+ M3), the
length of AB is 4.86/3 =1.62 units and B is thus located. [(M1
+ M2) is twice M3.] (You can check the calculations by using a
ruler, of course, with different length measurements.)
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National Engineering Center CEEC Seminar June 2 -4, 2004
The subtraction point [(M1 + M2+ M3) - M4] lies on the extension of the line joining M4 and B, on the side of B. This is
denoted by C. The length of M 4 B is 3.25 units. Since [M1 + M2+
M3]/M4 = 3, by the inverse rule, M 4 C/BC = 3 and M 4 B/BC = 2.
The length of BC therefore is 3.25/2 = 1.63 units and C is
located. The composition of the resulting mixtures is xa =
0.405 and xc= 0.475.
Example 6-3
100 kg. of pure salt, NaCl, is added to 500 kg of NaOH
solution containing 60% NaOH and 40% H2O. The mixture is
heated until 100 kg of H2O has evaporated. What is the
composition of the resulting mixture?
Solution:
Let the abscissa be xNaOH and the ordinate be xNaCl. In the
diagram, point A represents pure salt, point B, the 60% NaOH
solution, and point C, pure water.
To add A to B, connect point A to point B by a straight line.
1.0
A
DRAFT
0.9
0.8
0.7
0.6
xNaCl
0.5
0.4
0.3
F
0.2
E
D
0.1
C
0
B
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Figure 6-11
xNaOH
Denote the addition point as D. Since the ratio of the mass of
pure salt to the NaOH solution = 100/500, DB/AD =100/500,
DB/AB = 100/600. By direct measurement, AB is 11.83 units.
DB, therefore, is (1/6)(11.83) = 1.97 units and D is located.
An alternative way does not make use of direct measurement. The projection of the desired line on either the x or the
y-axis is used. Note that the scale xNaCl can be used by considering the triangle ACB. AC is scaled from 0 to 1.0 and by
considering similar triangles, point D can be located. AED is
similar to ACB. By the inverse lever arm rule, BC/AC = 1/6.
Since AC = 1.0, EC = (1/6)(1.0) = 0.167 (corresponding to xNaCl
of point D).
Graphical Solution
205
For the subtraction point, F is located at the extension of
CD on the side of D since the mass corresponding to D is
larger. CD is measured as 5.28 units. DF is, therefore,
(1/5)(5.28) = 1.06 units. The reader should go over the details
of arriving at this answer. He should also try the other method
used in the first part of this example.
The composition of the final solution is xNaOH = 0.6 and xNaCl =
0.2.
HEAT OF MIXING AND HEAT OF SOLUTION
When ideal substances are mixed, the molecules do not interact and no heat is evolved or absorbed. No heat results in
mixing them. The enthalpy of the mixture is equal to the sum
of the molar enthalpy of each pure component multiplied by
its corresponding mole fraction.
HT = y1H1 + y2H2 + y3H3 + ... +ynHn
where
Hn = the enthalpy of component n
yn = the mole fraction of component n
HT = enthalpy of the mixture
If the mixture is non-ideal, the interaction between the
molecules will yield a heat of mixing, ∆Hmix.
DRAFT
HT = Y1H1 + Y2H2 + Y3H3 + ... YnHn + ∆Hmix
The values of ∆Hmix are obtained experimentally by mixing
the substances and measuring the decrease or increase in
enthalpy. If heat is given off, the reaction is exothermic; if
absorbed, the reaction is endothermic. Most data for the heat
of mixing are available only for two component systems where
a solute is dissolved in a solvent (usually water). In this case
the heat of mixing is also referred to as the heat of solution.
The basis used is one mole of the solute. Table 6-1 gives the
heat of solution of some solutes dissolved in water. The data
for other substances at infinite dilution is given in the Appendix 6.
Table 6-1 Heat of Solution of Some Solutes in Water at 25oC
Solute
kmol of water
Hsoln, MJ/kmol of solute
-29.56
1
Ammonia
Ethanol
206
2
3
4
5
10
20
30
40
50
100
200
∞
1
2
3
4
-30.07
-32.78
-33.29
-33.62
-34.29
-34.46
-34.50
-34.50
-35.54
-34.58
-34.67
-34.67
-0.81
-1.72
-2.79
-3.76
National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
Sulfuric acid
5
7
10
25
50
100
200
1
2
5
10
25
50
75
100
200
500
1,000
5,000
10,000
100,000
∞
-4.70
-6.14
-7.37
-9.31
-9.98
-10.27
-9.60
-28.09
-41.95
-58.05
-67.07
-72.65
-73.39
-73.73
-74.04
-74.99
-76.79
-78.63
-84.49
-87.13
-93.70
-96.25
ENTHALPY-CONCENTRATION DIAGRAM
We can present the data on heat of mixing of two components in graphical form. We refer to these graphs as enthalpyconcentration diagrams. Fig. 6-12 is an example. This graph
gives the enthalpies of mixtures of water and sulfuric acid at
different concentrations and temperatures. The reference
temperature is 0°C (i.e., the enthalpies of liquid water and
pure sulfuric acid are set to zero). The experimental values of
the enthalpy of the mixture are plotted against the composition of sulfuric acid at different temperatures. Given the
concentration and the temperature of a mixture of water and
sulfuric acid, we can get the enthalpy of the mixture from the
graph. The graph is useful in determining the resulting
enthalpy, concentration, or temperature when two or more
solutions of sulfuric acid and water of different concentrations
are mixed together. The enthalpy and composition can be
calculated arithmetically. Since one of the coordinates of the
diagram is concentration, we can apply graphical methods of
addition and subtraction. This is illustrated in the following
example.
Example 6-4
Find the temperature and composition of the resulting mixture
when the following liquids are mixed:
50 kg 10% H2SO4 at 82oC
100 kg 70% H2SO4 at 15.5oC
100 kg of pure water at 25oC
Solve (a) arithmetically and (b) graphically.
Solution:
Fig. 6-12
Graphical Solution
207
DRAFT
(a) Arithmetic solution
Enthalpies of the solutions from Fig. 6-12:
10% H2SO4, 82°C: H = 240 kJ/kg
70% H2SO4, 15.5°C: H = -280 kJ/kg
pure water, 25°C: H = 105 kJ/kg
Total mass = 250 kg
Mass of H2SO4 = 0.1(50) + 0.7(100) + 0(100) = 75 kg
Mass % = (75/250) x 100 = 30%
Final enthalpy of the mixture =
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National Engineering Center CEEC Seminar June 2 -4, 2004
50(240) + 100(−280) + 105(100)
= -22 kJ/kg
250
From Appendix 8, the intersection of H = -22 kJ/kg and
concentration = 30% corresponds to the isotherm 58°C (read
by interpolation).
D
200
A
DRAFT
Enthalpy, kJ/kg
100
0
E (the addition point
T = 58 oC
-100
C
-200
30% acid
B
-300
0
50
Concentration, % H 2SO4
100
Fig. 6-13
(b) Using the graphical technique
The corresponding values of the three liquids are plotted on
the enthalpy concentration diagram. (See Fig.6-13.) Point A
represents 100 kg pure water point B 100 kg 70% H2SO4. The
addition point is C, which is the midpoint of line AB. Point D
represents 50 kg 10% H2SO4. The addition point of 200 kg and
50 kg is E with the ratio of the line segments DE/EC = 200/50
= 4. Point E lies at T = 58oC and composition = 30% sulfuric
acid.
PROBLEMS
6-1
The used mixed acids coming from three different sources
has the following compositions (by mass %):
(a) 30% HCl, 10% HNO3, and 60% H2O
(b) 50% HCl, 45% HNO3, and 5% H2O
(c) 5% HCl, 80% HNO3, and 15% H2O
Graphical Solution
209
If 120 kg of (1), 300 kg of (2) and 150 kg of (3) are mixed
together, what is the composition of the resulting mixture?
Solve graphically. Plot the composition of HNO3 in the
abscissa and that of HCl, on the ordinate.
6-2 100 kg of copra containing 40% oil and 60% solids is to be
treated with n-hexane to recover the oil. If the resulting
n-hexane and oil mixture contains 30% oil and 70% hexane,
how much n-hexane was used? Solve graphically. (Assume
that the n-hexane-oil mixture remaining within the copra is
also 30% oil and 70% n-hexane).
6-3
Five mixtures consisting of components A, B, and C have the
following compositions and masses:
DRAFT
M1, 263 kg
xa1 = 0.1
xc1 = O.1
M2, 137 kg
xa2 = 0.45
xc2 = 0.2
M3, 950 kg
xa3 = 0.05
xc3 = 0.35
M4, 810kg
xa4 =0.25
xc4 =0.65
M5,400 kg
xa5=0.05
xc5=0.85
Find the composition of the mixture that results after mixing
the 5 masses.
a) by adding the masses successively, that is, first, add M3
to M1 + M2, then add M4 to M1+ M2 + M3, and finally, add M5
to M1 + M2 + M3 + M4.
b) by grouping into [(M1 + M3) + (M2 + M4)] + M5.
6-4 A mixture weighing 850 kg and containing components A, B,
and C with xa = 0.2 and xc = 0.3 loses 100 kg C when the
mixture is heated. Upon cooling, 50 kg of solids containing
70% A and 30% B precipitate out. What is the composition of
the remaining solution? Solve graphically.
6-5 When pure furfural (solvent) is added to a mixture containing
0.2 mass fraction diphenylhexane and 0.8 mass fraction
docosane, two separate layers can be obtained. If the composition of the two layers are:
mass fraction
Upper layer
Lower layer
furfural
diphenylhexane
docosane
furfural
diphenylhexane
docosane
0.88
0.101
0.019
0.004
0.104
0.892
How much solvent was added to 500 kg of solution of
docosane and diphenylhexane? What are the amounts of the
layers? Solve graphically.
6-6 Three solutions each composed of KNO3, and H2O with the
following compositions:
% KNO3
%NaNO3
M1
M2
M3
6
25
29
57.5
35
5
are mixed together and then heated until some of the water
evaporates. The final solution contains 39% KNO3 and 25%
NaNO3. The original masses of M1 and M2 are equal. Find the
masses of each of the original solutions and the amount of
210
National Engineering Center CEEC Seminar June 2 -4, 2004
the water that is evaporated. Solve graphically. Plot xKNO3 on
the abscissa.
DRAFT
6-7 Two mixtures consisting of NaCl, NaOH, and water. The
composition of the first is 10% NaOH, 40% NaCl and 50%
water. The second consists of 50% NaOH, 15% NaCl, and
35% water. Both solutions are subjected to heat and lose
water. The first solution weighs 1500 kg and loses 500 kg of
water. When the two solutions are mixed, the resulting
mixture analyzes 35% NaOH, 40% NaCl, and 25% H2O. What
is the original mass of the second mixture? How much water
evaporates from it? Solve graphically. Plot xNaOH on the
abscissa and xNaCl on the ordinate.
Graphical Solution
211
DRAFT
212
National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
—Chapter 7—
Equilibrium and
Mass & Heat Balances
We now consider an important concept that is essential in
chemical engineering -- equilibrium. Equilibrium is a condition of balance between opposed forces, influences or actions.
Changes usually occur for any system that is not in an
equilibrium condition. The farther away a system is from
equilibrium, the greater is its tendency to change. Let us
consider some examples in everyday life:
y Bring a hot rod in contact with a cold rod. Heat transfers
from the hot rod to the cold rod. The temperature of the
hot rod decreases while the temperature of the cold rod
increases. After a while, heat stops transferring and the
temperature of the two rods is the same. Heat always
flows from a hot region to a cold one. When the temperature of the two regions becomes equal, a state of equilibrium is attained.
y Consider a hot cup of coffee on a table. You cannot drink
it all at once because it is too hot. You have to wait for it
to cool down. Heat has to transfer from the cup of coffee
to the surroundings. If you forgot about it, after a while
you will notice that its temperature is the same as that of
the surroundings.
y Consider a raisin. Place it in a glass of water. After a
while, water diffuses into the raisin. Equilibrium exists
when water stops diffusing in.
y Consider two tanks containing the same kind of gas. The
pressure of tank 1 is greater than that of tank 2. The two
tanks are connected by a pipe that has a valve. If we open
the valve, then the gas from tank 1 will flow to tank 2
until the two tanks have the same pressure.
y Place a wet cloth together with a dry cloth. Notice how
water diffuses towards the dry cloth until both cloths
have the same amount of moisture.
A system in equilibrium undergoes no measurable change
with time. In most instances, a system that appears to be
static may actually be in a dynamic state. Changes taking
effect towards one direction are being counterbalanced by
those towards the opposite direction. All factors that tend to
effectuate a change are balanced.
Whenever a difference in the value of a characteristic (called
potential difference) exists, transfer of appropriate properties
occurs.
Equilibrium 213
DRAFT
Equilibrium is the basis of the transport operations. A
knowledge of the difference in potentials (also as the driving
force) gives us an idea of the relative rate at which the transfer
of properties occurs. The higher the difference, the faster is
the rate. A potential difference or driving force causes changes
in a system. In heat transfer, it is the temperature difference -heat flows from the hot to the cold region. The heat transfer
rate is a function of the temperature difference. In diffusion
(transfer of mass), it is the concentration difference. In fluid
flow, the difference in pressure or head is the driving force.
The study of equilibrium is important since it is the criterion
that determines whether a potential or driving force exists. In
some cases, the attainment of equilibrium is regarded as the
ultimate condition that can be approached. Concepts of
efficiency compare actual conditions to the equilibrium conditions. A study of equilibrium in chemical engineering allows
the use of the theory of rate processes. It can also be a basis of
efficiency of operations.
In the extraction of oil from oil seeds using a solvent, the oil
seed is brought in contact with the solvent. Due to the difference in concentration, oil transfers to the solvent until such
time when the concentration difference is negligible, a condition that can be regarded as the ultimate obtainable. This,
however, requires an infinite time. Knowledge of the equilibrium condition enables us to choose a final condition that is
an acceptable fraction of the limiting value. This could be
attained in a finite time. Time and efficiency are important
factors in chemical process economics. A knowledge of equilibrium conditions is indispensable.
The equilibrium conditions for any system can be observed
and recorded. We can record the data in graphical or tabular
form or in computer databases. Whenever possible, theoretical
or empirical relationships are used.
SOME TERMS AND DEFINITIONS IN EQUILIBRIUM
We need to know some terms when dealing with equilibrium.
A phase is any amount of matter that is homogeneous
throughout in terms of state and composition. A phase can be
gaseous, liquid, or solid. Only one gaseous phase is possible in
a system since gases mix homogeneously. For instance, CO2
and N2 gases will form a homogeneous mixture and will only
be one phase. On the other hand, there can be several liquid
and solid phases. Water and oil will form two distinct phases
when brought in contact with one another. The existence of a
phase does not demand continuity. Some phases can be
dispersed. For instance, undissolved sugar in a water-sugar
solution is a distinct solid phase even though the sugar is in
particulate form.
The components of a system are the distinct substances
that make up that system. Water in an enclosed container
where liquid water and water vapor coexist is a
one-component system. A mixture of sugar and water is a
two-component system. In determining the number of components of a system, relationships among the substances
present should be checked. For instance, a mixture of H2SO4
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National Engineering Center CEEC Seminar June 2 -4, 2004
and H2O is a two-component system even though the H2SO4
exists as 2H+ and SO4= ions in H2O. The H+ and SO4= ions are
related by the formula H2SO4.
The value of an intensive variable does not depend on the
mass of the phase or phases present. Pressure, temperature,
density, and composition are examples. The value of an extensive variable is dependent on the quantity of the mass present.
The variance or degree of freedom of a system at equilibrium
is the number of variables that may be independently specified
to fix or to completely identify the system. For instance, to
characterize the boiling point of water, one has to specify only
one variable, say the pressure. At a pressure of one atm, the
boiling point is 100°C. Specifying a pressure in this case
completely defines the equilibrium between liquid water and
water vapor.
DRAFT
GIBB'S PHASE RULE
In 1875-78, J. Willard Gibbs formulated his theory involving chemical potential that could describe physical and chemical systems. In the discussion that follows, we show the
derivation of the phase rule for cases where no chemical
reaction occurs.
Consider a closed container initially void of anything. C
number of different materials are added into the container. No
chemical reaction occurs. The container is then brought to the
desired pressure and temperature until equilibrium is
attained. Now P phases will be found in the container.
Assuming each phase contains all the components, the total
number of concentrations that must be fixed to characterize
the system is PC. Since the intensive variables, pressure and
temperature are fixed, the total number of intensive variables
is (PC + 2).
However, we cannot fix (PC + 2) variables independently.
Each phase possesses its own equation of state (for instance,
recall the equation of state for gases). The number of these
relationships is P. Next, the concentration of a component in a
particular phase is controlled by the concentration of that
component in the other phases. Therefore, one component has
(P - 1) restrictions. For C components, C(P -1) restrictions
exist.
Therefore the number of intensive variables that can be
varied without causing a change in the number of phase is
F = P C + 2 - P- C(P - 1)
which reduces to
F = C - P + 2,
the well-known Gibb's Phase Rule
where
C = number of components
P = number of phases
F = degrees of freedom or variance.
Equilibrium 215
As an example, consider water that can exist in the gaseous
phase, in the liquid phase, and in several solid phases. If the
gas, liquid, and solid phases are in equilibrium, P = 3 and
substituting into the equation,
F=C+2-P=1+2-3=0
The degree of freedom is zero. This means that the condition
at equilibrium is fixed. For example, the temperature and
pressure are fixed. (On a plot of P vs.
T, this is a fixed point).
P
If equilibrium between two phases
occur, say water vapor and liquid
water, then
F=1+2-2=1
(To, Po )
T
The degree of freedom is one. This
means that to fix the system, we
have to specify either the temperature or the pressure. The plot of the
data on equilibrium of the two phases
is a line. Either P or T should be set P
to fix the condition.
