HANDBOOK ON - JustAnswer



HANDBOOK ON

MATERIAL AND ENERGY

BALANCE CALCULATIONS

IN

MATERIALS PROCESSING

[pic]

THIRD EDITION

By Arthur E. Morris

Gordon Geiger

and

H. Alan Fine

© by Arthur E. Morris 2009

Acknowledgements

This Handbook was prepared under Subcontract 00014529, with Bechtel BWXT Idaho, LLC. We are grateful for the assistance of Simon Friedrich of OIT-DOE in obtaining the contract.

Several people made substantial contributions to the Handbook. First, Chapter 3 is largely the work of two graduate students from Texas A and M University, Mr. Blair Sterba-Boatwright and Mr. Peng-lin Huang. Both are candidates for the PhD. in statistics, and made what was a very rough draft into a polished product. Second, Mr. Semih Perdahcioglu developed FlowBal and MMV-C, and revised FREED to include a reaction tool. Semih is now a research assistant at University of Twente, Netherlands. Third, Mr. Knut Lindqvist wrote the code for Super Goal Seek and Super Solver, and Robert Baron wrote U-Converter.

We are obligated to three faculty members for miscellaneous advice with various topics. Professor Eric Grimsey (WASM Kalgoorlie) provided valuable suggestions, encouragement, and help, especially with the application of the degree-of freedom concept. Professor David Robertson (Missouri University of Technology) cleared up a number of points regarding the material on continuously-mixed and unsteady-state processes. Professor Mark Schlesinger (Missouri University of Technology) permitted use of several of his examples and exercises.

Finally, no text is generated in isolation. Appendix C contains a number of general references that provided background information and material data, and other texts that influenced the structure of this edition of the Handbook. Some of the problem-solving strategies of those texts were modified to fit the computationally-intensive approach used in this edition.

Preface to the First Edition

We live in a day and age when realization of the “limits to growth” and the finite extent of all of our natural resources has finally hit home. Yet our economy and our livelihoods depend on successful operation of industries that require and consume raw materials and energy. This success depends, in turn, on efficient use of the available resources, which not only allows industry to conserve materials and energy, but also allows it to compete successfully in the world markets that exist today.

The duties of the metallurgical engineer include, among many other things, development of information concerning the efficiency of metallurgical processes, either through calculation from first principles, or by experimentation. The theory of the construction of material and energy balances, from which such knowledge is derived, is not particularly complicated or difficult, but the practice, particularly in pyrometallurgical operations, can be extremely difficult and expensive.

In this Handbook we have tried to review the basic principles of physical chemistry, linear algebra, and statistics which are required to enable the practicing engineer to determine material and energy balances. We have also tried to include enough worked examples and suggestions for additional reading that a novice to this field will be able to obtain the necessary skills for making material and energy balances. Some of the mathematical techniques which can be used when a digital computer is available are also presented. The user is cautioned, however, that the old computing adage “garbage in, garbage out” is particularly true in this business, and that great attention must still be paid to setting up the proper equations and obtaining accurate data. Nevertheless, the computer is a powerful ally and gives the engineer the tool to achieve more accurate solutions than was possible just twenty-five years ago.

It is hoped that readers, particularly those who are out of practice at these kinds of calculations will ultimately be able to perform energy balances in processes for which they are responsible, and as a result be able to improve process efficiencies. A bibliography of past work on this subject is presented in an appendix to provide reference material against which results of studies can be checked. Hopefully, results reported in the future will reflect increases in efficiency.

H. Alan Fine

University of Kentucky

Lexington, Kentucky

Gordon H. Geiger

University of Arizona

Tucson, Arizona

Preface to the Third Edition

Because the fundamental bases on which the laws of conservation of mass and energy depend remain the same, users of this edition will find an essential similarity between this edition and the slightly revised (second) edition of 1993. Two noteworthy changes in the professional engineer’s practice have occurred since 1993, however. First, in the last 20 years, a dramatic shift has occurred away from metallurgical engineering and the extractive industry towards materials engineering. A large and growing number of recent graduates are employed in such fields as semiconductor processing, environmental engineering, and the production and processing of advanced and exotic materials for aerospace, electronic and structural applications. Second, in the same time frame the advance in computing power and software for the desktop computer has significantly changed the way engineers make computations.

This edition of the text reflects these changes. The text now includes examples that involve environmental aspects, processing and refining of semiconductor materials, and energy-saving techniques for the extraction of metals from low-grade ores. But the biggest change comes from the computational approach to problems. The spreadsheet program Excel is used extensively throughout the text as the main computational “engine” for solving material and energy balance equations, and for statistical analysis of data. A large thermodynamic database (FREED) replaces the thermodynamic tables in the back of the previous Handbook. A number of specialized add-in Excel programs were developed specifically to enhance Excel’s problem-solving capability. And finally, on-line versions of two commercial programs for steam-table and psychrometric calculations were identified and demonstrated. These programs simplify the rather difficult calculations heretofore required. The use of Excel and the introduction of the add-in programs have made it possible to study the effect of a range of variables on critical process parameters. More emphasis is now placed on multi-device flowsheets with recycle, bypass and purge streams whose material and heat balance equations were previously too complicated to solve by the normally-used hand calculator. Appendix A has a brief description of these programs.

The Excel-based program FlowBal is the most important addition to this Edition. FlowBal helps the user set up material and heat balance equations for processes with multiple streams and units. FlowBal uses the thermodynamic database program FREED for molecular mass and enthalpy data. FlowBal’s purpose is to introduce the increasingly important subject of flowsheet simulation. FlowBal and all other software and supplementary reference material is on a CD included with the Handbook. A text file on the CD describes its contents.

Additional changes have been made throughout the text. There are now ten chapters instead of six, which reflects a desire to organize the material in non-reactive vs. reactive material and energy balance sections. The concept of degree-of-freedom analysis has been introduced to provide a basis for analyzing the adequacy of information presented in a flowsheet. The concepts of extent-of-reaction and the equilibrium constant are presented as ways to designate how far a given chemical reaction will (or can) proceed. The introduction of the equilibrium constant requires the Handbook user to have completed a course in Chemistry which contains lectures on thermochemistry, or at least to have available the chemistry text book used in the freshman year of their engineering education. Chapter 3 has been completely revised to emphasize the statistical analysis of experimental data, while de-emphasizing the descriptive material on chemical analysis and techniques for sampling process streams. A final chapter has been added on case studies, showing the application of computational techniques and software to more complex processes.

This edition frequently uses web citations and Wikipedia as references and suggestions for further reading. Wikipedia has well-written articles on many Handbook topics, and more are being added. Wikipedia is a work in progress, so readers are encouraged to search it for additional information even if a Wikipedia reference is not listed in the text. The web pages cited as Chapter references may change after publication, and other sites may appear.

Finally, a web page has been created where changes and additions to the Handbook are posted. The web page contains updates to the Handbook software, error corrections, references to new software, and links to other sites having useful information on material/energy balances and process simulation. We encourage all Handbook users to alert the authors to useful information and to submit material for posting on the page.



