# Wilfredo Jose.stoic

September 5, 2017 | Author: John Bryan Aldovino | Category: Distillation, Heat Exchanger, Chemical Engineering, Heat, Engineering

#### Short Description

Concepts for Chemical Engineering Calculations...

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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

1 1 1 3 4 4 4 4 5 5 5 6 6 7 7 8 8 13 13 14 15 16 23 27 29

32 32 35 36 36 37 39 41 42 44 45 48 48 49 54 54 56 56 58 59

i

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 ii

60 61 63 65 66 66 67 67 68 68 69 70 73 73 73 73 79 82 82 83 91 91 92 93 95 96 96 97 97 97 99 99 101 103 111 114 115 121 126 129 131 139 139 140 141 145 146 147

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

148 148 154 166 177 177 177 178 178 179 180 181 181 182 183 187 189 189 190 191 191 192 193 197 199 203 206 207 209 213 214 215 216 217 218 218 219 226 226 226 227 231 232 236 243 251 iii

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

iv

251 257 257 258 258 258 260 261 261 263 264 265 267 268 272 274 275

—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

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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 2

National Engineering Center CEEC Seminar, June 2 -4, 2004

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. 4

National Engineering Center CEEC Seminar, June 2 -4, 2004

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 6

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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 8

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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

.. .

M'3

.. .

.

Mn

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 where M1, M2, M3 … Mn M’1, M’2, M’3 … M’m

Eq. 1-3

= masses of input streams 1, 2, 3, … n, respectively = 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 10

Eq. 1-7

National Engineering Center CEEC Seminar, June 2 -4, 2004

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 14

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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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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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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 22

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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 24

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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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 28

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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

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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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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

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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

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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

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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

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

Symbol

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 3600 s = 0.887 ft 2 ft 2 in 2 0.229 cm % % % s (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 % 3600 s % % h ft 14.7 psi atm 1/6 −4 lbm K = 2.98 % 10 h \$ ft psi 2/3

2/3

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 2 F = mv 2gc

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 t

DRAFT

or

Eq. 2-14

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

then x 1 + x 2 + x 3 + ... + x n = 1

Eq. 2-16

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

xi = 1 i=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. 600 kg/h 3 Q= w ! = 1.161 kg/m 3 = 516.8m /h 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

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National Engineering Center CEEC Seminar, June 2 -4, 2004

600 kg/h 3 Q= w ! = 0.895 kg/m 3 = 670 m /h 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 2 600 kg/h 3 Q= w ! = 0.895 kg/m 3 = 670 m /h 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 V = K1 1 P

P 1 V 1 = P 2 V 2 = P 3 V 3 = .....P n V n

DRAFT

Eq. 2-22

PV = K 1 = constant

or

where

Eq. 2-21

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 V = K2T V 1 /T 1 = V 2 /T 2 = V 3 /T 3 = ... = V n /T n where

Eq. 2-24 Eq. 2-25 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

8.314 0.082 1.987 10.73

Values of R

kJ/(kmol⋅K) (atm⋅liter)/(mol⋅K) BTU/(lbmol⋅°R) (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

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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

T = O + 273 = 273 K

P = 101.325 kPa

(1)(8.314)(273) = 22.4 m 3 V = nRT P = 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 mols mols mols

CO2 CO O2 N2

= = = =

0.12 0.03 0.07 0.78

(100) (100) (100) (100)

= = = =

12 3 7 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 76

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 15.00% H2 1.98% O2 6000 cfm PT = 21 cm H20 head Partial pressure of H2O =

11.03% CO2 51.43% N2 0.21% CH4 T = 200°C Patm = 758 mm Hg 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) (0.1103)(1421) (0.1500)(1421) (0.5143)(1421) (0.0198)(1421) (0.0021)(1421)

= = = = = =

289.16 mm Hg 156.73 mm Hg 213.14 mm Hg 730.79 mm Hg 28.13 mm Hg 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.

 RTB o − A o − C o /T 2 RTb − a a 2 P = RT + + 6 + 3c 2 (1 + 2 )e −/v 2-54 v + 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

(6.25)(8.314)(233) = = 1211 kPa P = nRT V 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) 150 ft-lbforce to joule 1000 dynes to Newton 105 barrels to m3 (1 barrel = 42 gal) 55 miles per hour to meters per second 34 pints U.S. liq. to m3 21,000 in3 to m3 900 fluid ounces to m3 2100 lbm/ft3 to kg/m3 7150 Btu/second to watt Express the following quantities in standard SI units: (Follow the rules given in the text.)

b. c. d. e. f. g. h. i. j. 2-2

a. A woman's vital statistics of 36"-25"-37" b. The distance of the closest galaxy to our own, the Megallanic

DRAFT

c. d. 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) A radio frequency of 100,300,000 cycles per second A corrosion rate of 0.04 inch per year A viscosity of 2.4 × 10-5 lbm/(ft.s) 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?

j. k. l. m.