P2
The degree of freedom is useful in
determining the variables relevant in
the equilibrium of physical and P1
chemical
systems.
It
aids
in
constructing diagrams (known as
T1 T2
equilibrium phase diagrams) or in
T
setting up theoretical or empirical
relations. For this reason, the phase Figure 7-2
rule is extensively applied in chemical
engineering.
DRAFT
Figure 7-1
PHASE RULE FOR ONE COMPONENT SYSTEMS
Applying the phase rule to a one-component system provides
an ideal illustration. Considering the phase rule
F+P=C+2
with C = 1 and P = 1,
the degree of freedom is
F=2
For one-component systems, the suitable intensive variables
are pressure and temperature. The P-T equilibrium data are
usually determined experimentally. The resulting plot is a
phase diagram. Fig. 7-3 is an example. The plot is divided into
three distinct areas that correspond to the solid, liquid, and
gas phases. (Other solid or liquid phases can likewise be
represented.) In order to specify a point in any of these three
regions, both P and T should be specified (F = 2).
216
National Engineering Center CEEC Seminar June 2 -4, 2004
Line AB which separates the gaseous area from the liquid is
called the P-T liquid-vapor equilibrium line. Line AD is the
solid-vapor equilibrium line while line AC is the solid-liquid
equilibrium line. In order to fix a point along these lines, only
one variable, either P or T, should be specified (F = 1).
Point A is called the triple point and represents the condition where the gas, liquid, and solid phases are in equilibrium.
Since it is fixed, the degree of freedom is equal to 0.
C
liquid
B
Solid
P
A
Gas
D
DRAFT
T
Figure 7-3
EQUILIBRIUM VAPOR PRESSURE
Consider a pure liquid that is placed in an evacuated
container. The liquid does not fill up the volume. Some of the
molecules in the liquid have enough energy to escape and
form the vapor phase. As a gas, the molecules can bombard
the walls of the container resulting in gas pressure. Now, some
of the molecules in the vapor can return to the liquid phase.
At a fixed temperature, there comes a time when the rate at
which the molecules escape from the liquid equals the rate at
which the molecules return to the liquid. This is a state of
equilibrium. The pressure exerted by the system in the
container is called the vapor pressure of that liquid at that
particular temperature. If the temperature is raised, the vapor
pressure of the substances rises. Experimental data on the
vapor pressures of most substances are available in the literature and in various references.
For water, the Steam Table gives the vapor pressures at
different temperatures. (See Appendix 9.) For most
compounds, the experimental vapor pressure-temperature
relationship can be represented by the Antoine Equation,
ln P vo = A −
B
T+C
where
P vo is the vapor pressure
T is the temperature
Equilibrium 217
A, B, C are constants that are dependent on the type of
substance (see the Appendix for the values of the constant for
selected substances).
From the equation, it is obvious that if ln P vo is plotted
B in a rectangular graph a straight line results.
against
T+C
The P vo versus 1/T plots for common substances are given in
the Appendix.
SOME TERMS IN THE EQUILIBRIUM OF PURE SUBSTANCES
DRAFT
Here are the definition of some terms in the equilibrium of
pure substances. An equilibrium condition is also known as a
saturated condition. At equilibrium, the liquid phase is called
saturated liquid while the vapor phase is called saturated
vapor. At this condition, the temperature is known as
saturated temperature while the pressure is known as
saturated pressure. Thus, in the Appendix, the table for water
is known as the “Saturated Steam Table”. A vapor with a
temperature above the saturation temperature is known as a
superheated vapor while a liquid with a temperature below the
saturation temperature is known as a supercooled liquid.
Superheated vapor and supercooled liquid are not part of a
system in equilibrium.
At equilibrium conditions, the equilibrium temperature is
also known as the boiling point at the equilibrium pressure.
When the equilibrium pressure is one atmosphere, the equilibrium temperature is known as the normal boiling point of the
solution. Thus, the normal boiling point of pure water is
100°C.
However, a limit for the temperature that can be attained by
a liquid and vapor equilibrium exists. The system can only be
heated up to the temperature known as the critical
temperature, the temperature at which the vapor phase
becomes identical with the liquid phase. Above the critical
temperature, only one phase exists. The degree of freedom is
therefore equal to 1 + 2 - 1 = 2.
SOLID-VAPOR EQUILIBRIA
As in liquid-vapor equilibrium, the solid phase can be in
equilibrium with vapor state. Change of state from solids to
vapor is known as sublimation. Thus, solid CO2 or dry ice
sublimes at ordinary pressures.
Example 7-1
What is the boiling point of water at a pressure of 1.5 atm?
At an elevation of 15,000 feet?
Solution:
From the Appendix, referring to the vapor-liquid equilibrium
line for water at a pressure of 1.5 atm or (1.5)(760) = 1140 mm
Hg, the corresponding equilibrium temperature is 110oC. At
1.5 atm therefore, the boiling point of water is 110oC.
218
National Engineering Center CEEC Seminar June 2 -4, 2004
At an elevation of 15,000 ft the atmospheric pressure is
16.88" Hg (Critchfied, H.S., General Climatory, Prentice-Hall
Inc., New Jersey, 1960, p 42).
The equilibrium temperature
corresponding to 16.88" Hg (16.88 % 760 = 428.77 mm Hg)
29.92
is 85oC, the temperature at which water will boil. (This is
obtained also from the Appendix.)
Example 7-2
A nitrogen-acetone gas mixture consists of 20% acetone and
80% nitrogen.
The mixture is made by bubbling dry nitrogen gas through
acetone. Thus, the nitrogen becomes saturated with acetone.
The pressure is at 800 mm Hg. At what temperature is the
bubbling made?
Solution:
The nitrogen is 80% by mole while the acetone is 20% by
mole. At the bubbling condition, the total pressure is 800 mm
Hg. The partial pressure of acetone = 0.2(800) = 160 mm Hg.
Since the gas mixture is saturated, this is also the vapor
pressure of acetone. The Antoine equation for vapor pressure.
DRAFT
log 10 P = A -
B
+T
can be used, where T is in °C and P is in mm Hg.
A = 7.23157
θ = 237.23
B = 1277.03
Solving for T,
T=
=
B
−
A - log 10 P
1277.03
− 237.23= 16.78°C
7.23157 − log 10 160
16.78°C is the bubbling temperature. The graphical representation of vapor pressure in the Appendix can likewise be
used but is less accurate.
MIXTURE OF VOLATILE LIQUIDS; RAOULT'S LAW
When we mix two or more volatile liquids together in an evacuated closed container (and maintaining a head space), some
molecules vaporize and some recondense. When the rate of
vaporization equals the rate of condensation, then equilibrium is attained. With components A and B in two phases,
then
F=C+2-P=2+2-2=2
Equilibrium 219
The degree of freedom is 2. For this system, the intensive
variables can be pressure, temperature, and the composition
of the liquid phase and vapor phases. xA and xB are the composition of the liquid phase and yA and yB, of the vapor phase.
Phase diagrams involving this system can be constructed
using experimental data with any combination of the six
variables. The behavior of the substances can be represented
by theoretical or empirical formulas based on experimental
data. Raoult's Law is one such relationship. Ideal solutions
follow this law.
Let us consider an ideal mixture consisting of components A
and B. The mixture follows the following relationships:
and
where
P A = P vA x A
P B = P vB x B
PA is the partial pressure of component A
PB is the partial pressure of component B
xA is the mole fraction of A in the liquid phase
xB is the mole fraction of B in the liquid phase
DRAFT
We can explain the equilibrium relationship as follows:
Consider two volatile liquids A and B that are mixed and
placed in an evacuated container. The mixture does not fill up
the entire vessel. At a fixed temperature, the molecules of A
and B vaporize until equilibrium is reached. The composition
of the vapor is dependent on the composition of the liquid.
The surface of the liquid (liquid-vapor interface) consists of
molecules A and B. The area occupied by A is directly proportional to the mole fraction of component A while that of B, to
the mole fraction of component B. Both A and B have tendencies to escape. If only A is present, the pressure exerted is PvA
(the vapor pressure of A); if only B is present, the pressure is
PVB. If both A and B are present, the pressure exerted by A is
less than the vapor pressure of pure A. Likewise, the partial
pressure of B is less than the vapor pressure of pure B.
If no interaction exists between A and B, then their amounts
in the vapor phase are directly proportional to their respective
mole fractions in the liquid phase.
PA } xA
PB } xB
P A = k 1 x A = P VA x A
P B = k 2 x B = P VB x B
Vapors A and B
yA, yB
where PVA and PVB are the proportionality constants.
Mixtures of substances following
this law are "ideal mixtures" or
Liquids A and B
"ideal solutions." The behavior this
xA, xB
law predicts is independent of any
specific
characteristic
of
the
molecules which means that it has
a similarity with the ideal gas law.
Molecules of different substances
behave similarly if no form of Figure 7-4
220
National Engineering Center CEEC Seminar June 2 -4, 2004
interaction exists among them. In general, mixtures of
substances which are similar in structure or which are
members of a homologous series behave ideally. Heptane,
C7H16 and octane, C8H18 form ideal solutions and so with
methyl alcohol and ethyl alcohol, benzene and toluene, etc.
The total pressure in the vapor phase is
PT = PA + PB
and the mole fractions in the vapor phase (assuming ideal gas
behavior) are
yA =
P A P VA
=
x
PT
PT A
yB =
P B P VB
=
x
PT
PT B
In general, PVA and PVB are functions of temperature. Therefore,
temperature enters implicitly in the relationship.
DRAFT
Example 7-3
A liquid mixture containing 80% hexane and 20% octane is in
equilibrium with the vapor at a temperature of 100°C. What is
the total pressure of the system? What is the composition of
the vapor phase?
Solution:
This is a straightforward application of Raoult's law.
The partial pressure of hexane = Ph = Pvh xh
The partial pressure of octane = Po = Pvo xo
xh = 0.8 and xo = 0.2
The partial pressures Pvh and Pvo can be obtained from the
Appendix.
At 100°C, Pvh = 1700 mm Hg and Pvo = 340 mm Hg
Ph = (1700)(0.8) = 1360 mm Hg
Po = (340)(0.2) = 68 mm Hg
The total pressure, PT = Ph + Po = 1428 mm Hg.
Assuming ideal gas behavior, the vapor composition is the
same as the partial pressure fraction.
P
y h = P h = 1360 = 0.952
1428
T
P
y o = P o = 68 = 0.048
1428
T
Raoult's law can also be used to find the temperature of a
system in equilibrium given the other conditions. The use of
graphical or tabular data on vapor pressure leads to a
Equilibrium 221
trial-and error solution since the temperature enters implicitly
into the equation. If the Antoine equation is used, a direct
solution of the resulting system of simultaneous equations can
be executed. A particular use of solving for the temperature in
such system in equilibrium is in finding the "dew point" or the
"bubble point" of a gaseous or a liquid mixture respectively.
The "dew point" of a gaseous condensable mixture refers to the
equilibrium at which the liquid phase begins to appear when
the mixture is cooled. The "bubble point" of a volatile liquid
mixture, on the other hand, is the equilibrium temperature at
which the mixture begins to boil (bubbles begin to appear)
when the mixture is heated.
Raoult's law is also applicable to systems consisting of more
than 2 components. For a system of n components consisting
of components A, B, C, D, ... N, n equations can be written
P A = P VA x A
P B = P VB x B
P C = P VC x C
DRAFT
P D = P VD x D
:
:
P N = P VN x N
With such a system of operations, we could solve for the
unknown quantities. Raoult's law finds ample applications in
separation processes. A particular example is in the equilibrium flash distillation of a mixture of petroleum fractions. A
liquid petroleum fraction consisting of several components is
heated and is brought to a desired temperature and the
equilibrium vapor is allowed to form. This vapor can be
separated and condensed to give a product different from the
original liquid.
Example 7-4
(a) What is the bubble point of a liquid mixture of 45% toluene
and 55% benzene if the pressure is 1000 mm Hg? What is the
composition of the first vapor formed?
(b) What is the dew point of a gaseous mixture of 45% toluene
and 55% benzene at a pressure of 1000 mm Hg? What is the
composition of the liquid formed?
Solution:
The problem involves finding the equilibrium temperature as
well as the composition. This type of problem requires a trialand error solution if tabular or graphical data are to be used.
(a) With t representing toluene and b, benzene,
yt =
222
P vt
x,
PT t
yt =
P vt
(0.45)
1000
National Engineering Center CEEC Seminar June 2 -4, 2004
yb =
P vb
x ,
PT b
yb =
P vb
(0.55)
1000
The vapor pressures Pvt and Pvb are functions of temperature
but there is only one temperature that will make yt + yb = 1.
Therefore, several temperatures should be tried until the
relation is satisfied.
Assuming a temperature of 90°C, Pvt = 420 mm Hg and Pvb =
1000 mm Hg.
y t = 420 (0.45) = 0.19
1000
y b = 1000 (0.55) = 0.55
1000
y t + y b = 0.19 + 0.55 = 0.74 ! 1.00
Assuming T = 100°C, Pvt = 550 mm Hg and Pvb = 1300 mm
Hg.
y t = 550 (0.45) = 0.248
1000
DRAFT
y b = 1300 (0.55) = 0.715
1000
y t + y b = 0.248 + 0.715 = 0.963 ! 1.00
Assuming T = 105°C, Pvt = 640 and Pvb = 1500 mm Hg
y t = 640 (0.45) = 0.288
1000
y b = 1500 (0.55) = 0.825
1000
y t + y b = 0.288 + 0.8255 = 1.113 ! 1.00
T should be between 100°C and 105°C. With T = 102°C, Pvt =
580 mm Hg and Pvb = 1350°C.
y t = 580 (0.45) = 0.261
1000
y b = 1350 (0.55) = 0.743
1000
y t + y b = 0.261 + 0.743 = 1.004
The bubble point is close to 102°C with yt = 0.26 and yb =
0.74.
(b) The solution is similar to the solution of (a) but with the
composition of the liquid as the unknowns.
Equilibrium 223
P
y t = Pvt x t = 0.45
T
P
y b = Pvb x b = 0.55
T
Different temperature are assumed until xt + xb = 1 is satisfied.
A trial-and-error solution can be avoided if the Antoine
equation for benzene and for toluene is used. From the
Appendix,
log 10 P vb = 6.90565 − 1211.033
220.790 − T
and
log 10 P vt = 6.95464 − 1344.800
219.482 − T
for benzene
for toluene
However the resulting system of 3 equations in 3 unknowns
(the third equation being yt + yb = 1 for solution (a) is
non-linear and is quite difficult to solve manually. However,
we can easily solve them using one of the many available
computer methods of solution.
Example 7-5
DRAFT
What is the total pressure exerted by a liquid-vapor system
in equilibrium if the liquid phase consists of 10% propane,
10% butane, 30% pentane, 10% hexane, and 40% octane at a
temperature of 10°C?
Solution:
At 10°C, the vapor pressure of the components are:
(a)
propane: 4600 mm Hg
(b)
butane : 1000 mm Hg
(c)
pentane: 250 mm Hg
(d)
hexane : 68 mm Hg
(e)
octane : 5.5 mm Hg
PT = Pa + Pb + Pc + Pd + Pe
= P va x a + P vb x b + P vc x c + P vd x d + P ve x e
= 4600(0.1) + 1000(0.1) + 250(0.3) + 68(0.1) + 5.5(0.4)
= 644 mm Hg
Example 7-6
Application of Raoult's Law
A liquid mixture consists of 65% n-pentane and 35%
n-hexane by mole. It is placed in a closed container and is
heated to 100°C and a pressure of 4 atm. What is the resulting composition of the vapor and liquid phase in equilibrium?
What percent of the original liquid is vaporized?
Solution:
224
National Engineering Center CEEC Seminar June 2 -4, 2004
We can consider that the input is the liquid (65% n-pentane
and 35% n-hexane) and the outputs are the vapor and liquid
in equilibrium although they are actually in one closed
container. The most convenient basis is 100 mols of the initial
liquid mixture. Since the system has no "tie" component, we
should use algebraic method.
Required:
a) yp, yh, xp, xh
b) percent of original liquid that vaporized
Vapor
yp , yh
T = 100oC
P = 4 atm
Liquid feed
Liquid
65% n-pentane
35% n-hexane
xp , xh
Figure 7-5
DRAFT
Basis: 100 moles of liquid feed
Let
V = vapor formed
L = liquid remaining
Overall mass balance:
100 = V + L
(1)
(0.65)(100) = yp V + xp L
(2)
n-pentane balance:
Using Raoult's law (at 100°C),
Pvp = 4700 mm Hg, vapor pressure of n-pentane
Pvh = 1700 mm Hg, vapor pressure of n-hexane
For n-pentane
P vp
x = 4700 x
P T p 4(760) p
(3)
P
y h = Pvh x h = 1700 x h
4(760)
T
(4)
yp =
For n-hexane
The composition relationship are:
yp + yh = 1
(5)
xp + xh = 1
(6)
Substitute equations (3) and (4) in equation (5)
Equilibrium 225
4700 x + 1700 x = 1
3040 p 3040 h
Substitute equation (6),
4700 x + 1700 (1 − x ) = 1
p
3040 p 3040
( 4700 − 1700 )x p = 1 − 1700
3040 3040
3040
x p = 0.4467
x h = 1 − 0.4467 = 0.5533
y p = 4700 (0.4467) = 0.6906
3040
y h = 1 − 0.6906 = 0.3094
From the n-pentane balance:
0.65(100) = 0.6906V + 0.4467(100 − V)
DRAFT
V = 83.36
L = 100 − 83.36 = 16.63
% of the liquid feed that is vaporized = 83.36 (100) = 83.36%
100
HENRY'S LAW
Some mixtures do not follow Raoult's Law. However, we can
express their behavior by using Henry's Law.