Arthur E. Morris

Thermart Software

San Diego, California

Gordon H. Geiger

University of Arizona

Tucson, Arizona

Handbook Table of Contents

1. Dimensions, Units, and Conversion Factors

1.1 The SI System of Units -

1.2 The American Engineering System (AES) of Units -

1.3 Conversion of Units -

1.4 Unit Conversions Using the U-Converter Program -

1.5 Amount of Substance — the Mole Unit -

1.6 Density and Concentration -

1.7 Electrical Units -

1.8 Calculation Guidelines -

1.9 Summary -

References and Further Reading -

Chapter 1 Exercises -

2. Thermophysical and Related Properties of Materials

2.1 Properties and State of a Substance -

2.2 Gibbs Phase Rule -

2.3 The Gas Phase -

2.4 Condensed Phases -

2.5 Vapor Saturation -

2.6 Effect of Pressure on Phase Transformation Temperatures -

2.7 Steam and Air Property Calculations -

2.8 Properties of Solutions -

2.9 Summary -

References and Further Reading -

Chapter 2 Exercises -

3. Statistical Concepts Applied to Measurement and Sampling

3.1 Basic Statistical Concepts and Descriptive Tools -

3.2 Distributions of Random Variables -

3.3 Basic Applications of Inferential Statistics to Measurement -

3.4 Curve Fitting -

3.5 Experimental Design -

3.6 Summary -

References and Further Reading -

Chapter 3 Exercises -

4. Fundamentals of Material Balances with Applications to Non-Reacting Systems

4.1 System Characteristics -

4.2 Process Classification -

4.3 Flowsheets -

4.4 The General Balance Equation -

4.5 Material Balances on Simple Non-Reactive Systems -

4.6 Strategy for Making Material Balance Calculations -

4.7 Degree-of-Freedom Analysis -

4.8 Using Excel-based Calculational Tools -

4.9 Balances on Systems with Multiple Devices -

4.10 Extension of Excel’s Calculational Tools for Repetitive Solving -

4.11 Special Multiple-Device Configurations I — Recycle and Bypass -

4.12 Special Multiple-Device Configurations II — Counter-Current Flow -

4.13 Using FlowBal for Material Balance Calculations -

4.14 Continuously-Mixed Reactors -

4.15 Graphical Representation of Material Balances -

4.16 Measures of Performance -

4.17 Controllers -

4.18 Summary -

References and Further Reading -

Chapter 4 Exercises -

5. Stoichiometry and the Chemical Equation

5.1 Atomic and Molecular Mass -

5.2 Composition of Compounds and the Gravimetric Factor -

5.3 Writing and Balancing Chemical Equations -

5.4 Calculations Involving Excess and Limiting Reactants -

5.5 Progress of a Reaction -

5.6 Practical Indicators of the Progress of Reactions and Processes -

5.7 Parallel, Sequential and Mixed Reactions -

5.8 Practical Examples of Reaction Writing and Stoichiometry -

5.9 Use of Chemical Reactions in FlowBal -

5.10 Balancing Aqueous (Ionic) Reactions -

5.11 Summary and Conclusions -

References and Further Reading -

Chapter 5 Exercises -

6 Reactive Material Balances

6.1 Balances Using Molecular and Atomic Species

6.2 The Use of Excel-based Computational Tools in Reactive System Balances

6.3 Combustion

6.4 The Production of Gases of Controlled Oxygen and Carbon Potential

6.5 Gas-Solid Oxidation-Reduction Processes

6.6 Processes Controlled by Chemical Reaction Kinetics

6.7 The Reconciliation of an Existing Materials Balance

6.8 The Use of Distribution Coefficients in Material Balance Calculations

6.9 Time-Varying Processes

6.10 Systems Containing Aqueous Electrolytes

6.11 Summary

References and Further Reading

Chapter 6 Exercises

7 Energy and the First Law of Thermodynamics

7.1 Principles and Definitions

7.2 General Statement of the First Law of Thermodynamics

7.3 First Law for an Open System

7.4 Enthalpy, Heat Capacity, and Heat Content

7.5 Enthalpy Change of Phase Transformations

7.6 Enthalpy Change During Chemical Reactions

7.7 Thermodynamic Databases for Pure Substances

7.8 Effect of Temperature on Heat of Reaction

7.9 The Properties of Steam and Compressed Air

7.10 The Use of FREED in Making Energy Calculations

7.11 Heat of Solution

7.12 Summary

References and Further Reading

Chapter 7 Exercises

8 Energy Balances in Non-Reactive Systems

8.1 General Energy Balances

8.2 Heat Balances in Adiabatic Systems

8.3 Psychrometric Calculations

8.4 Energy Efficiency

8.5 Recovery and Recycling of Heat

8.6 Multiple-Device System Balances

8.7 Use of FlowBal for Heat Balance Calculations

8.8 Energy Balances Involving Solution Phases

8.9 Aqueous Processes

8.10 Graphical Representation of Energy Balance

8.11 Summary

References and Further Reading

Chapter 8 Exercises

9 Heat and Material Balances on Reactive Systems

9.1 Thermal Constraints on a Material Balance

9.2 Heat of Combustion of Fuels

9.3 Adiabatic Processes

9.4 Heat Balances Using FlowBal

9.5 Quality of Heat and Thermal Efficiency

9.6 System Balances for Reacting Systems With Heat Exchangers

9.7 Aqueous Processes

9.8 Electrolytic Processes

9.9 Summary

References and Further Reading

Chapter 9 Exercises

10 Case Studies

10.1 Material Balance Simulation of an H-Iron Process with Gas Tempering and Recycle

10.2 A Mass and Heat Balance Simulation for the Use of DRI in EAF Steelmaking

10.3 Natural Gas Combustion Control and the Wobbe Index

10.4 Reduction of Hematite to Magnetite

10.5 Conversion of Quartz to Cristobalite in a Fluidized bed

Appendix A Computational Tools for Making Material and Energy Balance Calculations

CD Table of Contents

1. Enhanced Excel Tools

Super Goal Seek

Super Solver

Unit Conversions

U-C.xla

U-C.rtf

FREED

2. FlowBal and MMV-C

Selected Examples

3. Statistical Analysis Examples: StatTools.xls

4. Supplementary Information

Reference data

Steam and air

AirPVT.xls

Documents

General References

1. General Chemistry

Zumdahl, S. S, and Zumdahl, S. A., Chemistry, 7th Edition, Houghton Mifflin, 2007.

2. Substance Properties and Thermodynamic Data

Haynes, William M., Ed., CRC Handbook of Chemistry and Physics, 91st Edition. CRC Press, 2010.

Speight, James, Lange's Handbook of Chemistry, 70th Anniversary Edition, McGraw-Hill Professional, 2005,

Green, Donald W. and Perry, Robert H., Perry’s Chemical Engineers’ Handbook. 8th Edition, McGraw-Hill Professional, 2007.

3. Encyclopedias

Encyclopedia of Materials Science and Technology, Elsevier, 2001.

Ullmann's Encyclopedia of Industrial Chemistry, 6th Edition, Wiley-VCH, 2002.

Kirk-Othmer Encyclopedia of Chemical Technology, 5th Edition, John Wiley & Sons, Inc, 2001.

4. Material and Energy Balances

Schlesinger, Mark E., Mass and Energy Balances in Materials Engineering, Prentice Hall, 1996.

Oloman, Colin, Material and Energy Balances for Engineers and Environmentalists (Advances in Chemical and Process Engineering), Imperial College Press, 2009.

Himmelblau, David M., and Riggs, James B., Basic Principles and Calculations in Chemical Engineering, 7th Edition, Prentice Hall, 2003.