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

T

2 = 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=

0.0682"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 Be´ 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. b. c. d.

200 psig 4 cm of Hg vacuum 29 mm of H20 head 15 mm of H20 draft 2-27 Convert the following pressures:

pressures

to

the

indicated

absolute

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 94

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 96

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, 100

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

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 x a2 + x b2 + x c2 = 1 x a3 + x b3 + x c3 + x d3 = 1 x a4 + x b4 + x c4 + x d4 = 1 x a5 + x b5 + x c5 = 1

3-20 3-21 3-22 3-23 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 102

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 + 666.67 200 = 0.9D + 0.5S + 0.006(666.67)

3-31 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 75 106

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 108

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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) % 110

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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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, 114

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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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 (100)(0.1) x p1 = = 0.136 73.33

3-55

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 (46.67)

x p2 =

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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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 120

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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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

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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 128

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

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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

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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

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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

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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

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 +

1 2

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: 148

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 Na2CO3 %

= 100 %

1 2 (32)

105.99

% 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 +

5 2C

d 3CaSiO3 + 2P +

5 2 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 60.08 152

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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 +

1 2 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. 154

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. 156

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: C2H6 +

7 2 O2

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 158

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 4-11 160

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 56.18 9.4 -----93.67

mol CO mol H2 mol air + CH4 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: 166

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 168

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 For every 10,000 kg of (a) the amount (b) the amount (c) the amount

→ CH3CHOHCH2OH

propylene glycol produced, calculate of make-up chemicals required of dipropylene glycol byproduct 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)

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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

Taking the differential dU = ØU ØV

T

dV + ØU ØT

V

dT

5-6

At constant volume,

dU = 0 + ØU ØT

V

dT

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 178

National Engineering Center CEEC Seminar June 2 -4, 2004

5-8

and

¶UU

dU = U2 − U1 = U = ¶ T1 Cv dt for 1 kmol 1 T2

2

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

Taking the differential,

dH = ØH ØP

T

dP + ØH ØT

P

dT

5-11

At constant pressure,

dH = 0 + ØH ØT

P

dT

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

2

1

dH = H2 − H1 = H = ¶ T1 Cp dt for 1 kmol T2

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.

DRAFT

For a solid, sensible heat =

¶TT

2

1

For a liquid, sensible heat =

¶TT

nC p,solid dT 2 1

nCp,liquid dT

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. 180

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 44

973 (27.437 + 42.3150 % 10 −3 T ¶ 298

− 1.9555 % 10 −5 T 2

+3.9968 % 10−9 T3 - 2.98732 % 10−13 T4)dT

−5 −9 −13 −3 = 100 27.437T + 42.3150 % 10 T 2 − 1.9555 % 10 T 3 + 3.9968 % 10 T 4 − 2.9872 % 10 T5 5 2 3 4 44 −5 −3 = 100 [27.437(973 − 298) + 42.3150 % 10 (973 2 − 298 2 ) − 1.9555 % 10 (973 3 − 298 3 ) 2 3 44 −9 −13 + 3.9968 % 10 (973 4 − 298 4 ) − 2.9872 % 10 (973 5 − 298 5]) 5 4

= 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. 182

National Engineering Center CEEC Seminar June 2 -4, 2004

973 298

n

Cp mix = y1 Cp 1 + y2 Cp 2 + y3 Cp 3 + ... + yn Cp n = y i C p,i i=1

where

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

5 473

4

= 10 29.36041T + 0.001306 T2 + 8.89052 % 10−6 T3 − 5.19315 % 10−09 T4 + 9.40236 % 10−13 T5

= −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

1273

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.

184

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 = HHo % 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 =

Cp

at 10 o C

+ 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 188

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 190

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 + rH0F R + ...) - (aH0F A + bH 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 + rHoC R + ...) - (aHoC A + bHoC 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

+ H oC

H2

− H oC butane

From Appendix 5, H oC H oC

butane =

-2879.01 MJ/kmol 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: 192

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 5-1 5-2

DRAFT

5-3

5-4 5-5

5-6

5-7 5-8 5-9 5-10 5-11 5-12

5-13

194

1400 grams of copper is heated from 30°C to 80°C? How much heat did it absorb? One kg of iron is initially at 35°C. It absorbs 500 joules of heat? What will be its final temperature? 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. 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? 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. 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. 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? 70 grams of pure ethanol is cooled from 30°C to 4°C. How much heat is removed? How much heat is needed to transform ice at -20° to steam at 130°C? Find the number of calories released as 300 grams of mercury cools from 86 to 31°C. What is the heat capacity of N2 at 500°C? 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. 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.