Pa = Haxa
where Ha is the Henry's Law constant, which is determined
experimentally. The values are available in references. Henry's
Law is similar to Raoult's Law wherein Ha replaces PVa.
MIXTURE OF IMMISCIBLE LIQUIDS
When two liquids are immiscible and vapor-liquid equilibrium
exists, the total pressure in the system is equal to the sum of
the individual vapor pressures of the components.
EQUILIBRIUM INVOLVING A CONDENSABLE AND A PERMANENT GAS; HUMIDITY AND SATURATION
Equilibrium in systems consisting of a condensable vapor
(substances which exist as liquids at ordinary temperatures)
and a permanent gas (substance that exists as a gas at
ordinary temperature and not easily liquefied by ordinary
means; its critical point is much lower than room
temperature) are frequently encountered. Examples of
condensable vapors are water, acetone, ether, ethyl alcohol,
226
National Engineering Center CEEC Seminar June 2 -4, 2004
benzene, hexane, etc.; while of permanent gases, air, CO2, N2,
O2, Ne, etc.
The air-water system is a common system. The air in the
atmosphere is in contact with bodies of water and we encounter air-water equilibrium. This is important in many aspects of
everyday life. In this system, air is nearly insoluble in water.
Applying Raoult's law,
P H 2 O = (P vH 2 O )(x H 2 O )
The fraction of water is nearly equal to 1.000 because of the
low solubility of air in water. Therefore,
DRAFT
P H 2 O = P VH 2 O
The partial pressure of water is equal to its vapor pressure
at the same temperature. The air is said to be saturated with
water vapor.
The system is commonly encountered in industry. An
example is in air conditioning. The same is true for the
weather. Extensive data for air-water system are available. In
our discussion, the principles as well as the terms in air-water
system are applicable to other
Air + Water vapor
systems involving a condensable
PH2O, Pair
vapor and permanent gas.
Liquid water
HUMIDITY AND SATURATION
xH2O = 1.000
We call air containing water,
"humid" air. "Humidity" is the
measure of the concentration of
Figure 7-6
water vapor in an air-water
system. Absolute molar humidity is the moles of water vapor
per mole of dry air. Absolute mass humidity is the mass of
water vapor per unit mass of dry air.
y H2O
, mole ratio of water to air
Y=
1 − y H2O
where Y is the absolute molar humidity
y H 2 O is the mole fraction of water
1 − y H 2 O = mole fraction of air
Assuming ideal gas behavior,
PH O
PH O
Y = P 2 = P − P2
air
T
H2O
where
P H 2 O = partial pressure of water
P air = partial pressure of air
PT = total pressure.
The mass humidity is
Equilibrium 227
Y∏ =
y H2O
% 18
1 − y H 2 O 29
where 18 is molecular weight of water and 29, that for air.
At equilibrium or saturation conditions, the saturation
humidity is
Y sat =
y H 2 O,sat
P VH 2 O
=
1 − y H 2O,sat P T − P VH 2 O
Instead of using absolute humidity, a common practice is to
express composition of humid air relative to composition at
saturation. These are percentage humidity and relative humidity. We can substitute the term "saturation" for "humidity"
Percentage humidity =
Y % 100
Y sat
Dry air has 0% humidity while saturated air has 100%
humidity
DRAFT
y H2O
P H2O
1 − y H2O
P T − P H2O
% 100 = y H2O,sat
% 100
Percentage Humidity =
P VH2O
1 − y H2Osat
P T − P VH2O
On the other hand,
Relative humidity =
P H2O
% 100
P VH 2 O
which is the ratio of the partial pressure of water to the
vapor pressure of water at the same temperature. This term is
widely used in weather reports and in industry.
In chemical engineering we prefer the use of percentage
humidity or saturation. Comparing,
Percentage humidity = relative humidity %
P T − P VH 2 O
P T − P H2O
These terms and the formulas are also applicable for
systems other than air-water, with percent saturation and
relative saturation being more appropriate.
Equilibrium or saturation conditions of a mixture of a gas
and a vapor give the maximum amount of vapor a gas can
hold at a particular temperature. The partial pressure of the
vapor component is equal to the equilibrium vapor pressure.
When a gas-vapor mixture is cooled, the "dew point" is the
temperature at which the vapor begins to condense. The dew
point refers to that temperature at which the vapor pressure is
equal to the partial pressure of the vapor. For example if a
benzene-air mixture at 40°C has a partial pressure of benzene
equal to 15 mm Hg, then the dew point is the temperature at
which the vapor pressure of benzene is equal to 15 mm Hg.
From the appendix, PVbenzene = 15 mm Hg corresponds to -7°C.
This is the dew point of the mixture.
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National Engineering Center CEEC Seminar June 2 -4, 2004
Example 7-7
200 grams of steam is injected into a large enclosed container
containing 50 kg of air. If the temperature is 30°C and the
pressure is 1.2 atm, calculate the
a) mass humidity
b) molar humidity
c) percentage humidity
d) percent relative humidity
e) dew point.
Solution:
a) The mass humidity,
Y∏ =
mass H2 O
mass dry air
=
200 g H2 O
(50)(1000) g air
= 0.004
g H2O
g dry air
DRAFT
b) The molar humidity,
Y=
=
mol H2 O
mol dry air
200/18
(50)(1000)/29
= 0.00644
mol H 2 O
mol dry air
c) The percentage humidity =
Y % 100
Y sat
At 30°C, the vapor pressure of water,
PT = 1.2 atm
y sat = P
P vH2O
30
=
= 0.0340
(1.2)(760) − 30
T − P vH 2 O
Percentage humidity = 0.00644 % 100 = 18.94%
0.0340
PH O
d) The percent relative humidity = P 2
VH 2 O
The partial pressure of water
= y H2O P T
200
18
=
(1.2)(760)
(50)(1000)
200 +
29
18
Equilibrium 229
= 5.84 mm Hg
Relative humidity = 5.84 % 100 = 19.46%
30
e) The dew point
Corresponding to a vapor pressure of 5.84 mm Hg (0.1130
psia), the equilibrium temperature is equal to 37.99°F (=
3.33°C), which is the dew point. The Steam Table was used in
getting the value.
Example 7-8
DRAFT
A gas-vapor mixture consists of benzene and nitrogen and
has a percentage saturation of 65%. The total pressure is 800
mm Hg and the temperature is 40°C.
a) What is the dew point and the percent relative saturation
of the mixture?
b) If the mixture is heated to 70°C, what is the value of the
percentage saturation, the relative saturation, and the
dew point with the total pressure maintained at 800 mm
Hg?
c) Up to what pressure can the mixture be compressed
without condensing any of the benzene if the temperature
is kept constant at 40°C
Solution:
a) At 40°C, the vapor pressure of benzene is equal to 170
mm Hg.
With the percentage saturation equal to 65%,
P benzene
P T − P benzene
= 0.65
P Vbenzene
P T − P Vbenzene
P benzene
170
= 0.65(
)
800 − P benzene
800 − 170
P benzene = 119.4 mm Hg
P benzene
% 100
Percent relative saturation = P
Vbenzene
= 119.4 % 100 = 70.22%
170
The dew point is the equilibrium temperature corresponding
to 119.4 mm Hg and is equal to 31°C.
b) At 70°C, the vapor pressure of benzene, PVbenzene = 540 mm
Hg. The partial pressure of benzene is still equal to 119.4 mm
Hg.
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National Engineering Center CEEC Seminar June 2 -4, 2004
P benzene
P T − P benzene
% 100
Percent saturation =
P Vbenzene
P T − P Vbenzene
119.4
800
− 119.4 % 100 = 8.45%
=
540
800 − 540
The dew point is still the same as in (a) (31°C) since the
partial pressure of benzene has not changed.
c) The N2 gas can hold all the benzene if the partial pressure
of benzene is less than the vapor pressure at 40°C. When the
partial pressure becomes equal to the pressure, the benzene
starts to condense.
The mole fraction of benzene in the uncompressed mixture,
y benzene =
P benzene 119.4
= 800 = 0.1492
PT
DRAFT
The maximum pressure, therefore, corresponds to the
partial pressure of benzene = 170 mm Hg.
P Vbenzene
PT
0.1492 = 170
PT
P T = 170 = 1140 mm Hg
0.1492
y benzene =
WET AND DRY BULB TEMPERATURES
Aside from pressure, humidity, temperature, we can characterize a vapor-gas mixture in equilibrium in terms of wet and
dry bulb temperatures. The dry-bulb temperature of a vaporgas mixture is the temperature registered by an ordinary
thermometer. The wet-bulb temperature corresponds to the
temperature registered by a thermometer with a wet bulb (with
water in the air-water system) but with the following
condition: the liquid should be in equilibrium with the
gas-vapor mixture. We can explain this using the air-water
system. When a thermometer with the bulb wet with water
(the bulb is covered by a wet wick) is in contact with an
air-water-vapor mixture, water evaporates from the wick. This
results in the cooling of the bulb since evaporation consumes
energy. At the same time, heat from the surroundings transfers to the wick since a temperature difference (between the
wick and the surroundings) exists as a result of the cooling of
the wick. When the rate of heat transfer from the surroundings becomes equal to the rate of heat usage for evaporation
(steady state), the thermometer registers the equilibrium
temperature. We call this temperature the wet bulb
Equilibrium 231
temperature. This temperature is also known as the adiabatic
saturation temperature.
We measure the wet- and dry-bulb temperature by using a
psychrometer. This instrument consists of two thermometers,
one having a wetted wick and the other, dry. To obtain the wet
and dry bulb temperatures of an air-water mixture, we swing
the instrument through the gas mixture for a reasonable
length of time until equilibrium is attained. We can read the
values directly. The wet and dry bulb temperatures are
functions of the water vapor content of air. They can be used
to indicate the moisture content.
THE HUMIDITY OR PSYCHROMETRIC CHART FOR THE
AIR-WATER SYSTEM
Y
%
70
(1 Y
00 sa
% t vs
H T
um sa
id t
ity
)
80
90
%
% Hum
idi
ty
DRAFT
We can represent the humidity data for the air-water system
at 1 atm in graphical form. The variables are humidity, dry
bulb temperature, wet bulb temperature, and percentage
saturation (or relative humidity). Fig. 7-7a shows a schematic
representation of Fig. 7-7b and 7-7c, the psychrometric chart
for air-water system at 1 atm. The molar humidity, Y is set on
the ordinate while the dry bulb temperature (or simply the
temperature) is set on the abscissa. The saturation humidity
versus the saturation temperature curve represents the conditions at 100% saturation. Curves for lower percentage saturations are similarly drawn.
%
60
Constant wet
bulb lines
T
Figure 7-7a. Humidity chart for air-water system
Parallel diagonal lines represent constant wet bulb
temperatures. Specifying any two of the four variables (humidity, percentage humidity, dry bulb temperature, and wet bulb
temperature) fixes the condition of an air-water mixture. For
example, given the dry bulb temperature and the percentage
humidity, the humidity and the wet bulb temperature can be
read from the graph. Given the wet bulb and the dry bulb
temperatures, the humidity and the percentage humidity can
be obtained from the graph.
232
National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
Equilibrium 233
DRAFT
234
National Engineering Center CEEC Seminar June 2 -4, 2004
We can trace the process undergone by an air-water mixture
on the humidity chart. From Fig. 7-8, heating an air-water
mixture from T1 to T2 at constant Y is represented by a
horizontal line with the arrow pointing to the right. Cooling is
represented by the other horizontal line from point 3 to point
4.
The dew point temperature is a variable that is implicitly
represented on the chart. If we cool an air-water mixture at
constant humidity, water begins to condense when the satura-
Y
Heating
2
1
4
3
Cooling
DRAFT
T1
T4
T1
T3
T2
T
Figure 7-8
tion value is reached. This corresponds to the temperature
known as the dew point. In Fig. 7-9a, this corresponds to
point B on the equilibrium line.
We can dehumidify humid air by chilling it below its dew
point. At the dew point, the air-water mixture is saturated
Y
Equilibrium
line
B
Y1
Y2
1
Air in
2
T1 Y1
Dew
point
T2
TB
(a)
T1
T
Chiller
Air out
T2 Y2
Condensed
H2O
(b)
Figure 7-9 Chilling of air-H2O below its dew point. (a) process on the
humidity chart (b) block diagram.
Equilibrium 235
with water vapor. Cooling it further makes the water in excess
of the saturation value to condense out.
If a mixture at T1 is cooled to T2, then the line representing
the process crosses the saturation line at B before reaching
point 2. Cooling below TB to T2 results in water condensing out
from the air-water mixture, i.e., the value of Y1 decreases to
Y2. In this process, the tie component is the bone-dry air.
Y1
mol H 2 O condensed
mol H 2 O
mol H 2 O
− Y2
= (Y 1 − Y 2 )
mol dry air
mol dry air
mol dry air
ADIABATIC HUMIDIFICATION AND DEHUMIDIFICATION
DRAFT
Keeping the wet bulb temperature constant, we can increase
or decrease the humidity of an air-water mixture. These operations are also known as adiabatic humidification and
dehumidification, respectively. On a humidity chart, these
processes follow paths parallel to the diagonal constant wet
bulb lines. In adiabatic humidification, the air-water mixture
picks up water while maintaining its wet bulb temperature
constant. In adiabatic dehumidification, the mixture loses
moisture. Fig. 7.10a shows these operations. Line AB represents adiabatic humidification while line CD represents
adiabatic dehumidification. Note that the dry bulb temperature is changing.
Fig. 7.10b represents a block diagram for a humidifier while
Fig. 7.10c represents a block diagram for a dehumidifier. In
TA
Adiabatic
Humidifier
YA
Y
YB
H2O in
B
YB
TB
(b)
YA
A
C
YC
TC
YD
YC
D
TC
TB
TD TA
Adiabatic
Dehumidifier
TD
YD
H2O out
T
(c)
(a)
Figure 7-10
humidification, the water picked up by air is
(Y A − Y B )
mol H 2 O
mol dry air
while for dehumidication, the water removal is
(Y C − Y D )
236
mol H 2 O
mol dry air
National Engineering Center CEEC Seminar June 2 -4, 2004
Some drying operations can be carried out at constant wet
bulb temperature or in other words, an adiabatic humidification process.
Example 7-9
Given the following conditions for an air-water mixture at 1
atm:
a) WB temp = 25°C, DB temp = 55°C, find the molar humidity and the percentage saturation.
b) Percentage saturation = 30%, DB temp = 38°C, find the
molar humidity, the WB temp, and the dew point.
c) Dew point = 15°C, WB temp = 30°C, find the molar
humidity, the percentage saturation, and the DB temp.
Solution:
The answers can be read from the psychrometric chart.
a) The point is represented by the intersection of the 25°C
diagonal wet bulb temperature line and the 55°C vertical dry
bulb line.
DRAFT
Reading from the chart,
Y = 0.013
kmol H 2 O
kmol dry air
Percentage saturation = 7%
b) The point is represented by the intersection of the 38°C
vertical dry bulb line and 30% saturation curve.
From the graph,
Y = 0.022
kmol H 2 O
kmol dry air
WB temp = 24°C
Dew point = 19°C
c) Dew point = 15°C is represented by the intersection of the
15°C dry bulb line and the 100% saturation curve. The 100%
kmol H2 O
. The condisaturation humidity is equal to 0.017
kmol dry air
tion of the air is, therefore represented by the intersection of
the horizontal line Y = 0.017 and the diagonal line WB = 30°C.
Reading from the chart,
% saturation = 4%
DB temp = 68°C
Example 7-10
Equilibrium 237
An air-water mixture at 60°C and at 20% humidity is cooled to
25°C. Trace the process on the humidity chart and find the
final humidity and the percentage saturation of the mixture.
Solution:
The bold line with arrows represents the cooling of the
100%
Humidity
Y
Condition 1
20% humidity
DBT = 60oC
Final
condition
DRAFT
25oC
32oC
T
60oC
Figure 7-11
mixture. At 32°C, the saturation value or the dew point is
reached and water vapor begins to condense out of the
mixture. Continuous cooling causes the condensation of
water. At the final condition, where the temperature equals
25°C,
kmol H 2 O
Y = 0.033
kmol dry air
and the mixture is 100% saturated with water vapor.
Example 7-11
Drying with conditioned air
A rotary drier dries a crystalline organic solid. The solid enters
with 20% water and leaves with a water content of 0.3%.
Conditioned air is used for drying.
Air at 28°C and 85% humidity is first chilled to 10°C and
then reheated to 88°C and fed to the drier counter-current
with the solids. It goes out of the drier at 40°C dry bulb and
32°C wet bulb.
a) For every 100 kg of dried product, what volume of fresh
air (28°C and 85% humidity) is required?
b) How much water condenses at the chiller?
Solution:
The humidity of the air at different conditions are needed.
We should trace the changes on a humidity chart.
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National Engineering Center CEEC Seminar June 2 -4, 2004
1
o
2
DBT = 28 C
85% humidity
Chiller
3
DBT
= 10oC
Condensed
H2O
4
DBT = 40oC
Heater
DBT =
88oC
Drier
WBT = 32oC
Solids in
20% H2O
Dried product out
100 kg
0.3% moisture
Figure 7-12
From the humidity chart:
Y 1 = 0.035
Y 2 = Y 3 = 0.013
mol H 2 O
(only heating with no humidity change)
mol dry air
Y 4 = 0.048
DRAFT
mol H 2 O
(at 28°C and 85% humidity)
mol dry air
mol H 2 O
(at 40°C DB and 32°C WB)
mol dry air
Equilibrium
curve
4
Y
1
2
10oC
28oC
3
T
40oC
88oC
Basis:
100 kg
Figure 7-13
of dried product
a) The bone-dry solid
and bone-dry air are the
key components. The
drier consists of
the
lower and upper parts.