Felder, Richard M. and Rousseau, Ronald W., Elementary Principles of Chemical Processes, 3rd Edition, Wiley, 2005 version, with Integrated Media and Study Tools and Student Workbook.

Reklaitis, G. V, Introduction to Material and Energy Balances, Wiley, 1983.

CHAPTER

1

Dimensions, Units, and Conversion Factors

M

ost science and engineering calculations are performed using quantities whose magnitudes are expressed in terms of standard units of measure or dimensions. A dimension is a property that can be measured, such as length, time, mass or temperature, or obtained by manipulating other dimensions, such as length/time (velocity), length3 (volume), or electric current/area (current density). Dimensions are specified by giving the value relative to some arbitrary standard called a unit. Therefore, the complete specification of a dimension must consist of a number and a unit. Convention, custom, or law can specify which units are used, such that the volume of a substance may be expressed in cubic feet, liters, or gallons.

There are two common systems of units used in engineering calculations. One is the American engineering system (AES) based on the foot (ft) for length, the pound-mass (lbm) for mass, degrees Fahrenheit (°F) for temperature, and the second (s) for time. The two main drawbacks of this system are the occurrence of conversion factors which are not multiples of 10, and the unit of force, which will be discussed later. The other is Le Systèm Internationale d’Unités or SI for short, which has gained widespread acceptance for all scientific and much engineering work. In 1991, the US Department of Commerce promulgated regulations for the required use of the SI system for all Federal agencies. Despite the nearly worldwide acceptance of the SI system, the AES system is still in use in many U.S. industries, and the last vestiges of its use may take decades to obliterate.

This text will emphasize the use of SI units with some exceptions. The calorie and atmosphere are used when dealing with thermodynamic data based on these units. Some non-SI units will be used in selected cases. Converting between units is made easier with a units conversion program (U-Converter, on the Handbook CD).

1.1 The SI System of Units

In 1960, the General Conference on Weights and Measures (CGPM, Conférence Général des Poids et Mesures) established conventions to be used for a set of basic and derived units. The National Institute of Standards and Technology (NIST) is the Federal agency assigned responsibility for publishing guides for SI use. Revisions were made since the first guide was issued in 1960, culminating in the publication of three important NIST documents (Butcher, 2006; Taylor, 2008; Thompson, 2008). These documents are described on the NIST web site. Another useful document is available from the U.S. Metric Association (Antoine, 2001).

There are three classes of SI units:

– base units – derived units – supplementary units

which together form what is called “the coherent system of SI units”. Table 1.1 gives the seven base quantities on which the SI is founded, and the names and symbols of their respective units, called “SI base units”. One of the SI base units — the candela for luminous intensity — is not used in this Handbook.

The quantity of a substance can be expressed in two ways: its mass (number of kilograms), or its number of atoms (moles). The amount unit is the mole (abbreviation mol), which specifies the amount of substance in a given mass. When defining quantity in terms of moles, the mole unit is defined as that amount of a substance containing as many elementary particles as there are atoms in 0.012 kg of the nuclide 12C. This number has been found to equal about 6.022 1367 x 1023, which is Avogadro’s Number (NA). Section 1.4 discusses the mole unit in more detail. The distinction between the mass and mole units and conversions between them is an important part of this Handbook, and will be covered in detail in later sections of this Chapter.

Table 1.1 SI base units.

|Base quantity |Name |Symbol |

|length |meter |m |

|mass |kilogram |kg |

|time |second |s |

|electric current |ampere |A |

|thermodynamic temperature |kelvin |K |

|amount of substance |mole |mol |

|luminous intensity |candela |cd |

Prefixes are used in SI to form decimal multiples and submultiples of SI units. The most common of these prefixes and their abbreviations are giga (G) for 109, mega (M) for 106, kilo (k) for 103, cent (c) for 10–2, milli (m) for 10–3, micro (μ) for 10–6, and nano (n) for 10–9. There are also some less-common prefixes.

1.1.1 Derived Units

There are two classes of derived units. First, those obtained by mathematical operations of multiplication or division. For example, velocity as meter per second (m/s), and current density as ampere per square meter (A/m2). Second are similarly derived units with special names, such as force (newton, or N, as m ( kg ( s–2) and energy (joule, or J, as m2 ( kg ( s-2 or as N ( m). NIST SP 811 (Thompson, 2008) gives a complete list of derived units. Table 1.2 and Table 1.3 list the derived units used in the Handbook.

Another group of derived units are those expressed with special names. For example, the molar entropy or molar heat capacity is better expressed as joules per mol kelvin, J/(mol ( K) rather than m2 ( kg ( s–2 ( K–1 ( mol–1. Table 1.4 lists some common examples of this type of unit. These special names exist for convenience, and such derived units can be expressed in different ways. Preference is given to customary use.

Table 1.2 Examples of SI derived units expressed in terms of SI base units.

| |SI derived unit |

|Derived quantity |Name |Symbol |

|area |square meter |m² |

|volume |cubic meter |m3 |

|speed, velocity |meter per second |m/s |

|acceleration |meter per second squared |m/s2 |

|mass density (density) |kilogram per cubic meter |kg/m3 |

|specific volume |cubic meter per kilogram |m3/kg |

|current density |amperes per square meter |A/m2 |

|amount-of-substance concentration (concentration) |mole per cubic meter |mol/m3 |

Table 1.3 Examples of SI derived units with special names and symbols.

| | SI derived unit |

|Derived quantity |Special name |Special symbol|Expression in terms of|Expression in terms of |

| | | |other SI units |SI base unit |

|frequency |hertz |Hz | |s–1 |

|force |newton |N | |m ( kg ( s–2 |

|pressure, stress |pascal |Pa |N/m² |m–1 ( kg ( s–2 |

|energy, work, heat |joule |J |N ( m |m² ( kg ( s–2 |

|power |watt |W |J/s |m2 ( kg ( s–3 |

|quantity of electricity |coulomb |C | |s ( A |

|electric potential, emf |volt |V |W/A |m3 ( kg ( s–3 ( A–1 |

|electric resistance |ohm |Ω |V/A |m2 ( kg ( s–3 ( A–2 |

|Celsius temperature |degree Celsius |°C | |K |

Table 1.4 Additional SI units without special names, expressed in terms of named units.

| | |SI derived unit |

|Derived quantity |Name |Symbol |Expression in terms |

| | | |of SI base units |

|heat flux density |watt per square meter |W/m2 |kg ( s-3 |

|heat capacity, entropy |joule per kelvin |J/K |m2 ( kg ( s–2 ( K–1 |

|specific heat capacity, entropy |joule per kilogram kelvin |J/(kg ( K) |m2 ( s–2 ( K–1 |

|molar entropy, heat capacity |joule per mole kelvin |J/(mol ( K) |m2 ( kg ( s–2 ( K–1 ( mol–11 |

|thermal conductivity |watt per meter kelvin |W/(m ( K) |m ( kg ( s–3 ( K–1 |

SI allows the expression of units as fractions, or expressed as negative exponentials. Thus it’s equally appropriate to express heat flux density as W/m2 or W · m-2.

1.1.2 Units Outside the SI

There are three categories of units outside the SI:

– those units that are accepted for use with the SI;

– those units that are temporarily accepted for use with the SI;

– those units that are not accepted for use with the SI, and are to be avoided.