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.

196

National Engineering Center CEEC Seminar June 2 -4, 2004

--Chapter 6--

Graphical Solution

DRAFT

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 a b c

6-1

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.

198

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

Rearranging 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

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, 200

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

xa1

D

E

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. 202

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

xc

M2

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.) 204

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 = 208

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 214

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 vb y b = P 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 (a) (b) (c) (d) (e)

vapor pressure of the components are: propane: 4600 mm Hg butane : 1000 mm Hg pentane: 250 mm Hg hexane : 68 mm Hg 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)

n-pentane balance:

(0.65)(100) = yp V + xp L

(2)

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. 228

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

P vH O 30 y sat = P P2 = = 0.0340 (1.2)(760) − 30 T − 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

YA

Y

YB

H2O in

B

YB

TB

(b)

YA

A C

YC

TC

YD

YC

D TC

TB

TD TA

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.

238

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 0.3% water

Solids in 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 % kmol bone-dry air (Y 1 ) 4 − Y 3 kmol water picked up = 1.368 % (

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. 240

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 % 1 = 30, 000 m3 % 30 + 273 22.4 m3 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 ya

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 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.

256

National Engineering Center CEEC Seminar June 2 -4, 2004

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 258

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 260

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 262

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 264

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: 266

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 268

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

273-810 273-1041

2 3

273-1173

2

273-1097

2

CuO Fe(a) FeO

2

5.44 + 0.001462T 2 10.87 + 0.003576T - (150600/T ) 4.13 + 0.00638T 2 12.62 + 0.001492T - (76200/T ) 2

Fe2O3

24.72 + 0.01604T - (423400/T )

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 4.38 5.05 16.2

0.00640T 0.00440T 0.00480T 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

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

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

2.9906E-05 -1.7825E-08

3.6934E-12

100

1500

1.8730E-04 -2.1643E-07

0.4746000 -3.1137E-04

33.933 -0.0084186

D

2.0130E-08

E

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 Sodium Sodium Sodium

Chloride Hydroxide Nitrate 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

206 179 157 137 120

278 889 232 734 917

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 93 82 73 65

379 784 848 333 038

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 51 45 40 36

791 447 883 994 690

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 29 26 23 21

894 540 571 940 602

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 17 16 14 13

523 671 018 540 218

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 10 10 9 8

032 968 011 149 372

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 7 6 5 5

671 037 463 943 471

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 kPa

Specific Volume

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

3 3 2 2 2

90 92 94 96 98

70.14 75.64 81.49 87.71 94.30

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

407 160 934 726 536

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

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

1 1 1 1

672.9 419.4 210.2 036.6 891.9

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

100 105 110 115 120

101.35 120.82 143.27 169.06 198.53

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 1.5898729 1.0550559 4.1868000 1.0550559 5.6782633 3.1545907 1.7307347 4.1868000 4.1868000 1.0000000 1.3332237 9.80638 1.0000000 1.0000000 1.0000000 1.0000000 2.9573530 3.0480000 3.0480061 2.98898 9.2903040 3.0480000 2.5806400 1.3558179 9.2903040 2.8316847 3.7854118 1.0000000 7.4569987 2.5400000 3.37685 2.48843 6.4516000 1.6387064 4.1868000 9.8066500 1.0000000 2.540000 1.6093440 4.4704000 1.3332237 2.8349523 2.9572530 4.7317647 1.0000000 1.3825495 4.4482216 4.7880258 4.5359237 1.6018463 1.4881639 6.8947573 9.4635295 1.4593903 1.0000000 1.0160469 9.0718474 1.3332237 1.000165 3.6000000 9.1440000

by E-01 E+03 E+03 E+03 E+00 E+00 E+00 E+00 E+03 E-02 E+03 E+01 E-05 E-07 E-05 E-07 E-05 E-01 E-01 E+03 E-02 E-01 E-05 E+00 E-02 E-02 E-03 E-03 E+02 E-02 E+03 E+02 E-04 E-05 E+03 E+00 E-06 E-05 E+03 E-01 E+02 E-02 E-05 E-04 E-01 E-01 E+00 E+01 E–01 E+01 E+00 E+03 E-04 E+01 E-04 E+03 E+02 E+02 E+00 E+03 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