Considering a balance
around the lower part of
the drier, the incoming
wet solids has the ratio,
Air in
3
4
Air out
Water evaporated
Solids out
Solids in
0.3% water
20% water
Figure7-14
1 kg wet solids in
(1 − 0.2) kg bone-dry solids
while the dried product has the ratio,
Equilibrium 239
1 kg wet solids in
(1 − 0.003) kg bone-dry solids
The difference,
(
kg water evaporated
1
1
−
), is
1 − 0.2 1 − 0.003
kg bone-dry solids
kg water evaporated =
100 kg product %
%(
(1 − 0.003) kg bone-dry solids
kg product
kg water evaporated
1
1
−
)
1 − 0.2 1 − 0.003 kg bone-dry solids
kg water evaporated = 24.62 kg.
Considering a balance around the upper portion,
kg water evaporated = kg water picked up by the air
= 24.62 kg
DRAFT
kmol water picked up by the air = 24.62 = (1.368)
18
kmol bone-dry air = 1.368 kmol H 2 O picked %
(Y 1
−Y )
4
3
= 1.368 % (
kmol bone-dry air
kmol water picked up
1
) = 39.02 kmol
0.048 − 0.013
= 39.09 kmol
Since dry air is a key component, it is also the amount of
dry air in the entering fresh air,
bone-dry air in entering fresh air = 39.09 kmol
humid air entering = (39.09) (1 + Y1) kmol
Assuming air to be an ideal gas, the volume of the entering
fresh air is:
(39.09)(1 + 0.035) kmol %
22, 400 liters 28 + 273
(
)
0 + 273
kmol
= 1.00 % 10 6 liters
b) The same amount of bone-dry air will pass through the
chiller and the water condensed will depend on the change in
humidity.
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National Engineering Center CEEC Seminar June 2 -4, 2004
Water condensed in chiller =
39.09 kmol bone-dry air % (0.035 − 0.013)
%
kmol water
kmol bone-dry air
18 kg 1000 g
= 15, 480 g
%
kmol
kg
Example 7-12 Drying with air recycling
A dryer is used to process 2000 kg/hr of a material containing
40% moisture to a final product containing 5% moisture. The
dryer consists of an air heater and a drying chamber. To save
energy in air heating, part of the exhaust air is recycled, mixed
with the fresh air, and the resulting air mixture heated before
feeding to the chamber. 30,000 m3 of fresh air per hour at
30°C dry bulb temperature and 65% saturation are used. This
is mixed with the recycled air and heated to 100°C dry bulb
temperature. Its wet bulb temperature is 42°C. The exhaust
air is at 60°C. Calculate
a) the humidity of the exhaust air.
b) the fraction of the exhaust air that is recycled.
DRAFT
Recycled air
Fresh air
30000 m3/hr
DBT = 30oC
65% saturation
PT
Heater
Dried
product
5% H2O
DBT = 100oC
WBT =
42oC
60oC
Drying
Chamber
Exhaust air
Wet feed
40% H2O
2000 kg/hr
Figure 7-15
Solution:
Required:
a) Humidity of the exhaust air
b) The fraction of the air from the dryer that is recycled.
As in Ex. 7-11, the arithmetic method will be used.
Basis:
1 hour
Fig. 7-14 shows the schematic diagram that includes the
recycle stream. First, consider the lower portion of the drier to
calculate the amount of water evaporated. This water is the
same water picked up by the air. Set up a mass balance
around the upper portion of the drier to determine the exit air
humidity. Next, set up a mass balance around the upper
portion of the drier within the recycle loop to determine the
total bone-dry air entering the drier. The bone-dry air in the
net exhaust air is the same as the bone-dry air in the fresh air
feed. The recycled bone-dry air is equal to the difference
between the gross bone-dry air entering the drier and the
bone-dry air in the exhaust air.
Consider the lower portion of the drier.
Water evaporated from the wet stock =
Equilibrium 241
kg H 2 O evap
( 0.4 − 0.05 )
1 − 0.4 1 − 0.05 kg bone-dry solids
Water evaporated
Dried product
5% moisture
Wet feed
40% water
2000 kg/hr
Dryingchamber
0.05 kg water
(1 - 0.05) kg bone-dry solids
0.4 kg water
(1 - 0.4) kg bone-dry solids
Figure 7-16
%
0.6 kg bone-dry solids
% 2000 kg wet feed
kg wet feed
= 736.8 kg H2 O µ
736.8 kg H2 O
= 40.94 kmol
18 kg/kmol
= water picked up by air
DRAFT
Consider the upper portion and set up a water balance
around the recycled loop:
At 30°C and 65% saturation, the molar humidity of the fresh
Recycled air
Fresh air
Heater
60oC
Drying
Chamber
Net exhaust air
40.94 kmol H2O
Figure 7-17
air,
kmol H 2 O
kmol bone-dry air
Bone-dry air balance:
The amount of bone-dry air in the fresh air
Y = 0.028
0 + 273 % 1 kmol %
= 30, 000 m3 % 30
+ 273 22.4 m3
1
1 + 0.028
= 1174 kmol
= amount of bone-dry air in the net exhaust air
total H2O in the incoming fresh air
= 1174 kmol bone-dry % 0.028
242
kmol H 2 O
kmol bone-dry air
National Engineering Center CEEC Seminar June 2 -4, 2004
= 32.87 kmol
Water balance:
total H2O in exit air = 32.87 + 40.94 = 73.81 kmol
Humidity of exhaust air =
73.81 kmol H2 O
= 0.063
1174 kmol bone-dry air
Considering a balance within the recycle loop,
The humidity of the heated air (DB = 100°C; WB = 42°C)
Dryingchamber
Yi = 0.044
DBT = 100oC
WBT = 42oC
Yo = 0.063
40.94 kmol water
picked up
Figure 7-18
DRAFT
= 0.044
kmol H2 O
kmol bone-dry air
The bone dry air passing through the drier =
40.94 kmol H2O picked up %
= (40.94)(
1
kmol H 2 O picked up
(Y o − Y i )
kmol bone-dry air
1
) = 2155 kmol
0.063 − 0.044
Balance around the point of splitting:
The amount of dry air in the recycle
=2155 - 1174 = 981 kmol
Fraction of air recycled = 981 = 0.455
2155
LIQUID-LIQUID EQUILIBRIUM
When we bring two different liquids in contact with one
another, they may mix in all proportions (completely miscible),
may not mix at all and form two layers (totally immiscible), or
may partially mix, still forming two layers (partially miscible).
Ethyl alcohol and water are completely miscible; oil and water,
totally immiscible, and nitrobenzene and water, partially
miscible. The mixing property is a function of the molecular
structure of the substances, temperature, and pressure. In
general, substances with similar structure are miscible.
In chemical engineering, we are particularly interested in
liquid systems consisting of three components. These find
applications in the extraction of valuable liquid substances.
Equilibrium 243
DRAFT
Applying the phase rule to this type of system: with C = 3 and
P = 2, the degree of freedom, F = C + 2 - P = 3 + 2 -2 = 3. In
this case, the temperature T (or pressure P) and the compositions (two) should be specified in order to fix the system. We
often desire to separate a valuable substance (called the
solute) in a pure form from a diluent. In order to accomplish
this, we bring another substance (a solvent) in contact with
the mixture. Mass transfers to the solvent. This transfer takes
place due to the concentration difference of the solute in the
two layers that are present. At equilibrium, the ultimate distribution of the solute is reached.
In these processes, the equilibrium diagram is essential. We
can determine this equilibrium diagram experimentally. For a
three-component system, the triangular graph is useful. As an
example, consider the system consisting of methyl isobutyl
ketone (MIK), acetone, and water. Acetone is considered the
solute or the valuable component and methyl isobutyl ketone
the extracting solvent. To obtain the equilibrium diagram
experimentally, measured amounts of the three components in
different concentrations are mixed thoroughly. The layers are
allowed to separate and reach equilibrium, and the concentrations (mass fraction) of each layer are measured. The concentration of the upper layer is denoted as y while the concentration of the lower layer as x. Acetone is component a; water,
component b; and MIK, c. Table 7-1 shows the equilibrium
composition of this system.
The concentration in the left group has its corresponding
equilibrium concentration on the right group. These are
plotted on a triangular diagram. See Fig. 7-19. The ordinate
represents the mass fraction of MIK (yc or xc) while the
abscissa represents the mass fraction of acetone (ya or xa).
First, the concentration of the upper layer (MIK rich) are
plotted, ya vs. yc and is now represented by the curve AP in the
diagram, xa vs. xc and is now represented by curve BP.
Table 7-1. Equilibrium composition of the system H2O-MIK-acetone
at 25°C, mass fraction1
Upper Layer
yb
yc
xa
Lower Layer
xb
xc
0.0460
0.0233
0.9320
0.0290
0.9500
0.0210
0.1895
0.0386
0.7730
0.1400
0.8400
0.0240
0.2440
0.0466
0.7100
0.1490
0.8140
0.0270
0.2890
0.0553
0.6650
0.1900
0.8310
0.0290
0.3760
0.0782
0.5470
0.2780
0.6840
0.0380
0.4320
0.1070
0.4620
0.3380
0.6060
0.0560
0.4700
0.1480
0.3830
0.4050
0.4950
0.1000
0.4830
0.1880
0.3280
0.4600
0.3700
0.1700
ya
ya or xa
ya or xa
ya or xa
plait point:
0.4850
0.2450
0.2700
1
Othmer, White, and Trueger, Industrial Engineering Chemistry, Vol. 33,
1941, p.1240.
The common point P is called the plait point at which only
one layer exists. The region within the envelop APB is called
244
National Engineering Center CEEC Seminar June 2 -4, 2004
the two-phase region. Only one phase exists outside this
envelop.
Whenever the composition of a three-component mixture is
1.0
A
0.9
xc,yc, mass fraction methyl isobutyl ketone
0.8
0.7
C
0.6
0.5
two phases
0.4
0.3
P
0.2
one phase
B
0.0
0.0
D
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.9
1.0
xa,ya, mass fraction acetone
0.7
xc,yc, mass fraction acetone in upper layer
DRAFT
0.1
0.6
x =y
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
xa, mass fraction acetone in lower (water) layer
within the two-phase region, two layers result. This points are
represented by a point on the upper equilibrium curve and
another on the lower equilibrium curve. A line joining two
equilibrium points is called a tie line. In the figure, line CD is
a tie line. To avoid drawing so many tie lines, an equilibrium
curve xa vs. ya is drawn directly below the triangle. With the
Equilibrium 245
aid of the equilibrium line ya vs. xa, corresponding equilibrium
points can easily be located. The next example illustrates this.
1.0
0.9
DRAFT
xc,yc, mass fraction methyl isobutyl ketone
0.8
0.7
A
0.6
0.5
0.4
0.3
0.2
0.1
D
0.0
0.0
0.1
0.2
0.4
0.3
0.5
0.6
0.7
0.8
0.9
1.0
0.7
0.8
0.9
1.0
xa,ya, mass fraction acetone
xc,yc, mass fraction acetone in upper layer
0.7
0.6
0.5
0.4
B
C
0.3
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
xa, mass fraction acetone in lower (water) layer
Figure 7-20
246
National Engineering Center CEEC Seminar June 2 -4, 2004
Example 7-13
For the system H2O-MIK-acetone, find the composition of the
lower layer in equilibrium with the upper layer that has a
composition, ya = 0.33 and yc = 0.6.
Solution:
Refer to Fig. 7-20 for the solution. The mixture containing ya =
0.33 and ya = 0.6 is located on the upper layer. This is represented by point A.
A vertical line is drawn from this point to a point B on the
line x = y on the graph below the triangle. A horizontal line
intersecting the equilibrium curve at C is drawn. From this
point a vertical line is traced until it intersects the envelop
representing the lower layer (point D).
Point D with xa = 0.23, and xc = 0.03 represents the mixture
in equilibrium with the mixture represented by point A. AD is
the line.
If the composition of the lower layer was given instead,
point A can be obtained by reversing the procedure.
DRAFT
Example 7-14
What results when 100 kg of pure methyl isobutyl ketone is
mixed intimately with 100 kg of a mixture containing 40%
acetone and 60% water? Solve graphically.
Solution:
This problem is solved using the equilibrium diagram. Pure
methyl-isobutyl ketone is represented by point A while the
mixture containing 40% acetone and 60% water, by point B.
(See Fig. 7-21)
The addition point of the resulting mixture is the midpoint C
of the line AB. Since C is within the two-phase region, it will
separate into two equilibrium layers. To find the composition
of this mixture, a tie line passing through C is located by trial
and error. This is line DE. Point E represents the lower layer
xa = 0.145 and xc = 0.03. Point D represents the upper layer,
ya = 0.23 and yc = 0.73. The relative amounts of the mixture E
and mixture D can be found graphically.
Mass D = CE = 4.6 cm = 2.091
Mass E
CD 2.2 cm
Mass D + Mass E = 200
Mass D
= 2.091
200 − Mass D
Mass D = 135.3 kg
Mass E = 64.7 kg
Equilibrium 247
1.0
A
xc,yc, mass fraction methyl isobutyl ketone
0.9
0.8
0.7
0.6
C
0.5
0.4
0.3
0.2
0.1
E
DRAFT
0.0
0.0
0.1
B
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.7
0.8
0.9
1.0
xa,ya, mass fraction acetone
xc,yc, mass fraction acetone in upper layer
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
xa, mass fraction acetone in lower (water) layer
Figure 7-21
Example 7-15
Acetone is desired to be recovered from a solution containing
40% acetone and 60% water by extraction with pure methyl
isobutyl ketone. 100 kg of the solution is to be processed and
248
National Engineering Center CEEC Seminar June 2 -4, 2004
100 kg of MIK is available. Calculate the percentage of acetone
that can be extracted if
a) 100 kg of MIK is used to extract the solution at once.
b) First, 50 kg of MIK is used to extract the solution and the
layer allowed to come to equilibrium. The water layer is then
re-extracted with the other 50 kg of MIK.
Solution:
100 kg MIK
40% acetone
Extract
Extractor
Raffinate
.
60% water
.
Figure 7-22
DRAFT
a) The conditions are identical to that of Example 7.14. The
results from that example can be used to calculate for the
percentage of acetone extracted. (Fig. 7-21)
mole fraction of acetone in upper layer = ya = 0.23
Total amount of upper layer = 135.3 kg
acetone extracted = 0.23(135.3) = 31.12 kg
acetone in original solution = 0.40(100) = 40 kg
% acetone = 31.12 % 100 = 77.8% extracted
40
b) 50 kg MIK is added to 100 kg solution containing 40%
acetone and 60% water. In Fig. 7-22, point A represents the
pure MIK and point B, the solution. The addition point of the
resulting mixture is C. AC = 100 = 2. Since C is inside the
50
BC
two-phase envelop, it will separate into two layers. To find the
composition and amounts of both layers, a tie line passing
through C is located by trial and error.
Upper layer (point D):
y a = 0.32
y c = 0.62
Lower layer (point E):
y a = 0.22
6 c = 0.035
The amount of D and E can be found graphically:
DC = 2.85 cm
CD = 2.85 cm
DE = 5.7 cm
Layer E = CD (150) = 2.85 (150) = 75 kg
5.7
DE
Layer D = EC (150) = 2.85 (150) = 75 kg
5.7
DE
Equilibrium 249
1.0
A
xc,yc, mass fraction methyl isobutyl ketone
0.9
H
0.8
0.7
D
0.6
0.5
F
0.4
C
0.3
0.2
0.1
E
DRAFT
G
0.0
0.0
0.1
0.2
B
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.7
0.8
0.9
1.0
xa,ya, mass fraction acetone
xc,yc, mass fraction acetone in upper layer
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
xa, mass fraction acetone in lower (water) layer
Figure 7-23
The acetone extracted in this stage = 75(0.315) = 23.6 kg
The layer E (lower or water layer) is re-extracted with 50 kg
MIK (point A). The addition point of A and E is point F. Again
the tie line (GH) is located by trial and error.
250
National Engineering Center CEEC Seminar June 2 -4, 2004
Upper layer:
Lower layer:
HF = 3.6 cm
ya = 0.17
yc = 0.79
ya = 0.095
yc = 0.03
FG = 3.8 cm
HG = 7.4 cm
Layer H = FG (125) = 3.8 (125) = 64.2 kg (upper layer)
7.4
HG
Layer G = HF (125) = 3.6 (125) = 60.8 kg (lower layer)
7.4
HG
The acetone extracted in the 2nd stage
= 0.17 (64.2) = 10.9 kg
Total amount of acetone extracted
= 23.6 + 10.9 = = 34.5 kg
% acetone extracted
DRAFT
= 34.5 % 100 = 86.25%
40
Notice that from a 77.8% extraction, using the same amount
of materials but a different manner of operation, the percentage extraction increased to 86.25%. This is a simple arrangement. Other types of arrangements can increase the percentage extraction even more. We will not discuss those because
they are beyond the scope of this book.
SOLID-LIQUID EQUILIBRIA
Another important application of equilibrium involves the
distribution of solids in some solvent. At a particular temperature and pressure, there is a definite amount of solute which a
solvent can hold in solution. This is an equilibrium on
saturated conditions. These conditions can be found experimentally and are known as solubility data presented in
tabular or graphical form. Solubility is commonly expressed in
terms of mass fraction, ( mass solute ); mass ratio,
mass solution
mass
solute
(
); or in concentration units, ( mass solute )
mass solvent
volume solution
PROBLEMS
7-1 Find the degrees of freedom for the following systems in
equilibrium:
a. salt and sugar (with undissolved quantities) in water.
b. hexane, heptane and octane in equilibrium with the
vapor.
c. oil and water in a closed vessel at 100°C.