The accepted units include minute, hour, or day (min, h, d) instead of second; liter (symbol L) instead of m3; and metric ton (t) instead of 103 kg or 1 Mg. (The metric ton is called the tonne in many countries). The concentration term % is acceptable in place of 0.01. For example, it is preferable to state, “the mass fraction of B is 0.02”, or “wB = 0.02”, but acceptable to state, “the mass fraction of B is 2 %”, or “wB = 2 %”. The temporarily accepted units include the pressure unit bar (bar), which is equivalent to 105 Pa. Some thermodynamic tables list the standard pressure as 1 bar. The standard atmosphere is approximately 1.013 bar.

Unacceptable units of course include those of the AES, such as ft and lb. Other “metric” unacceptable units are the dyne and erg (left over from the CGS system), the torr and atmosphere (atm) as units of pressure, the kilogram-force (kgf) as a unit of force, and the calorie (cal, in various dimensions) as a unit of energy. Similarly, concentration terms such as ppm or ppb are unacceptable unless required by law. As mentioned earlier, some thermodynamic data may be available only in units of cal and atm, in which case, they will be used in this Handbook with no further explanation.

1.1.3 Comments on Some Quantities and Their Units

Temperature. The quantity Celsius temperature (symbol t) is used in addition to the thermodynamic temperature expressed in the unit kelvin. It is defined by the equation:

T – TO [1.1]

where TO = 273.15 K by definition, and is exactly 0.01 K below the triple point of water. The degree unit Celsius is equal in magnitude to the degree unit kelvin. An interval or difference of °C can be expressed as well in the unit kelvin. Note that the centigrade temperature scale is obsolete; the degree centigrade is almost, but not quite, equal to the degree Celsius. This Handbook uses an upper-case T to denote temperature in unit kelvin (K), and a lower-case t to denote temperature in degree Celsius. Furthermore, to avoid confusion, we will spell out the word tonne instead of using t to designate the metric ton.

Weight. In science and technology, we define the weight of a body in a particular reference frame as the force that gives the body acceleration equal to the local acceleration of free fall. We define the acceleration of gravity as exactly 9.806 65 m/s2 at the standard location of sea level and 45° north latitude. Thus, the SI unit of the quantity weight is the newton (N). In commercial and everyday use, weight is usually used as a synonym for mass. Thus, the SI unit used in this way is the kilogram (kg). In order to avoid confusion, the term weight should be avoided, and mass used instead.

Amount of Substance, Concentration, Molality, etc. Concentration and fractional amount terms require special attention for SI usage (section 8.6, Thompson, 2008). In particular, the terms “molarity” and “normal” are not acceptable for SI, and “amount-of-substance fraction of B” is preferred to “mole fraction of B”. Section 1.7 discusses this issue in more detail.

Printing and Using Symbols and Numbers. This Handbook follows NIST recommendations (Thompson, 2008) in naming and formatting the various symbols used. Digits for numbers should be separated in groups of three. For example, the conversion factor for cubic feet (ft3) to cubic meters (m3) is 2.831 685 ( 10–2. However, the conversion factor from kilo-calorieth to joule is 4184, which is an exact factor. Try to avoid using a number like 2400 because it is unclear if the last two zeros are significant figures. Instead, use 2.4 ( 103 or 2.400 ( 103 as appropriate to indicate the number of significant figures. The exception to this policy occurs in the use of Excel, which does not accept spaces between digits.

1.2 The American Engineering System (AES) of Units

As much as we would like to see units such as horsepower, Btu and °F disappear, they have not. The AES is still used industrially, and is still used in some documents from Federal agencies intended for use by industry. The policy of this Handbook is to illustrate AES use in selected cases, point out some of the difficulties in its use, and illustrate the use of conversion factors from the AES to SI. A brief list of the more common conversion factors is given on the inside cover, and a more extensive list is in NIST SP811 (Thompson, 2008). The U-Converter program has the most comprehensive list.

A notable difference in SI and the AES system is the derived unit of force. In SI, the derived force unit is the newton (N), based on the natural force unit of kg ( m/s2. In the AES, a choice can be made to select an arbitrary unit of force or an arbitrary unit of mass. Newton’s law automatically fixes the other unit:

Force = mass ( acceleration [1.2]

If the pound is chosen as the unit of mass (lbm), it may be expressed in terms of the kilogram; the lbm has 0.4536 times the mass of a kg. Then the fundamental derived unit of force is that which will produce an acceleration in 1 lbm of 1 ft/s2. This unit is the poundal, with the dimensions of ft/s2.

If the pound is chosen as the fundamental unit of force, the lbf is the unit of force that will give a lbm an acceleration of 32.174 ft/s2. It is also the force of gravity between the lbm and earth existing at sea level and 45° latitude (the standard location). When the lbf is selected as the unit of force, the derived unit of mass is that mass which will be accelerated at the rate of 1 ft/s2 when acted on by 1 lbf. This derived unit of mass is known as the slug, with a mass of 14.5939 kg, and units of lbf (·sec2/ft. A lbf gives a lbm an acceleration of 32.174 ft/s2.

Unfortunately, it has been common for engineers to select the lbf as the unit of force, and the lbm as the unit of mass. When these are substituted in Equation [1.2], the resulting equation is neither algebraically or dimensionally correct. To avoid this incongruity, Equation [1.2] must be rewritten as:

Force = mass ( acceleration/gc [1.3]

where gc is a constant equal to 32.174 lbm ( ft/(lbf ( sec2), and is independent of location.

We define the weight of a body as the force of gravity existing between the body and earth, and since weight is a force, we express it in terms of the lbf when the AES is used. Fortunately, the variations in weight produced by latitude or elevation are small, rarely exceeding 0.25%. Thus, Equation [1.3] is rewritten as:

F = W ( a/g [1.4]

where F = force acting on a body in any direction in lbf;

W is the weight of the body in lbf;

g = acceleration of gravity at the location, in ft/s2;

a = acceleration of the body in the direction of the force, in ft/s2.

Note that the weight of a body at standard location, expressed in lbf, is numerically equal to the mass of the object, expressed in lbm. Thus at that location, a kilogram of water weighs 2.2046 lb (i.e., lbf), and the mass of water is 2.2046 lb (i.e., lbm). This numerical equality is a close approximation at other locations.

EXAMPLE 1.1 — Mass and Weight of Aluminum.

Calculate the mass of a block of aluminum with a volume of 0.1500 m3 (5.297 ft3) at the standard location. Calculate the gravitational force acting on the block, and the stress in a 0.500 cm (0.1969 in) diameter wire suspending the block. Make the calculations in SI and AES units.

Data. The mass density of aluminum is 2702 kg/m3 (168.7 lbm/ft3). Stress has the units of force per unit area (or pressure), expressed as Pa or lbf/in2. Assume the mass of the wire and air buoyancy can be neglected.

Solution. The SI mass of the block is (2702 kg/m3)(0.15 m3) = 405.3 kg. The weight of the block in SI units is:

W = m ( g

Weight (SI) = (405.3 kg)(9.8066 m/s2) = 3975 kg ( m/s2 = 3975 N.

The mass of the body in AES units is (5.297 ft3)(168.7 lbm/ft3) = 893.6 lbm. According to Equation [1.4], the numerical value of the weight in lbf = the weight in lbm when the acceleration of gravity equals gc so:

Weight (AES) = (5.297 ft3)(168.7 lbm/ft3) = 893.6 lbm = 893.6 lbf.