Equilibrium 251
DRAFT
d. carbon, iron and silicon.
e. coffee, milk, and sugar.
7-2 Calculate the boiling point of
a. benzene at 30 in Hg.
b. water at 10 mm Hg.
c. toluene at 10 atm.
d. propane at 50 psig.
7-3 What is the vapor pressure of
a. water at 80°F?
b. n-octane at 250°C?
c. n-pentane at 125°C?
d. mercury at 200°C?
7-4 Given that glycerol at mm Hg boils at 167.2°C and at 760
mm Hg, 290°C, calculate the constants A and B in the
equation
Log P V = A - B
T+ C
where PV is vapor pressure in mm Hg
T, temperature in K, and C = 230
What is the vapor pressure of glycerol at 200°C?
7-5 A flue gas on a wet basis contains 10% CO2 , 1% CO, 5% O2,
5% H2O and 79% N2. It is at 200°C and 758 mm Hg. To what
temperature should it be cooled to make it saturated with
water vapor, keeping the total pressure constant at 758 mm
Hg?
7-6 In a room measuring 4m × 5m × 4m was left an open tank
containing ethyl alcohol. Considering that the room is sealed
from the outside, calculate the amount of alcohol that has
evaporated into the room if the temperature is 30°C and the
atmospheric pressure is 755 mm Hg. Assume that the room
becomes completely saturated with alcohol.
7-7 The atmospheric pressure varies with elevation according to
the equation
P = 29.92 - 0.00111h
where P is pressure in in Hg and h is the altitude in ft.
What is the elevation where water boils at 95°C?
7-8 A liquid mixture at 0°C consists of 30% ethane, 40% propane,
and 30% n-butane. Find the composition of the vapor
mixture in equilibrium with this liquid. What is the total
equilibrium pressure?
7-9 A gaseous mixture consists of 61% n-hexane, 26% n-octane
and 13% n-decane at a temperature of 50°C. Find the equilibrium pressure and the composition of the liquid in equilibrium with this mixture.
7-10 Find the dew point of a gaseous mixture of 40% ethyl alcohol
and 60% water at a total pressure of 2 atm. What is the
composition of the first liquid that appears?
7-11 Find the bubble point of a mixture of 50% n-pentane and
50% n-butane at a pressure of 4 atm. What is the composition of the first vapors formed?
252
National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
7-12 The composition of a liquefied petroleum gas (LPG) is 0.5%
ethane, 0.1% ethylene, 16.4% propane, 2.1% propylene,
74.0% butane, and 6.9% butene. If the room temperature is
30°C, what is the pressure inside the tank and the composition of the gas that first issues from the mixture?
7-13 A liquid mixture of 50% mole n-butane and 50% mol
n-pentane was placed in an evacuated container and a
temperature of 60°C and a pressure of 2590 mm Hg resulted.
At equilibrium, find the composition of both the liquid and
vapor phases. What fraction of the original liquid was vaporized? If the mixture is heated further to 80°C, what is the
resulting pressure?
7-14 A benzene-toluene gaseous mixture containing 65% benzene
and 35% toluene is cooled from 150°C to 90°C at a pressure
of 800 mm Hg. What fraction of the original vapor condenses?
What is the composition of this condensed liquid if equilibrium is attained?
7-15 A natural gas consisting of 1% mole ethane, 13% propane,
8% isobutane, 27% n-butane, 6% isopentane, 15%
n-pentane, and 30% hexane is flash distilled at 43°C and 3.5
atm. Set up the complete set of equations (including the
vapor pressure data) that will be used to find the amounts
and the resulting compositions of the vapor and the liquid
fractions that are formed.
7-16 A helium-toluene mixture contains 30% mole toluene at
100°C and 760 mm Hg. Calculate the percentage saturation,
relative saturation, dew point and the molar humidity of the
mixture.
7-17 200 grams of benzene is allowed to evaporate in a 10 m3 tank
containing CO2 gas at 40°C and 750 mm Hg. If the mixture is
brought to 60°C and 1.2 atm, what is the percentage saturation and the relative saturation of the mixture?
7-18 A nitrogen-acetone mixture containing 20% acetone and 18%
nitrogen is at a temperature of 30°C and a pressure of 755
mm Hg. To what temperature should the mixture be chilled
so that when heated back to 30°C at the same pressure, the
percentage saturation will be 40%?
7-19 A mixture of benzene-nitrogen contains 15% benzene and is
at 40°C and 780 mm Hg. What will happen if the mixture is
compressed until the final pressure becomes 1500 mm Hg
with temperature kept at 40°C? What is the final percentage
saturation and concentration, (kg benzene)/(kg benzene-free
nitrogen)?
7-20 Dry air is flowing at a rate of 3 m3/s at a temperature of 40°C
and 1.1 atm. 1 kg/min of ethyl alcohol is bled into the
stream. If the final temperature of the mixture is 35°C and
the pressure is 765 mm Hg, what is the percentage saturation of the mixture?
7-21 In the Philippines, it is common to have 32°C dry bulb
temperature and 29°C wet bulb temperature. What is the
Equilibrium 253
percentage humidity, the relative humidity and the dew point
of the air-water mixture?
7-22 Air at DBT = 65°C and WBT = 38°C (condition 1) is cooled at
constant humidity to its dew point (condition 2), then cooled
further to 21°C (condition 3), and finally heated to 54°C at
constant humidity (condition 4). Draw the changes undergone by the air on a graph similar to the psychrometric chart.
Make a tabulation of DBT, WBT, % humidity, humidity and
dew point for the air-water system at conditions 1 to 4.
7-23 For the following conditions given for the air-H2O system fill
DRAFT
Dry bulb T
Wet bulb T
% saturation
Relative humidity
Molar humidity
Dew point
up the blanks.
1
40°C
20°C
2
3
4
35°C
40°C
50%
60%
0.06
25°C
7-24 A room measuring 15m × 60m × 3m is to be supplied with
conditioned air at 24°C and 35% saturation. The air in the
room is to be changed completely every 15 minutes.
Outside air at 38°C and 85% saturation is first chilled to
condense out some of the water vapor and is then reheated to
24°C.
a. To what temperature must the outside air be chilled?
b. What is the volumetric flow rate of the humid air (at pt.
Air conditioner
1
2
38oC
Chiller
Heater
85% saturation
T=?
3
24oC
35% saturation
15 m x 60 m x 3 m
Room
4
Figure 7-24P
1) entering the chiller in ft3/min?
7-25 A drier is processing a solid containing 30% by weight
moisture to a final product containing 5% moisture. Air at
80°C and 100 kPa with an absolute humidity of 0.012 kmol
H2O/kmol dry air is used as the drying medium. It goes out
of the drier at 40°C and 750 mm Hg with a humidity of 0.065
kmol H2O/kmol dry air. If the water evaporated in the drier is
30 kg /min, calculate
(a)
the kg of product per minute
(b)
the volume of the outgoing air in m3/min
7-26 A continuous drier turns out 500 kg/h of a product containing 2.5% moisture from wet solids containing 35% H2O. The
254
National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
air entering the drier at 105°C dry bulb and 40°C wet bulb
temperatures. It goes out at 45°C with the wet bulb temperature remaining constant. What is the volumetric flow rate of
the inlet air if its pressure is 102 kPa?
7-27 A drier is used to dry inorganic crystals from 45% moisture to
2% moisture. The air enters at 150°C, molar humidity = 0.05
mol water/mol dry air, and discharges at 90°C, WBT = 47°C.
The entering air is prepared by recycling a portion of the
discharged air with ambient air at 25°C, molar humidity =
0.018 mol water/mol dry air. Calculate the percentage of
discharge air to be recycled and the volumetric flow rate of
fresh air (P = 100 kPa).
7-28 Brick clay is dried in a continuous drier from a moisture
content of 55% to 15%. The product rate is 5,000 kg per
hour. Air with a 25°C DBT and a 10°C WBT at a flow rate of
46,000 m3 per hour (P = 99 kPa) is mixed with recycled air
and is heated before entering the dryer. The exhaust air goes
out at 80% humidity. One half of the exhaust air is recycled
back. What is the condition of the exhaust air? If the WBT of
the mixed air is 5 degrees lower than that of the exhaust air,
to what temperature is the mixed air heated?
7-29 To condition a process air, atmospheric air humidity by direct
injection of steam, and then heating to the desired temperature. If the air used is at 30°C and 80% saturated, and the
final air is at 90°C and 45% saturated, find the amount of the
steam injected per 100 kg of humid air saturated (99 kPa). At
what minimum temperature is the steam leaving the injection
chamber heated? This is a function of the steam pressure
and temperature, but in this case, assume that the condition
of the steam cam be varied so that this lowest temperature
can result.
7-30 It is desired to recover benzene from a nitrogen-benzene
mixture at 50°C, 120 kPa, and 90% saturation. This mixture
is passed through a cooler from which it leaves at 15°C and
saturated. Calculate the kg benzene recovered for every 1,000
m3 of the original mixture.
7-31 To recover acetone from a plastic in a processing plant, warm
nitrogen is blown countercurrent to the material. The gas on
leaving the drying chamber passes through a chiller and then
recirculates back to the drier through a heater which brings
it back to the original temperature. The nitrogen enters at
45°C and the percentage saturation of acetone is 9%. It
leaves the drier at 20°C and 95% saturated. 600 kg/h of
plastic containing 12% acetone is processed per hour. To
what temperature should the nitrogen be cooled in the
chiller? What is the volume of the air that is recycled
(m3/min)?
7-32 In the following mixing procedure find out whether 1 or 2
phases will form. Give the composition (mass fraction) of the
layers after equilibrium is attained.
Equilibrium 255
a. 100 kg acetone, 150 kg water and 200 kg methyl isobutyl
ketone are mixed.
b. 162 water, 189 kg MIK, and 549 kg acetone are mixed.
c. 400 kg of MIK and 536 kg of water are mixed.
d. 700 kg water, 1000 kg acetone, 2000 kg MIK are mixed.
e. 300 kg MIK and 200 kg acetone are mixed.
7-33 One hundred kg of a mixture containing 30% acetone and
70% water is extracted in a single-stage extraction unit with
200 kg of methyl isobutyl ketone containing 1% acetone.
After equilibrium is attained, how much extract and raffinate
phases are formed? What is the percentage recovery of
acetone?
7-34 Five metric tons per hour of a solution containing 35%
acetone and 65% water is to be extracted in a single stage
extraction unit using pure methyl isobutyl ketone. If the
raffinate phase contain 2% acetone, calculate the flow rate of
the solvent and the percentage recovery of acetone.
DRAFT
7-35 Find the amount of MIK solvent necessary to extract 77.2%
of the acetone in 100 kg of a mixture containing 40% acetone
and 60% water in a single-stage extraction unit.
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DRAFT
—Chapter 8—
Energy Balance
Energy is important in chemical engineering. In almost every
phase of an industrial process, energy is involved. To keep
track of the operation, changes in energy are constantly
monitored. Most chemical reactions require energy initially
while some produce energy that can be used in a process.
Most often, energy in an industrial process is associated with
high temperature. However, we also consider other important
forms of energy. This chapter is an introduction to thermodynamics, a core subject of chemical engineering.
Today there is much concern over the energy needs of every
nation. Energy is now an important aspect of the world political scene. We are now experiencing a crisis in the price of
petroleum. The nonrenewable sources of energy such as petroleum and coal will ultimately be exhausted. Therefore
non-conventional sources like geothermal energy, solar
energy, energy from waste, nuclear energy, etc., are being
tapped. The concept of energy balance becomes relevant.
Whether the objective is for accounting, recovery, or loss
prevention, a complete view of the energy balance of any
system is indispensable.
We presented the energy balance equation in Chapter 3.
Energy inputs = Energy outputs + Accumulation
This is also known as the first law of thermodynamics. The
energy accumulation can have positive, zero, or negative
values. As in the mass balance equation, energy balance
equation involving positive or negative accumulation result in
unsteady state processes. When the energy accumulation is
zero, the process involves steady state operation. The discussion in this chapter will cover steady state operations only.
The energy inputs and outputs can be any of several types of
energy: kinetic energy, potential energy, work, heat, internal
energy, electrical energy, magnetic energy, sound energy
nuclear energy, chemical energy, etc.
In Chapters 3 and 4, we discussed energy in the form of
heat. We mentioned that energy balances are usually carried
out together with a mass balance using the same basis. In this
chapter, we show that we can carry it out also on a basis of a
unit mass.
UNITS OF ENERGY
While we prefer to use the SI units, we still employ some of the
customary units. The unit of energy, work, and heat in the SI
is joule. A joule used in connection with work is the same as a
Energy Balance
257
joule used in heat transfer. The mechanical equipment of heat
is therefore not necessary as a conversion factor. The units
used are:
Energy units: joules, J; kilojoule, kJ; megajoule, MJ
Power units: watt W, kilowatt kW, megawatt MW
In the English system, heat is in Btu (British thermal unit)
and work, in ft-lbf. The mechanical equipment of heat is 778
ft-lbf/Btu. In the old metric system, heat is in calorie and
work is in joule with a mechanical heat equivalent of 4.18
joule/calorie.
DEFINITION OF SOME TERMS
DRAFT
We have to define some terms used in thermodynamics. Other
concepts will be explained in appropriate sections of this
chapter.
In thermodynamic problems, only a portion of a large extent
has to be considered. This is known as a system which refers
to the process or an apparatus in which analysis and calculations are made. Everything outside the system is known as the
surroundings. The system and the surroundings comprise the
universe. To identify the system under consideration, a boundary is provided. The boundary can be physical separation or
any convenient divider (actual or hypothetical) that will isolate
the system. The system may be closed or open. In a closed
system, mass neither enters nor leaves the system. It is also
called a non-flow system. In an open system mass enters and
leaves the system and is known as a flow system.
To describe thermodynamic systems, variable or properties
are needed (see Chapter 2). These properties can be extensive
properties (dependent on the amount of mass, e.g., volume) or
intensive (independent of mass, e.g., temperature, pressure,
density, concentration, etc.). A set of properties describing a
system at a particular time or portion of the system is known
as the state of the system. For instance, at time 1 (or at state
1), if the properties P1, V1, T1 completely describe the system
then the combination of the properties (also known as state
variables) refers to the state 1 of the system. At time 2 (or
state 2), a corresponding set of properties P2, V2, T2 represents
state 2.
TYPES OF ENERGY
Although any form of energy may be involved, the five forms
frequently encountered in chemical engineering are: potential
energy, kinetic energy, internal energy, work, and heat.
Occasionally, other forms such as electrical, light, or sound
energy may be involved, but their inclusion poses no problem.
KINETIC ENERGY
Kinetic energy is the energy possessed by a body due to its
motion. A moving car, a falling stone, and a rushing stream
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National Engineering Center CEEC Seminar June 2 -4, 2004
possess this form of energy. This energy is given by the
formula
K.E. = 12 mv2
8-1
where
K.E. = kinetic energy, J
m
= mass, kg
v
= velocity, m/s
In SI units no problem is posed by the influence of changes in
acceleration due to gravity. For a unit mass of 1 kg, specific
kinetic energy KE, is used.
Motion in the universe is relative. Everything moves and the
kinetic energy that is determined for one body is based on the
motion of that body relative to a datum body considered at
rest. In our planet's case, the earth is considered to be the
body at rest.
DRAFT
Example 8-1
Bunker oil flowing at 95 kg per minute (specific gravity = 0.94)
passes through the piping system shown below. What is the
specific kinetic energy of the oil at A, B, and C?
Solution:
Finding the specific kinetic energy is as simple as finding the
v2 . The flow
velocity at each point of interest since KE =
2
A
10 cm dia
B
5 cm dia
C
15 cm dia
Figure 8-1
velocity at point A
kg
1 m3
%
% 1 min % 1
60 s
min 0.94(1000) kg
(0.1 m) 2
4
= 0.214 m/s
= 95
KEA =
(0.214) 2
= 0.023 J/kg
2
The flow velocity at point B
kg
1 m3
1
%
% 1 min % 60 s
min 0.94(1000) kg
(0.05 m) 2
4
= 0.858 m/s
= 95
KE B =
(0.858) 2
= 0.368 J/kg
2
The flow velocity at point C
Energy Balance
259
kg
1 m3
1
%
% 1 min % 60 s
min 0.94(1000) kg
(0.15 m) 2
4
= 0.095 m/s
= 95
KE C =
(0.095) 2
= 0.005 J/kg
2
POTENTIAL ENERGY
Attraction between two masses always exists. The force of
attraction is a function of the distance between the two
masses. Considering the earth as one of the bodies and any
object on earth as the second body, the product of the force of
attraction and the distance between the object and the earth is
the energy that is pending or "potential." The potential energy
is given by the formula
P.E. = m z g
8-2
where P.E. = potential energy, J
m = mass, kg
z = distance or height, m
g = acceleration due to gravity, m/s2.
DRAFT
The height z is referred to a reference point designated as zero.
The absolute value of the potential energy due to the gravitational attraction of the earth however is not used. Instead, an
arbitrary reference elevation where the potential energy is
considered zero is used. Any convenient reference may be
used and sea level is commonly chosen.
For a unit mass of 1 kg, specific potential energy,PE, is
used.
Example 8-2
A dam has its water level 100 meters from the river bed. If 300
m3 of water flows down every second, what is the maximum
power that could be generated (in megawatts, MW)? Take the
acceleration due to gravity equal to 9.80 m/s2, and the specific
gravity of water = 1.00.
Solution:
The specific potential energy
PE = zg
= 100 m % 9.80 m/s2
= 980 J/kg
The power generated
3
kg
= 980 J % 300 ms % 1000 3 % 1 W % 1 MW
m
J/s 10 6 W
kg
= 294 MW
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National Engineering Center CEEC Seminar June 2 -4, 2004
HEAT (See Chapters 3 and 4.)
Closely related to temperature changes is the transient form of
energy called heat. We know that whenever a difference in
temperature exists between two bodies, energy is transferred
from the body at a higher temperature to the body at a lower
temperature. This manifestation of energy transfer is known
as heat energy. A body with a high temperature has its
molecules moving at a higher velocity than those of a body at a
lower temperature, thus contact of these two bodies allows
transfer of energy. Heat is a form of energy used in transferring energy from one system to another. Hence it is classified
as a transit energy. In mathematical equations, heat given off
by a body (exothermic process) is denoted by a negative
quantity while heat taken on by a body (endothermic process)
is positive. The heat quantity unit is joule and the symbol is Q.
WORK
Another form of energy in transit is work which is defined as
the product of a force acting over a distance, or
W = ¶ x 1 Fdx
DRAFT
x2
where
8-3
W = work, J
F = force, N
x = distance, m
The force, F, acting over a distance x1 to x2 is either a
variable or a constant. By convention, the work done by a
system is regarded as a positive quantity while the work done
on a system is negative.
The above definition of work denotes mechanical work.
However, in engineering systems, work is related to the expansion or contraction of a gas. Consider a contained gas in a
volume V at a pressure P and a temperature T. Changing the
volume results in work done either by the gas or by another
system on the gas. Picture the gas as though it were in a cylinder with a piston having a cross-sectional area, A. The length
of the cylinder is x1.
The pressure inside is
Final position
P and the force applied
on the piston is F = PA.
If the final position is at
x2, then xthe work on the
2
gas is x F dx, with the
¶
A
1
force acting over the
distance x2 - x1. Since F
= PA and dx = dV/A,
then
x2
x1
Figure 8-2
W = ¶ V 1 (PA)( dV ) = ¶ V 1 PdV
A
V2
V2
8-4
Energy Balance
261
In compression, work is being applied while in expansion,
work is being done by the gas. Work can be solved graphically
by obtaining the area under the curve F bounded by vertical
lines passing through S1 and S2 (interval S1S2) and the x-axis
as shown in Fig. 8-3.
Likewise, for gases, the P vs. V curve can be used to calculate work.
Example 8-4
How much work does 100
kmol of a gas at 35°C and
1 atm do when heated to
100°C
at
constant
pressure? Assume ideal
gas behavior.
F, force
F
S1
Solution:
The work done by the gas
in expansion is
S2
S, distance
DRAFT
Figure 8-3
P = f(V)
P
V1
V2
V
Figure 8-4
W = ¶ V1 PdV
V2
Using the ideal gas relation,
PV = nRT
and differentiating both sides:
PdV = nRdT
Substituting in the equation,
W = ¶ T1 nRdT
T2
The value of R = 8.31 x 103 joule/(K⋅kmol)
T1 = 35 + 273 = 308 K
T2 = 100 + 273 = 373 K
n = 100 kmol
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National Engineering Center CEEC Seminar June 2 -4, 2004
W = ¶ 308 100(8.31)(10 3 )dT
373
= 100(8.31)(10 3 )(373 − 308)
= 54.02 MJ
The sign is positive since the work is done by the gas.
WORK IN PUSHING FLUIDS IN AND OUT OF THE SYSTEM
("PRESSURE ENERGY")
DRAFT
Normally, work is “shaft work” -- work provided by pumps or
work harnessed by turbines. In systems involving continuous
flow of fluid, we encounter another type of work. Refer to Fig.
8-5. Here, a boundary is set across the flowing fluid. The fluid
to the left of the boundary is doing work by pushing the fluid
on the right; and the fluid on the right is being pushed and
work is being done on it. Knowing the pressure of the fluid
and considering a unit volume, then the work is simply
pressure multiplied by the volume.
In Fig. 8-6, referring to the inlet, the surroundings performs
work on the fluid by pushing it into the system.
W 1 = −P 1 V 1
At the exit, the fluid within the system performs work by
pushing the fluid outside.
Boundary
Flow
Figure 8-5
W2 = P2V2
For a flowing fluid, a change in conditions of P and V means
a
change
in
energy, ∆(PV) =
A1
P2V2 - P1V1, also
known
as
P1, V1
pressure
energy.
A
2
P2, V2
Example 8-4
A chemical plant
gets its bunker oil
from
a distributor
Figure 8-6
50
km
away
through a pipeline 10 cm in diameter. The pressure at the
Energy Balance
263
pump discharge is 1379 kPa and the plant receives the oil at
the end of the line at 137.9 kPa. If the oil flow rate is 100
m3/hr, what is the power requirement of the pump? The
specific gravity of oil is 0.94. Assume the specific volume of the
oil remains constant.
Solution:
Basis:
1 kg oil flowing
The work done by the pump per kg oil (consider the pipe
after the pump as the system) is
W = (PV) = P 2 V 2 − P 1 V 1
P1 = 1379 kPa
P1 = 1379 kPa
Pump
System boundary
P2 = 137.9 kPa
50 km
Figure 8-7
DRAFT
P2 = 137.9 kPa
m3
1
V2 = V1 =
0.94(1000) kg
Work done on the oil
W = V(P 2 − P 1 )
=
m3 (1.379 % 10 5 − 1.379 % 10 6 ) N
1
m2
(0.94)(1000) kg
= −1320 J/kg
Work done by the pump = 1320 J/kg (+)
Power requirement of the pump
3
kg
= 1320 J % 100 m % 940 3 % 1 hr % 1 W % kW
m
3600 s J/s 1000 W
kg
hr
= 34.47 kW
POINT OR STATE FUNCTION
We classify thermodynamic quantities in two general types:
those that depend on the path of a process and those that do
not depend on the path. In physical terms, let us consider that
we are traveling between two places using a land vehicle. The
straight line distance between the two places is always the
same and independent of the road traversed. On the other
hand, the gasoline consumption is definitely dependent on the
path. Different routes are available but the respective mileage
may not be the same. In going to Quiapo from Quezon City,
the fare is dependent on which route is taken but the straight
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National Engineering Center CEEC Seminar June 2 -4, 2004
line distance between the starting point and the destination
remains the same.
Some properties that are path independent are: internal
energy change, enthalpy change, temperature, pressure, and
specific volume, among others. We refer to such properties as
state functions. They depend only on the initial and final
states. An infinitesimal change in property G, dG, when
integrated from state 1 to state 2, will yield a finite differential
change ∆G = G2 - G1.
Since state functions are dependent only on the initial and
final states, represented by points in Fig. 8-8, we can refer to
them as point functions. In thermodynamics, ∆U and ∆H, are
used and absolute values of U and H are not required. Infinitesimal changes dU and dH are exact differentials.
Work and heat on the other hand, are quantities that are
G1
DRAFT
G
Path 3
1
Path 4
∆G = G2-G1
Path 1
G2
2
Path 2
T
Figure 8-8 The ∆G is independent of the 4 paths.
path dependent. Infinitesimal quantities of these two are not
perfect differentials so that when integrated, δW or δQ, yield
¶ W = W and ¶ Q = Q that are quantities rather than changes.
V
As Fig. 8-9 shows, the work is ¶ V 12 PdV.
We can trace numerous paths in going from point 1 to point
2. In traveling path 1, the work done is the area under the
curve and above the abscissa bounded by vertical lines
passing through point 1 and 2. Similarly, the work done in
traveling path 2 is the area under the second curve.
Obviously, the area under path 2 is smaller than that of the
other. We clearly see the dependence of work on the path.
THERMODYNAMIC PROCESSES
Path 1
1
Thermodynamic processes are P
described in terms of the changes
in the properties of a system and
the manner in which these
changes occur. A system at state
V1
1 with properties, T1, P1, V1, etc.
can undergo a change to state 2
with new set of properties, T2, P2,
V2, etc. The manner in which the Figure 8-9
Path
2
2
V2
V
Energy Balance
265
DRAFT
properties change can be of any sort  it can be controlled or
it can be dependent on the prevailing environment. It is possible to observe and analyze the reproducibility of these
changes. If we are interested in the change in property G, ∆G,
given the initial and final states above, we can device any path
for which the change can be calculated. We can present a
complicated process in terms of a combination of simplified
changes. The following are some simple changes.
Isobaric or constant pressure process. The pressure is kept
constant during the process. This process is commonly used
when carrying out experiments at atmospheric pressure. At
constant pressure, ∆P = O.
Isometric or constant volume process. The system is contained
in a constant volume. Processes carried out in rigid containers
are examples of constant volume process. ∆V = O
Isothermal or constant temperature process. The change is
carried out at constant temperature during which heat may
either be added or removed to maintain the temperature
constant. ∆T = 0.
Adiabatic process. This is a process where heat is neither
allowed to leave nor enter the system.
Consider a gas undergoing a change. Initially, it has the
following properties: Ti, Pi, Vi; then it finally changes to Tf, Pf,
and Vf.
Let us say that we are interested in the change in the
property G of the above process. If it is quite difficult to evaluate this property directly, we can break process in a series of
simple changes. For example, this process can be expressed in
terms of three changes in series: constant pressure, constant
volume, and constant temperature steps.
The system with initial state Ti, Pi, Vi is first subjected to a
Initial
Final
Pi Ti Vi
Pf Tf Vf
Figure 8-10
constant pressure and Ti allowed to change to T2 and Vi to V2.
The change in property G is (G2 - Gi) = ∆G1. Next, the system is
subjected to a constant volume process and the temperature
and pressure allowed to change to T3 and P3, respectively. The
change in property G is (G3 - G2) = ∆G2. Lastly, the system is
subjected to a constant temperature process and the final
Initial
State
Ti
Pi
Second
State
Constant
Pressure
∆P = 0
Vi
Gi
∆G1
T2
Third
State
Constant
Volume
P2
T3
P3
V2
∆V = 0
V3
G2
∆G2
G3
Final
State
Constant T = T
3
Temperature f
Pf
∆T = 0
Vf
∆G3
Gf
Figure 8-11
pressure and volume are attained. The final temperature Tf is
equal to T3. The change in G is (Gf - G3) = ∆G3. It is easily seen
that the overall change in G is given as follows:
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National Engineering Center CEEC Seminar June 2 -4, 2004
GT = Gf - Gi
= (G2 - Gi) + (G3 - G2) + (Gf - G3)
= ∆G1 + ∆G2 + ∆G3
For a state function therefore, the total change in a process
is equal to the sum of the changes for the different steps of the
process. The following are the expressions, quantities, and
changes for different simple changes:
Work in a constant volume process:
V
Work = ¶ V 12 PdV
since dV = 0
then work = 0
Work in a constant pressure process:
V
Work = ¶ V 12 PdV
since P is constant,
V
work = P ¶ V 12 dV = P(V 2 − V 1 )
Internal energy or enthalpy change in a constant temperature
process: (See Chapter 3.)
DRAFT
T
Since E = ¶ T 12 nC v dT
and dT = 0,
E = 0 at constant T
T
Similarly, since H = ¶ T 12 nC p dT
and dT = 0,
H = 0 at constant T
THE ENERGY BALANCE EQUATION
We can state the energy balance equation as:
(Energy inputs to the system) - (Energy outputs from the
system) + (Energy generation within the system) - (Energy
utilization within the system) = (Energy accumulation)
The inputs, outputs, and the accumulation are familiar
terms. An example of energy generation within a system, is the
provision of an internal heat source such as an electric heater.
In some system, as in a chemical reactor, some of the energy
is used up by the reaction. For simple cases, the balance
considered is:
Energy input - Energy output = Energy accumulation
For positive or negative accumulation, unsteady state exists.
In the chapter, only the steady state process will be considered
(accumulation = O),
Energy input = Energy output
Energy Balance
267
ENERGY BALANCE FOR FLOW OR OPEN SYSTEMS
Consider a typical system where the fluid continually flows in
and out as shown in Fig. 8-12. It includes a heat exchanger
that provides Q amount of heat and a turbine that does an
amount of work W. A boundary, represented by the broken
lines, isolates the system from the surroundings. The horizontal line at the bottom is the reference line and represents zero
elevation. The point at which the fluid enters and passes
through the boundary is designated as point 1 and the outlet
as point 2.
At the inlet and outlet points, appropriate properties and
quantities as well as some forms of energy are given. The fluid
with a specific volume V1 enters with a velocity v1 and
pressure P1. Point 1 is z1 m above the reference line and the
cross-sectional area of the inlet pipe is A1. Similar data for
point 2 are given.
Applying the law of conservation of energy and using a basis
Q
DRAFT
Poi
nt 1
Heat
Exchanger
Turbine
w1
v1
P1
V1
KE1
PE1
U1
z1
W
Poi
nt 2
Zero reference elevation
w2
v2
P2
V2
KE2
z2 PE2
U2
Figure 8-12 Energy balance in an open system.
v: average fluid velocity
V: specific volume
KE: specific kinetic energy
PE: specific potential energy
P: pressure
U: internal energy
W: work done
Q: heat transferred
of 1 kg of fluid flowing,
Energy Input =
Energy Output
PE 1 + KE 1 + P 1 V 1 + U 1 + Q = PE 2 + KE 2 + P 2 V 2 + U 2 + W
or in more specific terms
g z1 +
v 21
v2
+ P1V1 + U1 + Q = g z2 + 2 + P2V2 + U2 + W
2
2
In differential form,
2
g dz + dv + d(PV) + d U = Q − W
2
As a difference equation
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National Engineering Center CEEC Seminar June 2 -4, 2004
g z +
(v) 2
+ (PV) + U = Q − W
2
Since H = E + PV,
g z +
(v) 2
+ H = Q − W
2
Thus, for a flow system the energy balance states that the
change in potential energy plus the change in kinetic energy
plus the change in enthalpy equals the difference of Q and W.
In an isothermal process, since H = 0, the equation reduces
to
g z +
(v) 2
=Q−W
2
In cases where the kinetic and potential energy factors are
negligible, the equation reduces to
DRAFT
H = Q − W
The energy balance equation is useful in solving for any of
the terms that is unknown. We usually determine which of the
terms are negligible and come up with the suitable final
equation.
Example 8-5
Water at 30oC is to be pumped from a river to a storage tank
at a rate of 0.43 m3/min. The piping system consists of 8 cm
diameter pipes before the pump and 6 cm diameter pipes after
the pump. The discharge into the tank (6 cm dia.) is 30 meters
above the water level of the river. The pressure inside the tank
is kept at 200 kPa. If no heat is added to the water, how many
kW of energy is required for pumping? Neglect the energy lost
by the water in moving past the walls of the pipes. Take the
specific volume of water as 0.001 m3/kg.
Solution:
First, a diagram of the system is drawn with the system
boundary defined. Then the entrance (point 1) and the outlet
(point 2) are identified.
The system boundary chosen is represented by the broken
lines, although the pipe walls can also be considered to be a
Energy Balance
269
convenient boundary. The choice of points 1 and 2 is also
arbitrary and depends on convenience.
2
P2 = 200 kPa
6 cm diameter pipe
30 m
System boundary
8 cm
diameter pipe
P1 = 101.325 kPa
1
Reference elevation
Figure 8-13
DRAFT
Point 1 is chosen as the water level on the river. With this
choice, the velocity at point 1 can be regarded as zero, and the
pressure as that exerted by the atmosphere. Point 2 is the
water discharge at the outlet pipe in the tank. (Another choice
may be the inlet of the pipe for point 1 and the outlet for point
2. If point 1 is so chosen then the velocity of the fluid at the
entrance is not considered zero and the pressure is different
from the exerted atmosphere.) The reference line for elevation
is the water level at the river. (This could be any conveniently
chosen level. In this particular case, the choice lead to z1 being
zero.)
Applying the energy balance equation for 1 kg of water
flowing,
g(z 2 − z 1 ) + (v 22 − v 21 )/2 + (P 2 V 2 − P 1 V 1 ) + (U 2 − U 1 ) = Q − W
z 2 = 30 m
z1 = 0
3
1
v 2 = 0.43 m %
% 1 min
m
60 s
min
(6 cm %
)2 100 cm 4
= 2.53 m/s
v1 = 0
P2 = 200 kPa
P1 = 101.325 kPa
V 1 = V 2 = 0.001 m3/kg (Since water is incompressible)
Q=0
270
(No heat is added or removed from the system)
National Engineering Center CEEC Seminar June 2 -4, 2004
E1 = E2
(Since the temperature remains constant)
Substituting in the energy equation,
9.8 % (30 − 0) +
(2.53 ) 2 − 0 2
+ (200 − 101.325) % 0.001 +0 = 0 − W
2
294 + 3.20 +0.099 = -W
W = -297.3 J/kg
3
1 kg
% 1 min
Flow rate = 0.43 m %
60 s
min 0.001 m3
= 7.17 kg/s
Power requirement
= 297.3 J/kg x 7.17 kg/s
= 2131 J/s
= 2.13 kW
Nitrogen gas is heated from 25°C to 90°C in a vertical heat
exchanger 2 m high. The inlet velocity of the nitrogen is 5 m/s
while the outlet velocity is 7 m/s. Calculate the amount of
heat necessary. Assume that the heat capacity of nitrogen is
constant at 26.2 kJ/(mol⋅K). Consider the pressure essentially
constant.