Stress has units of force/area. The SI area of the wire is π(0.25/100)2 = 1.964 ( 10–5 m2. The AES area of the wire is π(0.09845)2 = 0.03043 in2. The stress σ is calculated as:

σ = F/A

σ (SI) = 3975 N/1.964 ( 10–5 m2 = 2.024 ( 108 Pa.

σ (AES) = 893.6 lbf/0.03043 in2 = 2.937 ( 104 lbf/in2.

Assignment. Calculate the force (in SI and AES) that would accelerate the aluminum block to a speed of 10 ft/s if applied over a time of 5 s.

EXAMPLE 1.2 — Kinetic Energy.

Calculate the kinetic energy (SI and AES units) of 1 mole of oxygen traveling at a linear speed of 101 m/s (331.4 ft/s).

Data. The mass of a mole of O2 is 0.0320 kg (0.07055 lbm). The kinetic energy (Ek) is ½ the product of the mass times the velocity squared.

Solution. Ek (SI) = ½(0.0320 kg)(101 m/s)2 = 163 J

Ek (AES) = ½(0.07055)(331.4 ft/s)2/32.174 lbm ( ft/lbf ( sec2 = 120 ft ( lbf

Assignment. Calculate the time required to accelerate the mass of O2 from rest to the stated velocity if the force applied is 2.00 N.

EXAMPLE 1.3 — Energy of Lifting.

Calculate the work in kW ( h (commonly designated kWh) required to raise 1 tonne of iron ore (1000 kg) a distance of 30 m at the standard location.

Data. Work is the product of force and distance. The force in this case is the acceleration of gravity times the mass. This is equivalent to the weight in newtons.

Solution. Work = (1000 kg)(9.8066 m/s2)(30 m) = 294 200 N ( m = 294 200 J

From Table 1.3, a joule is a watt ( second. Therefore,

Work = (294 200 W ( s)/[(3600 s/hr)(1000 W/kW)] = 0.0817 kWh

This value is the same as the change in potential energy. If the mass of iron ore were released to fall under the force of gravity, after falling 30 m its potential energy would decrease by 294 200 J, and its kinetic energy would increase by a like amount.

Assignment. Calculate the electric power required (assume 100 % efficiency in conversion) for an electric motor to lift the tonne of ore 30 m in 10 seconds.

1.3 Conversion of Units

The previous examples illustrated the need to be able to express dimensions and units in more than one system, and to express an answer in different units in the same system. Multiple unit conversions are more prevalent in later chapters where more complex problems are encountered. We already know certain relationships between common units such as time (60 seconds in 1 hour) and length (12 inches in 1 foot). One way to express the relationship between units is to use a conversion factor. The Handbook has a conversion factor table inside the front cover, and many more conversion factors are listed by NIST and other handbooks (Butcher 2006, Thompson 2008). Section 1.4 and Appendix A describe the Excel unit conversion program U-Converter (on the Handbook CD).

Two types of conversions are required. In the first, one unit is converted into another by multiplication or division by the conversion factor. In the second, it is necessary to use the conversion factor plus addition or subtraction of an additional term.

1.3.1 Conversion Factor Tables

The relationship between the units used in different systems is determined by convention or from the basic definition of the units. Conversion factors have been compiled into tables to assist the scientist or engineer. All of the conversion factors listed in this Handbook and on the CD are consistent with NIST (Thompson, 2008). Conversion factors may be exact or rounded off. Thompson lists exact conversion factors with boldface type. Conversion factors used in Handbook examples are generally taken to four significant figures unless a higher accuracy is appropriate.

Be aware that some units have multiple definitions. For example, there are eight different calories and six different British thermal units. There are two different types of pound, and three different types of ton (the short ton = 2000 lb, the long ton = 2200 lb, and the metric ton, designated tonne in this Handbook, = 1000 kg). This Handbook uses the thermochemical calorie and thermochemical Btu exclusively. The calth has a value of 4.184 J (exact), and the Btuth a value of 1054.35 J (also exact).

1.3.2 The Dimension Table

Simple conversions between the SI units for length, mass, and force were carried out in examples 1.1, 1.2, and 1.3. A procedure using a dimension table (sometimes called a conversion table) is useful where a single conversion factor is not available, and several factors must be combined to obtain the desired result. The dimension table contains both the numerical value of each unit and the units themselves. Conversion from one set of units to another is accomplished by multiplication of a series of conversion factors. The dimension table is a way of organizing the conversion process and reducing the chance for error.

A simple example will show how to develop a conversion factor to convert velocity from m/s to km/h. The method uses a series of equations. The first step is to convert m to km, through the substitution m = km/1000, and then s to h through the substitution s = h/3600, which gives the conversion

[pic].

Thus m/s is a larger unit than km/h, since 1 m/s is equivalent to 3.6 km/h.

If a dimension table is used, this substitution can be systematically represented through a series of cells, namely:

|1 m |km |3600 s |

|s |1000 m |h |

The first column in the dimension table represents 1 m/s and thus contains the numerical value of 1 and the units of length m in the upper cell, and the dimension of time s in the lower cell. The middle column contains the conversion factor from m to km. The rightmost column contains the conversion factor from s to h. The units in the conversion factors must cancel such that the final unit is km/h. Multiplying the numerical values gives the conversion factor of 3.6. In other words, multiply m/s by 3.6 to get km/h.

EXAMPLE 1.4 — Units of Energy.

The heat of formation of compounds is frequently listed in units of calth/mole. Use a dimension table to calculate the conversion factor for changing the heat of formation of CO2 from calth/mol to kWh/kg.

Solution. The molar mass of CO2 = (12.01 + 32) = 44.01 g. The dimension table is:

|cal |mol |1000 g |4.184 J |W ( s |h |kW |

|mol |44.01 g |kg |cal |J |3600 s |1000 W |

The conversion factor is then 2.6408 ( 10–5. This should be multiplied times the listed value in calth/mole to obtain a value in kWh/kg. (Note: the kW ( h is commonly designated as kWh).

Assignment. Calculate a conversion factor for the heat of formation of CO2 from calth/mol to Btuth/lbm.

EXAMPLE 1.5 — Dimensions for Flowrate.

A furnace produces 1025 short tons of metal per day. Use a dimension table to calculate the production rate in kg/s.

Solution. The dimension table below includes the initial numerical value of the unit (1025).

|1025 ton |907.185 kg |d |

|1 d |ton |8.64 ( 104 s |

Multiplying the numerical values and canceling units gives 10.76 kg/s. For general use, the above dimension table gives a conversion factor of 1.050 x 10–2.

Assignment. A power plant burns 25 Mg of coal per day. Each Mg of coal contains 1 mg of mercury. Calculate the largest mass of mercury that the plant could emit if it operated steadily for a year.

These examples show how to use a dimension table to create a new conversion factor from a combination of factors. It can also be used to convert a value given in one set of units to a value in another set. You have a choice of using base or derived units, and the order of multiplication is arbitrary. The clear and concise manner of the dimension table helps eliminate errors and its use is strongly recommended, especially for persons of limited experience in making such conversions. The U-Converter program can also generate conversion factors.

1.3.3 Conversion Equations — Temperature and Pressure

Some unit conversions cannot be accomplished by use of a dimension table. For this case, the addition or subtraction of a quantity must accompany the multiplication of conversion factors. Two examples of dimensions that often require conversion equations are temperature and pressure. Conversion equations are required here because there are units with different sizes and different zero points.