Solution:
Basis:
1 kg N2
7 m/s
90oC
Q
2m
Reference
elevation
Point 2
Heat Exchanger
DRAFT
Example 8-6
25oC
5 m/s
Point 1
Nitrogen gas
Figure 8-14
Applying the energy balance equation involving enthalpy,
(H2 - H1) + (z2 - z1) + (v22 - v12)/2 = Q - W
z2 = 2 m
z1 = 0
Energy Balance
271
g(z2 - z1) = (9.8 m/s2) × 2 m
= 19.6 J
v2 = 7 m/s
v1 = 5 m/s
(v22 - v12)/2
= (49 - 25)/2
= 12 J
W=0
Q = 58500 + 19.6 + 12
= 58532 J, the heat requirement.
This example shows that the kinetic and the potential
energy terms become negligible when heat quantities are
involved
CLOSED OR NON-FLOW SYSTEMS
DRAFT
In systems where materials remain enclosed in a container,
kinetic energy of flow or change in potential energy are not
involved. A batch chemical reactor (Fig. 8-15) is a typical
example. Here, the reaction is promoted after the reactants are
placed in the vessel. A motor-driven stirrer and a heating or
cooling coil are usually provided.
Since the tank is stationary
Motor-driven
∆v2 = 0 and ∆z = 0. The energy
stirrer
balance equation reduces to
E = Q − W
In closed system, the heat
involved in a constant volume
process is
Heating or
cooling coil
= ∆U + W
Figure 8-15
= ∆U + ∫PdV
= ∆U + 0
= ∫CvdT
for every mole of material or mass depending on units of Cv.
Qv
The heat involved in a constant pressure process is
Qp
= ∆U + W
= ∆U + ∫PdV
= ∆U + P∆V
= ∆H
= ∫Cp dT for every mole of material
Example 8-7
One cubic meter of oxygen at 138 kPa and 20oC is heated at
constant volume to 150oC. At this new condition, the gas is
then cooled to 10oC while maintaining the pressure constant.
What is ∆U, ∆H, Q, and W for the constant volume process,
the constant pressure process, and for the entire process?
272
National Engineering Center CEEC Seminar June 2 -4, 2004
Take the heat capacity of oxygen at constant pressure as 29.8
kJ/(kmol⋅K).
Condition 2
Condition 1
Oxygen
V1 = 1 m3
P1 = 138 kPa
T1 = 20oC
Process I:
Constant
Volume
P2 = ?
T2 = 150oC
Condition 3
Process II:
Constant
Pressure
V3 = ?
T3 = 10oC
Figure 8-16
Solution:
Basis: 1 m3 of oxygen at 138 kPa
kmol O 2 = 1 m 3 %
138 % 273 % 1 kmol
101.325 20 + 273 22.4 m 3
= 0.0567 kmol
Cv = Cp - R = 29.8 - 8.31 = 21.49 kJ/(kmol⋅K)
Process 1
HI = n ¶ 20+273 Cp dT
DRAFT
150+273
= 0.0567 (29.8) (150 - 20)
= 219.7 kJ
UI = n ¶ 20+273 Cv dT
150+273
= 0.0567(21.49)(150-20)
= 158.4 kJ
QI = ∆U = 158.4 kJ
W = ∫PdV = 0
Process II
HII = ¶ 150+273 Cp dT
10+273
= 0.0567(29.8)(10-150)
= -236.6 kJ
UII = ¶ 150+273 Cv dT
10+273
= (0.0567)(21.49)(10-150)
= -170.6 kJ
QII = ∆H = -236.6 kJ
WII = ¶ V2 PdV
V3
= P 2 (V 3 − V 2 )
Energy Balance
273
P 2 = (138) 150++ 273 = 199.2
20 273
V 3 = (1 m 3 ) 10 + 273 = 0.669
150 + 273
WII = 199.2 × (0.669 - 1) = -65.9 kJ
(As a check, WII = QII - ∆UII = -236.6 -(-170.6) = -66.0 kJ)
For the entire process:
E T = U I + U II
= 158.4 + (-170.6)
= -12.2 kJ
H T = H I + H II
= 219.7 + (-236.6)
= -16.9 kJ
DRAFT
Q T = Q I + Q II
= 158.4 + (-236.5)
= -78.2 kJ
W T = W I + W II
= 0 + (-65.9)
= -65.9 kJ
PROBLEMS
8-1 What is the kinetic energy of 10 kg of water flowing at a
velocity of 1.5 m/s?
8-2 What is the kinetic energy of 1,000 kmol/h of a natural gas
flowing in a pipe 0.5 m in dia. The average molecular weight
of the gas is 20.
8-3 The potential energy of a 10 kg fluid is 1,000 J. How high is it
above the reference point.
8-4 Find the potential energy of 10 Mg of water on a mountain
top 1,090 m above sea level.
8-5 Find the work done by 1 kg of dry ice (solid carbon dioxide)
when it sublimes (at -79°C) against a constant external
pressure of 100 kPa.
8-6 How much work is done when 7.5 kmol of ammonia is vaporized at -33°C (the normal boiling point) and 1 atm?
8-7 Calculate Q, W and ∆U for the reaction,
CH4 + 2O2 → CO2 + 2H2O
using a datum temperature of 25°C.
274
National Engineering Center CEEC Seminar June 2 -4, 2004
APPENDICES
Appendix 1
Normal Boiling Points, Critical Temperatures, Critical
Pressures
Appendix 2
Appendix 3
Heat Capacity of Selected Inorganic Solids
Heat Capacity of Some Liquids
Appendix 4
Heat Capacity of Some Gases
Appendix 5
Standard Heat Formation and Standard Heat of
Combustion
Appendix 6
Standard Heat of Formation and Integral Heat of
Formation
Appendix 7
Properties of Saturated Steam
Appendix 8
Vapor Pressure - Antoine Equation Coefficients
Appendix 9
Vapor Pressure of Various Liquids and Solids
Conversion Factors to SI for Selected Quantities
Atomic Numbers and Atomic Masses
APPENDIX 1
NORMAL BOILING POINTS, CRITICAL TEMPERATURES, CRITICAL PRESSURES
Substance
Acetaldehyde
Acetic Acid
Acetone
Acetylene
Ammonia
Argon
Benzene
1, 3-Butadiene
n-Butane
I-Butane
Carbon Dioxide
Carbon Tetrachloride
Chlorine
Cyclohexane
Ethane
Ethyl Alcohol
Ethene (Ethylene)
Helium
n-Heptane
n-Hexane
Hydrogen
Hydrogen Bromide
Hydrogen Chloride
Hydrogen Sulfide
Mercury
Methane
Methanol
Nitric Oxide
Nitrogen
Nitrogen Tetroxide
Nitrous Oxide
Oxygen
n-Octane
n-Pentane
Phenol
Phosphine
Propane
1-Propanol
Propane (Propylene)
n-Propionic Acid
Sulfur Dioxide
Sulfur Trioxide
Toluene
Normal
Boiling Point
o
C
Critical
Temperatures
o
C
Critical
Pressures
atm
19.8
118.1
56.5
-84.9
-33.4
-185.7
80.1
-0.3
-13.4
-78.5 (subl)
-76.8
-34.6
80.75
-88.6
78.4
-103.9
-268.9
98.4
64
-252.7
-66.9
-85
-59.4
356.9
-161.4
64.7
-151
-195.8
21.3
-89.5
-183
125.7
36.3
181.2
-87.4
-42.2
97.19
-47.7
140
-10.02
44.7
110.8
188
321.6
256
36
132.4
-122
288.5
152
153
134
31.1
283.1
144
281.0
32.1
243.1
9.7
-267.9
266.8
234.8
-239.9
90
51.4
100.4
>1550
-82.5
240
-94
-147.1
158
36.5
-118.8
296
197.2
419
51
95.6
263.7
92.3
339.5
157.2
218.3
320.6
57.2
47
62
111.5
48
47.7
42.7
36
37
73
45
76.1
40.4
48.8
63.1
50.9
2.26
26.8
29.5
12.8
84
81.6
88.9
>200
45.8
78.7
65
33.5
99
71.7
49.7
24.6
33.0
60.5
64
43
49.9
45
53
77.7
83.6
41.6
APPENDIX 2
Compound
HEAT CAPACITY OF SELECTED INORGANIC SOLIDS
o
C p = cal/(mol)( C)
Al
Equation (T = K)
4.80 + 0.00322T
Al2O3
22.08 + 0.008971T - (522500/T )
273-1973
3
BaCl2
17.0 + 0.00334T
273-1198
?
BaSO4
21.35 + 0.0141T
273-1323
5
CaCl2
16.9 + 0.00386T
273-1055
?
273-1033
3
CaCO3
Useful Range Uncertainty, %
273-931
1
2
2
19.68 + 0.01189T - (307600/T )
2
CaO
10.0 + 0.00484T - (108000/T )
273-1173
2
CaSO4
18.52 + 0.02197T - (156800/T )
273-1373
5
C (graphite)
Cu
2.673 + 0.002617T - (116900/T2)
273-1373
273-1357
2
1
CuO
Fe(a)
10.87 + 0.003576T - (150600/T )
273-810
273-1041
2
3
273-1173
2
FeO
2
5.44 + 0.001462T
2
4.13 + 0.00638T
2
12.62 + 0.001492T - (76200/T )
2
Fe2O3
24.72 + 0.01604T - (423400/T )
273-1097
2
Fe3O4
41.17 + 0.01882T - (979500/T )
273-1065
2
FeS2
10.7 + 0.01226T
273-773
?
Pb
PbO
5.77 + 0.00202T
10.33 + 0.00318
2
6.20 + 0.00133T - (67800/T )
273-600
273-544
2
2
273-923
273-991
1
?
Mg
MgCl2
2
17.3 + 0.00377t
10.86 + 0.001197T - (208700/T2)
MgO
Ni
NH4Cl (a)
4.26 + 0.00640T
9.80 + 0.0368T
273-2073
273-626
273-457
2
2
5
P(red)
K
KCl
KNO3
0.21 + 0.0180T
5.24 + 0.00555T
10.93 + 0.00376T
6.42 + 0.0530T
273-472
273-336
273-1043
273-401
10
5
2
10
Si
5.74 + 0.000617T - (101000/T2)
273-1174
2
SiC
8.89 + 0.00291T - (284000/T )
273-1629
2
2
2
SiO2 (a quartz)
10.87 + 0.008712T - (241200/T )
273-848
1
Ag
AgCl
AgNO3(a)
5.60 + 0.00150T
9.60 + 0.00929T
18.83 + 0.0160T
273-1234
273-728
273-433
1
2
2
Na
NaCl
NaNO3
5.01+ 0.00536T
10.79 + 0.00420T
4.56 + 0.0580T
273-371
273-1074
273-583
1.5
2
5
S(rhombic)
S(monoclinic)
Sn
SnCl2
3.63 + 0.00640T
4.38 + 0.00440T
5.05 + 0.00480T
16.2 + 0.00926T
273-368
368-392
273-504
273-520
3
3
2
?
Zn
ZnCl2
5.25 + 0.00270T
15.9 + 0.00800T
273-692
273-638
1
?
ZnO
11.40 + 0.00145T - (182400/T2)
273-1573
1
Appendix 4
HEAT CAPACITIES- GASES
Empirical Constants for Molal Heat Capacities of Gases
at Constant Pressure (P=O)
C p = A+BT +CT 2+DT 3+ET 4
where T is in K, C p in kJ/(kmol K)
Compound
Formula
A
B
C
D
E
T MIN
T MAX
Acetone
CH3COCH3
35.918
0.0938960
6.3174E-11
100
1500
Acetylene
C 2 H2
19.360
0.1151900 -1.2374E-04
7.2370E-08
1.6590E-11
200
1500
Ammonia
NH3
33.573 -0.0125810
8.8906E-05 -7.1783E-08
1.8569E-11
100
1500
Benzene
C 6 H6
-31.368
0.4746000 -3.1137E-04
8.5237E-08 -5.0524E-12
200
1500
n-Butane
C4H10
20.056
0.2815300 -1.3143E-05 -9.4571E-08
3.4149E-11
200
1500
Carbon dioxide
CO2
27.437
0.0423150 -1.9555E-05
3.9968E-09 -2.9872E-13
50
5000
Carbon monoxide
CO
29.556 -0.0065807
1.2227E-08
2.2617E-12
60
1500
Chlorine
Cl2
27.213
0.0304260 -3.3353E-05
1.5961E-08 -2.7021E-12
50
1500
Cyclohexane
C6H12
13.783
0.2074200
5.3682E-04 -6.3012E-07
1.8988E-10
100
1500
Ethane
C 2 H6
28.146
0.0434470
1.8946E-04
1.9028E-07
5.3349E-11
100
1500
Ethyl alcohol
C2H5OH
27.091
0.1105500
1.0957E-04
1.5046E-07
4.6601E-11
100
1500
Ethyl ether
(C2H5)20
35.979
0.2844400 -1.2673E-06 -1.0128E-07
3.4529E-11
200
1500
Ethylene
C 2 H4
32.083 -0.0148310
2.4774E-04 -2.3766E-07
6.8274E-11
60
1500
n-Heptane
C7H16
26.984
0.5038700 -4.4748E-05 -1.6835E-07
6.5183E-11
200
1500
n-Hexane
C6H14
25.924
0.4192700 -1.2491E-05 -1.5916E-07
5.8784E-11
200
1500
Hydrogen
H2
25.399
0.0201780 -3.8549E-05
3.1880E-08 -8.7585E-12
250
1500
Hydrogen chloride
HCl
29.244 -0.0012615
1.1210E-06
4.9676E-09 -2.4963E-12
50
1500
Hydrogen sulfide
H2S
33.878 -0.0112160
5.2578E-05 -3.8397E-08
9.0281E-12
100
1500
Methane
CH4
34.942 -0.0399570
1.9184E-04 -1.5303E-07
3.9321E-11
50
1500
Methyl alcohol
CH3OH
40.046 -0.0382870
2.4529E-04 -2.1679E-07
5.9909E-11
100
1500
Methylcyclopentane
C5H9CH3
-9.939
0.4252800
1.2521E-05 -1.8864E-07
6.4751E-11
200
1500
Nitric oxide
NO
33.227 -0.0236260
5.3156E-05 -3.7858E-08
9.1197E-12
50
1500
Nitrogen
N2
29.342 -0.0035395
1.0076E-05 -4.3116E-09
2.5935E-13
50
1500
Oxygen
O2
29.526 -0.0088999
3.8083E-05 -3.2629E-08
8.8607E-12
50
1500
n-Pentane
C5H12
26.671
0.3232400
4.2820E-05 -1.6639E-07
5.6036E-11
200
1500
Propane
C 3 H8
28.277
0.1160000
1.9597E-04 -2.3271E-07
6.8669E-11
100
1500
Sulfur dioxide
SO2
29.637
0.0347350
9.2903E-06 -2.9885E-08
1.0937E-11
100
1500
Sulfur trioxide
SO3
22.466
0.1198100 -9.0842E-05
2.5503E-08 -7.9208E-13
100
1500
Toluene
C 7 H8
-24.097
0.5218700 -2.9827E-04
6.1220E-08
1.2576E-12
200
1500
Water
H2O
33.933 -0.0084186
2.9906E-05 -1.7825E-08
3.6934E-12
100
1500
1.8730E-04 -2.1643E-07
2.0130E-08
APPENDIX 5
STANDARD HEATS OF FORMATION AND STANDARD HEATS OF COMBUSTION
o
for organic compounds at 25 C
o
For ∆Η c, the standard states of products are H2O(l), CO2(g), N2(g), SO2(g) and HCl(aq)
Compound
Formula
State
∆HoF, MJ/kmol ∆HoC, MJ/kmol
Acetaldehyde
CH3CHO
g
-166.470
-1193.15
Acetic Acid
CH3COOH
l
-487.340
-872.28
Acetone
Acetylene
Benzene
C3H6O
C2H2
C6H6
n-Butane
Isobutane
C4H10
C4H10
1-Butene
Carbon disulfide
C4H8
CS2
Carbon monoxide
Carbon tetrachloride
CO
CCl4
Chloroethane
Cumene
C2H5Cl
C6H5CH(CH3)2
Cyclohexane
C6H12
Cyclopentane
C5H10
Ethane
Ethyl acetate
Ethyl alcohol
C2H6
CH3COOC2H5
C2H5OH
-248.360
226.900
49.069
82.982
-126.230
-158.550
-134.610
1.172
87.923
115.350
-110.600
-139.840
-106.760
-105.090
-41.232
3.936
-156.340
-123.220
-105.930
-77.288
-87.724
Ethyl benzene
C6H5C2H5
Ethyl chloride
Ethyl ether
Ethylene
Formaldehyde
n-Heptane
C2H5Cl
C2H5OC2H5
C2H4
H2CO
C7H16
-1791.07
-1300.50
-3269.81
-3303.72
-2879.01
-2850.92
-2870.68
-2720.40
-1075.88
-1103.31
-283.18
-352.40
-385.23
-1422.09
-5218.93
-5264.11
-3922.53
-3955.65
-3293.09
-3321.77
-1560.92
-2255.68
-1367.83
-1410.18
-4567.92
-4610.21
n-Hexane
C6H14
Methane
Methyl alcohol
CH4
CH3OH
Methyl chloride
Methyl cyclohexane