The SI unit of temperature is the kelvin, designated K, which is an absolute scale. Its zero point is absolute zero. The absolute scale for the AES is the Rankin scale, designated R, whose zero point is also absolute zero. The conversion factor for these two scales is:

K = °R/1.8 [1.5]

Temperature scales such as Fahrenheit and Celsius are relative temperature scales because the zero points of these scales are fixed at different arbitrary standards. For Celsius, the zero point is fixed at 273.15 K, which is 0.01 K below the triple-point of water. For Fahrenheit the zero point is 255.372 K. The numerical value of a given temperature interval or temperature difference whose value is expressed in the unit degrees Celsius (°C) is equal to the same numerical value or difference when its value is expressed in the unit kelvin (K). Similarly, the numerical value of a given temperature interval or temperature difference whose value is expressed in the unit degrees Fahrenheit (°F) is equal to the same numerical value or difference when its value is expressed in the unit degrees Rankin (°R). Thus, temperature intervals or differences may be expressed in either the degree Celsius or the kelvin, and similarly for the degree Fahrenheit and degree Rankin:

1.0 ΔK = 1.0 Δ°C; 1.0 Δ°R = 1.0 Δ°F [1.6]

Also,

1.0 Δ°C = 1.8 Δ°F; 1.0 ΔK = 1.8 Δ°R [1.7]

Converting between kelvin, °C, °R, K, and °F requires more than a single conversion factor, except for the conversion between K and °R already listed in Equation [1.5].

T(K) = t(°C) + 273.15 [1.8]

T(°R) = t(°F) + 459.67 [1.9]

t(°F) = 1.8t(°C) + 32 [1.10]

The main point in dealing with temperature conversions is that a degree is both a temperature and a temperature interval. It’s very important to recognize this difference when conversions are being made.

EXAMPLE 1.6 — Conversion of Temperature.

The melting point of gold at 101.325 kPa is 1337.58 K on the International Practical Temperature Scale.* Convert this value to a) °C, b) °F, and c) °R.

Solution. a) Insert the value of 1337.58 in rearranged Equation [1.8] to get:

t(°C) = 1337.58 – 273.15 = 1064.43 °C

b) Combine Equation [1.8] and [1.10] to get:

t(°F) = 1.8(K – 273.15) + 32

Insert 1337.58 for K to get:

t(°F) = 1.8(1337.58 – 273.15) + 32 = 1947.97 °F

c) Insert 1337.58 into Equation [1.5] to get:

1.8(1337.58) = 2407.64 °R

Assignment. Calculate the temperature at which the numerical value of t(°C) = t(°F).

EXAMPLE 1.7 — Conversion Formula.

The molar heat capacity of copper is a nearly linear function of temperature over a short range. From 300 to 600 K, Cp is given by the following expression, where the upper-case T refers to the unit kelvin, and the lower-case t refers to the unit Celsius:

Cp = 22.30 + 0.00720(T) J/(mol ( K)

Calculate a formula to give Cp as a function of °C.

Solution. The T in the expression is the temperature in K, so we can use Equation [1.8]:

Cp = 22.30 + 0.00720(°C + 273.15)

Cp = 24.27 + 0.00720(t) J/(mol ( °C)

Assignment. Calculate a formula to give Cp for copper in terms of cal/(g ( K).

Pressure is defined as force per unit area (in the SI system, N/m2). The SI unit for pressure is the pascal (Pa). A pascal is a very small unit of pressure. To get an idea of the intensity of a Pa, the force created by a column of water 1 mm high is 10 Pa. Another reference point is that 1 atm is approximately 100 kPa, and 1 bar ( 100 kPa.

Pressure is similar to temperature in that it can be expressed in absolute or relative scales, with several different units. Absolute scales base pressure readings on a perfect vacuum or a completely evacuated reference point for zero pressure. Relative scales have the same units, but with the zero point being 1 standard atm (1.013 5 × 105 Pa). Conversion factors for the more common absolute pressure units are found on the inside cover, and more conversion factors can be generated with the U-Converter program on the Handbook CD.

Of all the units, pressure is probably the most confusing to deal with, partly because of the multiplicity of units and scales, and the reluctance of engineers to adopt the Pa as a pressure term. Particularly common are units of psig and psia, which refer to gage and absolute pressures in terms of lbf/in2. The height of a column of mercury in a barometer is a traditional way to measure atmospheric pressure. A column of mercury 760 mm (0.760 m) high at 0 °C exerts a pressure of 1.013 5 × 105 Pa (one atm) at its base at the standard location; this value is very slightly affected by the temperature of the mercury.

Commonly, the gage pressure is taken to mean the pressure difference between the ambient and the measured fluid, hence the need to specify the ambient pressure if the gage pressure is to have complete meaning. Flow measurements in ducts using Pitot tubes are read from a scale that may be calibrated in inches of water; this is a gage-type pressure unit. Pressures in evacuated chambers are often expressed in terms of mm Hg (1 mm Hg pressure is 1 torr), or microns of Hg. A vacuum gage is always calibrated in absolute pressure units. Figure 1.1 shows the relationship between various pressures.

Because the gage pressure is the differential pressure of the fluid in comparison to that of the atmosphere, it does not, in itself, have any thermodynamic significance. It is useful, however, in determining the absolute pressure. As illustrated in Figure 1.1, the absolute pressure equals the algebraic sum of the gage pressure and the atmospheric pressure.

[pic]

Figure 1.1 Relations of various pressures.

EXAMPLE 1.8 — Conversion of Pressure.

A vacuum processing system removes dissolved gases from steel. The dial on the thermocouple vacuum gage inserted into the vessel reads 18.3 microns Hg pressure. Calculate the absolute pressure inside the vessel in Pa if the surrounding atmospheric pressure is 100 000 Pa (1 bar).

Solution. From the context of the problem, the thermocouple gage is really indicating the absolute pressure in the system. This indicates the necessity of clarifying the meaning of pressure read by a “gage”. As mentioned earlier, a column of mercury 760 mm (0.760 m) high at 0 °C exerts a pressure of 1.013 25 ( 105 Pa at its base at the standard location. The dimension table for converting µm of Hg to Pa is:

|18.3 μm Hg |1 m |1.013 25 ( 105 Pa |

| |106 μm |0.76 m Hg |

Note that we should never use gage (relative) pressure to define vacuum conditions.

Assignment. The Weather Channel® covers the hurricane news each fall. The severity of a hurricane is given in terms of the pressure of the “eye” in millibars. What is a pressure difference of 1 millibar equivalent to in Pa and atm?

EXAMPLE 1.9 — Conversion of Pressure.

A blower produces 10 psig air pressure on a day when the barometric (ambient atmosphere) pressure is 29.03 in Hg. What is the absolute pressure produced by the blower in Pa and psia?

Solution. The key to the problem is to convert the gage and barometric pressure to the same units so they can be added together. Starting with the barometric pressure:

|29.03 in Hg |1 m |1.013 25 ( 105 Pa |

| |39.37 in |0.76 m Hg |

Next, the gage pressure:

|10 lbf |1 in2 |4.448 N |

|in2 |6.4516 ( 10-4 m2 |1 lbf |

= 6.894 ( 104 N/m2 = 68 94 kPa.

Adding the gage and atmospheric pressure gives 1.6725 ( 05 Pa. Using the conversion factor from psi to Pa (inside cover of Handbook):

|1.6725 ( 105 Pa |psia |

| |6895 Pa |

Assignment. A manometer gives a reading of 3.5 in H2O as a differential pressure between the inside of a duct and ambient. What is the pressure differential in Pa?