CH3Cl
C7H14
Methyl cyclopentane
C6H12
Nitrobenzene
n-pentane
C6H5NO2
C5H12
Propane
Propene
n-Propyl alcohol
n-Propyl benzene
Styrene
Toluene
C3H8
C3H6
C3H8O
C6H5CHCH2
C6H5CH3
m-Xylene
C6H4(CH3)2
o-Xylene
C6H4(CH3)2
p-Xylene
C6H4(CH3)2
l
g
l
g
g
l
g
g
l
g
g
l
g
g
l
g
l
g
l
g
g
l
l
g
l
g
g
l
g
g
l
g
l
g
g
l
g
g
l
g
l
g
l
l
g
g
g
g
l
g
l
g
l
g
l
g
l
g
-277.820
-235.470
-12.464
29.810
-105.090
-272.980
52.318
-115.970
-224.540
-187.940
-198.960
-167.300
-74.898
-238.800
-201.380
-81.978
-192.290
-154.870
-138.460
-106.760
-173.170
-146.540
-103.920
20.427
-255.390
-38.426
147.460
12.016
50.032
-25.435
17.240
-24.455
19.010
-24.440
17.960
-2732.26
-1411.91
-563.84
-4820.14
-4856.73
-4165.91
-4197.56
-890.95
-727.04
-764.47
-767.15
-4568.34
-4603.76
-3940.36
-3672.10
-3094.04
-3511.89
-3538.51
-2221.52
-2059.85
-2069.95
-5221.74
-4441.78
-3912.56
-3950.58
-4554.90
-4597.61
-4555.91
-4599.37
-4555.91
-4598.32
APPENDIX 6
STANDARD HEATS OF FORMATION and INTEGRAL HEATS OF SOLUTION
AT INFINITE DILUTION of INORGANIC COMPOUNDS (25oC and 1 atm)
o
o
Compound
Formula
State ∆Hf , MJ/kg mol ∆Hs ,MJ/kg mol
Aluminum chloride
AlCl3
s
-695.8
-332
Aluminum sulfate
Ammonia
Ammonium chloride
Ammonium nitrate
Ammonium Sulfate
Barium Chloride
Calcium Carbide
Calcium Carbonate
Calcium Chloride
Calcium Hydroxide
Calcium Oxide
Calcium Phosphate
Carbon, graphite
Carbon, amorphous
Carbon dioxide
Carbon disulfide
Al2(SO4)3
NH3
NH4Cl
NH4NO3
(NH4)2SO4
BaCl2
CaC2
CaCO3
CaCl2
Ca(OH)2
CaO
Ca3(PO4)
C
C
CO2
CS2
Sulfur Dioxide
Sulfur Trioxide
Sulfuric Acid
Water
CO
Cr2O3
CuO
CuSO4
Cu2O
HBr
HCl
HF
HI
H2S
Fe3O4
FeSO4
PbCl2
PbO2
PbSO4
MgCO3
MgCl2
Mg(OH)2
HNO3
KCl
KOH
KNO3
K2SO4
SiO2
AgNO3
Na2CO3
Na3CO3 10H2O
NaCl
NaOH
NaNO3
Na2SO4
Na2SO4 10H2O
SO2
SO3
H2SO4
H2O
Zinc Oxide
Zinc Sulfate
ZnO
ZnSO4
-6359.3
-46.22
-315.6
-365.38
-1180.1
-860.64
-62.8
-1707.7
-795.49
-987.2
-635.97
-4129
0
10.89
-393.776
87.92
115.35
-110.6
-1129
-115.3
-770.4
-166.8
-36.26
-92.373
-269
25.96
-20.16
-1117.9
-923.19
-359.4
-276.8
-94.37
-1114
-642.3
-925.28
-173.35
-436.16
-426.13
-493.04
-1434.6
-860
-123.2
-1131.7
-4084.6
-411.28
-427.01
-425.13
-1385.4
4327
-297.1
-395.4
-811.86
-286.03
-241.99
-348.2
-979.21
-318.7
-34.7
15.16
25.79
6.196
-13.23
Carbon monoxide
Chromic Oxide
Cupric Oxide
Cupric sulfate
Cuprous oxide
Hydrogen Bromide
Hydrogen Chloride
Hydrogen Flouride
Hydrogen Iodide
Hydrogen Sulfide
Iron Oxide
Iron Sulfate
Lead Chloride
Lead Dioxide
Lead Sulfate
Magnesium Carbonate
Magnesium Chloride
Magnesium Hydroxide
Nitric Acid
Potassium Chloride
Potassium Hydroxide
Potassium Nitrate
Potassium Sulfate
Silicon Dioxide
Silver Nitrate
Sodium Carbonate
s
g
s
s
s
s
s
s
s
s
s
s
s
s
g
l
g
g
s
s
s
s
g
g
g
g
g
s
s
s
s
s
s
s
s
l
s
s
s
s
s
s
s
s
s
s
s
s
s
g
g
l
l
g
s
s
Sodium Chloride
Sodium Hydroxide
Sodium Nitrate
Sodium Sulfate
-82.98
-16.24
-81.48
-19.43
-73.31
-84.74
-75.19
-60.54
-81.94
-19.18
-64.9
25.96
-146.8
-33.36
17.23
-55.35
35
23.78
-23.45
69.08
3.89
-42.9
-21.4
-2.34
78.92
41.45
-226.6
-96.25
-81.43
APPENDIX 7 Properties of Saturated Steam
o
Values in table based on zero internal energy of liquid water at 0.01 C
Temp
o
C
Absolute
Pressure
kPa
Specific
Volume
of Vapor
cm3/kg
Enthalpy, kJ/kg
Liquid
Vapor
Latent Heat of
Evaporation
kJ/kg
0
2
4
6
8
0.6109
0.7056
0.8131
0.9349
1.0724
206 278
179 889
157 232
137 734
120 917
0.02
8.37
16.78
25.20
33.60
2501.3
2505.0
2508.7
2512.4
2516.1
2501.4
2496.7
2491.9
2487.2
2482.5
10
12
14
16
18
1.2276
1.4022
1.5983
1.8181
2.064
106 379
93 784
82 848
73 333
65 038
42.01
50.41
58.80
67.19
75.58
2519.8
2523.4
2527.1
2530.8
2534.4
2477.7
2473.0
2468.3
2463.6
2458.8
20
22
24
26
28
2.339
2.645
2.985
3.363
3.782
57 791
51 447
45 883
40 994
36 690
83.96
92.33
100.7
109.07
117.43
2538.1
2541.7
2545.4
2549.0
2552.6
2454.1
2449.4
2444.7
2439.9
2435.2
30
32
34
36
38
4.246
4.759
5.324
5.947
6.632
32 894
29 540
26 571
23 940
21 602
125.79
134.15
142.50
150.86
159.21
2556.3
2559.9
2563.5
2567.1
2570.7
2430.5
2425.7
2421.0
2416.2
2411.5
40
42
44
46
48
7.384
8.208
9.111
10.098
11.175
19 523
17 671
16 018
14 540
13 218
167.57
175.91
184.27
192.62
200.97
2574.3
2577.9
2581.4
2585.0
2588.5
2406.7
2401.9
2397.2
2392.4
2387.6
50
52
54
56
58
12.349
13.628
15.019
16.529
18.166
12 032
10 968
10 011
9 149
8 372
209.33
217.69
226.04
234.41
242.77
2592.1
2595.6
2599.1
2602.6
2606.1
2382.7
2377.9
2373.1
2368.2
2363.4
60
62
64
66
68
19.940
21.860
23.934
26.170
28.590
7 671
7 037
6 463
5 943
5 471
251.13
259.49
267.86
276.23
284.61
2609.6
2613.1
2616.5
2620.0
2623.4
2358.5
2353.6
2348.7
2343.7
2338.8
APPENDIX 7, continued
Temp
o
C
Absolute
Pressure
Specific
Volume
kPa
70
72
74
76
78
31.19
33.99
36.99
40.22
43.68
of Vapor
cm3/g
5 042
4 652
4 297
3 973
3 677
80
82
84
86
88
47.39
51.36
55.60
60.14
64.98
90
92
94
96
98
Enthalpy, kJ/kg
Liquid
Vapor
Latent Heat of
Evaporation
kJ/kg
292.98
301.36
309.74
318.13
326.51
2626.8
2630.2
2633.6
2637.0
2640.3
2333.8
2328.9
2323.9
2318.9
2313.8
3 407
3 160
2 934
2 726
2 536
334.91
343.30
351.70
360.10
368.51
2643.7
2647.0
2650.3
2653.6
2656.9
2308.8
2303.7
2298.6
2293.5
2288.3
70.14
75.64
81.49
87.71
94.30
2 361
2 200
2 052
1 915.0
1 789.1
376.92
385.33
393.75
402.17
410.61
2660.1
2663.3
2666.5
2669.7
2672.9
2283.2
2278.0
2272.8
2267.6
2262.3
100
105
110
115
120
101.35
120.82
143.27
169.06
198.53
1 672.9
1 419.4
1 210.2
1 036.6
891.9
419.04
440.15
461.30
482.48
503.71
2676.1
2683.8
2691.5
2699.0
2706.3
2257.0
2243.7
2230.2
2216.5
2202.6
125
130
135
140
145
232.1
270.1
313.0
361.3
415.4
770.6
668.5
582.2
508.9
446.3
524.99
546.31
567.69
589.13
610.63
2713.5
2720.5
2727.3
2733.9
2740.3
2188.5
2174.2
2159.6
2144.7
2129.6
150
155
160
165
170
475.8
543.1
617.8
700.5
791.7
392.8
346.8
307.1
272.7
242.8
632.20
653.84
675.55
697.34
719.21
2746.5
2752.4
2758.1
2763.5
2768.7
2114.3
2098.6
2082.6
2066.2
2049.5
175
180
185
180
195
200
892.0
1002.1
1122.7
1254.4
1397.8
1553.8
216.8
194.05
174.09
156.54
141.05
127.36
741.17
763.22
785.37
807.62
829.98
852.45
2773.6
2778.2
2782.4
2786.4
2790.0
2793.2
2032.4
2015.0
1997.1
1978.8
1960.0
1940.7
CONVERSION FACTORS TO SI FOR SELECTED QUANTITIES
To Convert from
barrel (for petroleum, 42 gal)
British thermal unit (Btu)
Btu/lbm-deg F (heat capacity)
Btu/second
Btu/ft2⋅hr⋅deg F, (heat trans. coef.)
Btu/ft2-hour (heat flux)
Btu/ft-hr-deg F (thermal conduc.)
calorie (International Table)
cal/deg C
centimeter
centimeter of mercury (Oo C)
centimeter of water (4o C)
centipoise
centistoke
dyne
erg
fluid ounce (U.S.)
foot
foot (U.S. survey).
foot of water (39.2o F)
foot2
foot/second2
foot2/hr
foot-pound force
foot2/second
foot3
gallon (U.S. liquid)
gram
horsepower (550 ft-lbf/s)
inch
inch of mercury (60o F)
inch of water (60o F)
inch2
inch3
kilocalorie
kilogram-force (kgf)
micron
mil
mile (U.S. Statute)
mile/hr
millimeter of mercury (0o C)
ounce-mass (avoirdupois)
ounce (U. S. fluid)
pint (U.S. liquid)
poise (absolute viscosity)
poundal
pound-force (lbf avoirdupois)
pound-force-second/ft2
pound-mass (lbm avoirdupois)…
pound-mass/foot3
pound-mass/foot-second
psi…
quart (U.S.
slug…
stoke (kinematic viscosity)
ton (long, 2240 lbm)
ton (short, 2000 lbm)
torr (mm Hg, 0o C)
watt (International of 1948)…
watt-hr
yard
To
meter3 (m3)
joule (J)
joule/kilogram-kelvin (J/kg-K)
watt (W)
joule/meter2-second-kelvin (J/m2⋅s⋅K)
joule/meter2-second (J/m2 s)
joule/meter-second-kelvin (J/m s K)
joule (J)
jole/kilogram-kelvin (J/kg K)
meter (m)
pascal (Pa)
pascal (Pa)
pascal-second (Pa-s)
meter2/second (m2/s)
newton (N)
joule (J)
meter3 (m3)
meter (m)
meter (m)
pascal (Pa)..
meter2 (m2)
meter/second2 (m/s2)
meter2/second (m2/s)
joule (J)
meter2/second (m2/s)
meter3 (m3)
meter3 (m3)
kilogram (kg)
watt (W)
meter (m)
pascal (Pa).
pascal (Pa)
meter2 (m2)
meter3 (m3)
joule (J).
newton (N)
meter (m)
meter (m)
meter (m)
meter/second (m/s)
pascal (Pa)
kilogram (kg)
meter3 (m3)
meter3 (m3)
pascal-second (Pa s)
newton (N)
newton (N)
pascal-second (Pa s)
kilogram (kg)
kilogram/meter3 (kg/m3)
pascal-second (Pa s)
pascal (Pa)
meter3 (m3)
kilogram (kg)
meter2/second
kilogram (kg)
kilogram (kg)
pascal (Pa)
watt (W)
joule (J)
meter (m)
Multiply by
1.5898729 E-01
1.0550559 E+03
4.1868000 E+03
1.0550559 E+03
5.6782633 E+00
3.1545907 E+00
1.7307347 E+00
4.1868000 E+00
4.1868000 E+03
1.0000000 E-02
1.3332237 E+03
9.80638
E+01
1.0000000 E-05
1.0000000 E-07
1.0000000 E-05
1.0000000 E-07
2.9573530 E-05
3.0480000 E-01
3.0480061 E-01
2.98898
E+03
9.2903040 E-02
3.0480000 E-01
2.5806400 E-05
1.3558179 E+00
9.2903040 E-02
2.8316847 E-02
3.7854118 E-03
1.0000000 E-03
7.4569987 E+02
2.5400000 E-02
3.37685
E+03
2.48843
E+02
6.4516000 E-04
1.6387064 E-05
4.1868000 E+03
9.8066500 E+00
1.0000000 E-06
2.540000
E-05
1.6093440 E+03
4.4704000 E-01
1.3332237 E+02
2.8349523 E-02
2.9572530 E-05
4.7317647 E-04
1.0000000 E-01
1.3825495 E-01
4.4482216 E+00
4.7880258 E+01
4.5359237 E–01
1.6018463 E+01
1.4881639 E+00
6.8947573 E+03
9.4635295 E-04
1.4593903 E+01
1.0000000 E-04
1.0160469 E+03
9.0718474 E+02
1.3332237 E+02
1.000165
E+00
3.6000000 E+03
9.1440000 E–01
Atomic Numbers and Atomic Masses
Element
Actinium
Aluminum
Americium
Antimony
Argon
Arsenic
Astatine
Barium
Berkelium
Beryllium
Bismuth
Boron
Bromine
Cadmium
Calcium
Californium
Carbon
Cerium
Cesium
Chlorine
Chromium
Cobalt
Copper
Curium
Dysprosium
Einsteinium
Erbium
Europium
Fermium
Fluorine
Francium
Gadolinium
Gallium
Germanium
Gold
Hafnium
Helium
Holmium
Hydrogen
Indium
Iodine
Iridium
Iron
Krypton
Lanthanum
Lawrencium
Lead
Lithium
Lutetium
Magnesium
Manganese
Mendelevium
Symbol
Ac
Al
Am
Sb
Ar
As
At
Ba
Bk
Be
Bi
B
Br
Cd
Ca
Cf
C
Ce
Cs
C1
Cr
Co
Cu
Cm
Dy
Es
Er
Eu
Fm
F
Fr
Gd
Ga
Ge
Au
Hf
He
Ho
H
In
I
Ir
Fe
Kr
La
Lr
Pb
Li
Lu
Mg
Mn
Md
Atomic
number
89
13
95
51
18
33
85
56
97
4
83
5
35
48
20
98
6
58
55
17
24
27
29
96
66
99
68
63
100
9
87
64
31
32
79
72
2
67
1
49
53
77
26
36
57
103
82
3
71
12
25
101
Atomic
weight
227.03
26.98
243
121.75
39.95
74.92
210
137.33
247
9.01
208.98
10.81
79.90
112.41
40.08
251
12.01
140.12
132.90
35.45
51.99
58.93
63.55
247
162.50
254
167.26
151.96
257
19.00
223
157.25
69.72
72.59
196.97
178.49
4.00
164.93
1.01
114.82
126.90
192.22
55.85
83.80
138.91
260
207.2
6.94
174.97
24.31
54.94
258
Element
Mercury
Molybdenum
Neodymium
Neon
Neptunium
Nickel
Niobium
Nitrogen
Nobelium
Osmium
Oxygen
Palladium
Phosphorus
Platinum
Plutonium
Polonium
Potassium
Praseodymium
Promethium
Protactinium
Radium
Radon
Rhenium
Rhodium
Rubidium
Ruthenium
Samarium
Scandium
Selenium
Silicon
Silver
Sodium
Strontium
Sulfur
Tantalum
Technetium
Tellurium
Terbium
Thallium
Thorium
Thulium
Tin
Titanium
Tungsten
Uranium
Vanadium
Xenon
Ytterbium
Yttrium
Zinc
Zirconium
Symbol
Hg
Mo
Nd
Ne
Np
Ni
Nb
N
No
Os
O
Pd
P
Pt
Pu
Po
K
Pr
Pm
Pa
Ra
Rn
Re
Rh
Rb
Ru
Sm
Sc
Se
Si
Ag
Na
Sr
S
Ta
Tc
Te
Tb
TI
Th
Tm
Sn
T1
W
U
V
Xe
Yb
Y
Zn
Zr
Atomic
number
80
42
60
10
93
28
41
7
102
76
8
46
15
78
94
84
19
59
61
91
88
86
75
45
37
44
62
21
34
14
47
11
38
16
73
43
52
65
81
90
69
50
22
74
92
23
54
70
39
30
40
Atomic
weight
200.59
95.94
144.24
20.18
237.05
58.70
92.91
14.01
259
190.2
16.00
106.4
30.97
195.09
244
209
39.09
140.91
145
231.04
226.03
222
186.2
102.91
85.45
101.07
150.4
44.96
78.96
28.09
107.89
22.99
87.62
32.06
180.95
97
127.60
158.93
204.37
232.04
168.93
118.69
47.90
183.85
238.03
50.94
131.30
173.04
88.91
65.38
91.22