EXAMPLE 1.10 — Pressure in a Liquid.

Calculate the gage pressure (metallostatic head) 3 m below the surface of molten iron at 1873 K, in atm, Pa, and psig.

Data. The mass density of molten iron is a function of temperature:

ρFe = 8300 – 0.836T (kg/m3)

Solution. At 1873 K, the density of molten iron is 6730 kg/m3. The pressure at depth is the force acting on an area, so that the pressure exerted by the weight of iron (expressed as kg in the density term) must be converted to force. The force per meter of depth times depth then gives the gage pressure. The dimension equation is:

|6730 kg weight |9.807 N |3 m |

|m3 |kg weight | |

Using the same conversion factor as was used in Example 1.9, we obtain 28.7 psig. The pressure in atm is obtained my dividing the pressure in Pa by 1.013 ( 105, or 1.95 atm.

Assignment. Calculate the depth of water at which the gage pressure is 1 atm.

1.4 Unit Conversions Using the U-Converter Program

Conversion factors and equations are most convenient where only a few conversions are necessary. However, when many conversions are required, or conversions are required for lesser-used units, it is helpful to have a computer program to make the calculations. An Excel program called U-Converter (U-C for short) has been developed for making all sorts of unit conversions. U-C uses values from NIST (Thompson 2008) as described earlier in this chapter. U-C is on the Handbook CD in the folder Unit Conversions. U-Converter is an Excel add-in (U-C.xla) which can be opened in any Excel worksheet. More information on U-C and a sample calculation is given in Appendix A. The U-C User’s Guide (U-C.rtf) explains how to use the program.

1.5 Amount of Substance — the Mole Unit

A material balance requires clear and complete specification of the quantity of every substance present. Quantity can be expressed in mass and amount units. The mole is the SI unit for “amount of substance”. The mole unit specifies the amount of substance in a given mass. The CGPM established the following definition of the mole (abbreviation mol, symbol: n, also v):

1. The mole is the amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kg of carbon 12.

2. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or other specified groups of such particles.

This definition requires, at the same time, specification of the nature of the quantity whose unit is the mole. In special cases, SI accepts the unified atomic mass unit (symbol μ) which is equal to 1/12 of the mass of an atom of the nuclide 12C; 1 μ = 1.660 540 2 ( 10–27 kg. The only allowed name is “unified atomic mass unit”, and the only allowed symbol is μ. Terms such as “a.m.u” are no longer acceptable. Similarly, the terms “atomic weight” and “molecular weight” are obsolete, and have been replaced by the equivalent but preferred terms “relative atomic mass”, symbol Ar, and relative molecular mass, symbol Mr. Like atomic weight and molecular weight, relative atomic mass and relative molecular mass are quantities of dimension one, and are expressed simply as numbers.

Examples: Ar (Si) = 28.0855. Mr (H2) = 2.0159 Ar (12C) = 12 (exactly)

The term “molar mass” is used to describe the amount of a substance. For example, the relative atomic mass of a fluorine atom is Ar (F) = 18.9984. The relative molecular mass of a fluorine molecule is Mr (F2) = 2Ar(F) = 37.9968. The molar mass of F2 is then 37.9968 ( 10–3 kg/mol = 37.9968 g/mol. The amount of substance of 0.100 kg of F2 is then n(F2) = 100 g/(37.9968 g/mol) = 2.63 moles. The preferred molar mass unit (symbol M) is kg/mol, but g/mol (and its equivalent, kg/kmol) are acceptable. The number of atoms in 0.012 kg of 12C is given a special name: Avogadro’s Number (NA), with a value of 6.022 136 7 ( 1023.

Terms such as gram-mole (symbol g-mol) and kilogram-mole (symbol kmol, or sometimes kg-mol) are also obsolete, but since they are still in common use, we will use them in this book. The AES amount of substance unit is the pound-mole (symbol lb-mol). These units are defined as the amount required to have a mass (in grams, kilograms, or pounds) numerically equal to the mass per atom (molecule) in Ar or Mr.

. A convention used in some thermodynamic tables is the gram formula weight (symbol gfw) to designate the atomic or molecular mass in grams. For example, the gfw for F is 18.9984, while the gfw for F2 is 37.9968.

Values of Ar for 95 elements are listed on the inside cover of the Handbook (Coursey 2005). We recall that the mole is a defined quantity, based on the mass of 12C. However, when we examine the table of relative atomic mass on the inside cover of the Handbook, the value listed for carbon is not 12, it is 12.0108. The relative atomic mass listed for an element is for the naturally occurring entities, and naturally occurring carbon is composed of more than one isotope of carbon. Although natural carbon is mainly 12C, it has about 1.11% 13C and a trace of 14C. Even though naturally occurring carbon does not contain a single atom with mass 12.0108 g, for stoichiometric purposes we consider carbon to consist of only one type of atom with an atomic mass of 12.0108 g. We do this so we can make a material balance on systems containing natural carbon in both mass or amount-of-substance units, and convert accurately between the different units.

Another thing to note on the listing of Ar is that some elements (such as antimony and iodine) have their values listed to seven significant figures, while other elements have values listed to only four or five significant figures. The difference comes from the variation in isotope concentration in an element from one raw material to another. An element extracted from seawater may have a different isotopic concentration than if extracted from bedrock. The higher the variation in isotope concentration, the less accurate is the listing of a single number for the relative atomic mass of the naturally occurring species. For the vast majority of material balance calculations, using relative atomic mass expressed to five (or sometimes even four) significant figures is acceptable (see Section 1.8 for discussion on significant figures).

EXAMPLE 1.11 — The SI and AES Mole.

How many moles of a substance are there in one lb-mol?

Solution. Let X be the molar mass of any substance. The units of X are g/mol, and lbm/lb-mol. The dimension table is:

|X lbm |453.5924 g |1 g-mol |

|1 lb-mol |1 lbm |X g |

Canceling the units and X gives a conversion factor of 453.5924 g-mol per lb-mol. The mass of a g-mol is then 0.0022046 times the mass of a lb-mol.

Assignment. Calculate the number of atoms in 1 lbm of ozone.

For engineering calculations, the conversion factor for mass between a g-mol and a lb-mol is often approximated to 454. In other words, there are 454 g-mol in a lb-mol.

1.6 Density and Concentration

The mole unit specifies the amount of substance in a given mass. However, it is often necessary to know how much material is contained in a given volume, or the concentration of one substance in a mixture of several substances. The SI system specifies the way these quantities should be expressed, as shown in Figure 1.2. These units require some additional description.

The terminology listed in Figure 1.2 is very important, and (along with the symbols) will be used throughout this Handbook. A copy of Figure 1.2 is reproduced inside the back cover.

1.6.1 Density

The mass density of a substance is its mass per unit volume, with units of kg/m3, although g/cm3 is acceptable, and is the unit most often cited in reference books. Weight units are often indiscriminately substituted for mass in calculating the density. Sometimes the density is expressed without units, and is called the specific gravity. This is equal to the ratio of the density of the material to the density of a reference material at the temperature of interest. Water is most frequently used as the reference material for the specific gravity of solids or liquids. For most calculations, below 45 °C water has a density of 1.00 Mg/m3, or 1.00 g/cm3. For gases, air is commonly used as the reference material. The mass density of dry air is adequately given by:

[pic] [1.11]

where t = degrees Celsius, and P is in atmospheres.*

[pic]

Figure 1.2 Summary descriptions of nine quantities that are quotients involving amount of substance, volume, and mass. The volume fraction φB refers to substance B in a mixture of A, B, C, etc, and V*m refers to the molar volume of the pure substance.

The specific volume (v) may also be reported instead of the mass density. This is the volume per unit mass of a substance, and has the units of m3/kg. The specific volume is obviously the reciprocal of the density. A similar unit, the molar volume (Vm), is expressed as m3/mol.

EXAMPLE 1.12 — The Density of a Slurry.

A slurry is a solid-liquid mixture, where the solid phase is usually made up of fine particles. A fluid of high apparent density can be created by suspending an extremely fine dense solid in water. How many kg of magnetite (Fe3O4) should be suspended in 1000 kg of water to give slurries of density between 1100 and 1300 kg/m3?

Data. The density of magnetite is 5180 kg/m3.

Solution. This problem asks for a range of results, which indicates that we need several results over the specified range. A spreadsheet is the easiest way to make multiple calculations. First, we’ll do one calculation by hand to outline the solution.

Suppose we mix 100 kg of magnetite with 1000 kg (1 Mg) of water. The resulting slurry will weigh 1100 kg, and have a volume that is the sum of the volume of the water plus the magnetite. The volume of the water is very close to 1.00 m3. The specific volume of magnetite is 1.93 ( 10-4 m3/kg, so 100 kg of magnetite will occupy 1.93 ( 10-2 m3. The total volume of slurry will then be 1.0193 m3. The density of the slurry is:

ρslurry = 1100/1.0193 = 1079 kg/m3

The above procedure was used to calculate the density of several slurry compositions. A chart was prepared showing the density and wFe3O4 vs. the selected input variable. The results are shown in Table 1.5 and Figure 1.3.

Table 1.5 Results of Excel calculations on slurry density vs. kg Fe3O4/Mg water.

|kg Fe3O4 |m3 water |kg slurry |m3 Fe3O4 |m3 slurry |kg/m3 slurry |wFe3O4 |

|150 |1 |1150 |0.0290 |1.0290 |1118 |13.0% |

|200 |1 |1200 |0.0386 |1.0386 |1155 |16.7% |

|250 |1 |1250 |0.0483 |1.0483 |1192 |20.0% |

|300 |1 |1300 |0.0579 |1.0579 |1229 |23.1% |

|350 |1 |1350 |0.0676 |1.0676 |1265 |25.9% |

|400 |1 |1400 |0.0772 |1.0772 |1300 |28.6% |

An additional column of data was calculated to represent the wFe3O4 in the slurry. This was obtained by dividing the kg of Fe3O4 (first column) by the kg of slurry (third column), and expressing the result in mass %. The two Y variables plotted on Figure 1.3 appear at first glance to be linear over the range of X values, but a closer inspection indicates the relationships are slightly curved. If they were linear, the intervals of values in the density and wFe3O4 columns would be constant. Nevertheless, for many engineering applications, and considering the slightly variable density of magnetite*, assuming a linear relationship may be adequate.

Assignment. Calculate the amount of magnetite to add to 1000 kg of water to obtain a slurry density of 1210 kg/m3. Then, calculate the amount of water to add to the slurry to decrease the density to 1170 kg/m3.

[pic]

Figure 1.3 Chart showing relationship between input variable of kg Fe3O4/tonne water, slurry density, and wFe3O4 in the slurry.

The actual volume occupied by a granular solid is always greater than its theoretical density would predict. In the above example, 100 kg of magnetite was calculated to occupy 0.0193 m3, but dry magnetite in a hopper would occupy much more space. The term bulk density is used to describe the actual amount of space occupied by granular solids. The bulk density depends on the amount of void space in a container of the solid. The mathematical relationship is:

ρbulk = (1 – ω) ρtheoretical

where ω is the void space. The void space depends on the size and shape of the pieces of material, and how tightly they were packed or pushed together. The void space of bulk material is a minimum when particles of a range of sizes are present because the smaller particles tend to fill in the voids between the larger ones.

EXAMPLE 1.13 — Bulk Density of a Solid.

Dry magnetite has an as-received bulk density of 60% of theoretical. Calculate the required storage space for 150 tonnes (150 Mg) of magnetite.

Solution. The bulk density of magnetite is 0.6(5180) = 3108 kg/m3. The dimension table is:

|1 m3 |150 000 kg Fe3O4 |

|3108 kg Fe3O4 | |

This gives a bulk volume requirement of 48.3 m3.

Assignment. A container with a volume of 1.00 m3 is filled with dry (bulk) magnetite. Calculate the total mass in the container if the temperature is 25 ºC and the pressure is 1.00 atm. Hint: the void space is not empty.

1.6.2 Composition and Concentration

In most operations, the feed and product materials in process streams are not pure but rather mixtures or solutions of two or more substances. The concentration of components is based on mass or amount of substance per unit volume. Figure 1.2 specifies these as mass density (kg/m3), and amount-of-substance concentration (moles per m3).

The composition of components is defined as a part-per-part dimension. Figure 1.2 specifies these as mass fraction w (kilograms of component B divided by the total kilograms), amount-of-substance fraction x (moles of component B divided by the total number of moles),* molality b (moles of solute B in solution divided by the mass of the solvent in kg), or volume fraction φ (m3 of component B divided by the total m3). Terms such as normality, molarity, weight fraction or part per million (ppm) are no longer acceptable, despite their extensive use in process technology. Although b is specified by SI to designate molality, most chemistry books use m This can cause an unfortunate confusion with mass.

The difference between concentration and composition can be illustrated for example by the terms used to describe the properties of a dusty gas. The dusty gas concentration might be given in units such as grains of dust per cubic foot of gas (gr/ft3) or grams of dust per cubic meter of gas (g/m3). However, the dust composition would be expressed as mass fraction of its constituents, and the gas composition as volume fraction of its constituents.

The symbol % (percent) is an acceptable and recognized way to express composition. Because the symbol % represents a number, one should avoid attaching information to it. The preferred forms to use are, “the mass (or volume) fraction of B is 0.10” or “the mass (or volume) fraction of B is 10 %”, or “wB = 10 %”. Thus terms like “% (by volume)”, “% (by weight)”, or “at. %” are obsolete for SI use, but are still often encountered.

Conversion between amount-of-substance fraction (mole fraction) of B (xB) and mass fraction of B (wB) can be accomplished by using the following equations:

[pic] [1.12]

[pic] [1.13]

where M is the molar mass of a substance (see Figure 1.2 for definition of units). You may wish to copy these two equation to the inside cover for handy use.

EXAMPLE 1.14 — Mass and Mole Fraction.

Calculate the mass of chromium, nickel, and iron required to produce 1 kg of an alloy containing xCr = 20 %, xNi = 10 %.

Solution. This requires application of Equation [1.13] using the atomic mass values printed inside the front cover of this Handbook. The denominator of Equation [1.13] is given by:

xNi ( 58.69 + xCr ( 52.00 + xFe ( 55.85 = 55.36

The individual w values are calculated from Equation [1.13] as follows:

wNi = (0.1 ( 58.69)/55.36 = 0.106

wCr = (0.2 ( 52.00)/55.36 = 0.188

wFe = (0.7 ( 55.85)/55.36 = 0.706

Thus for 1 kg (1000 g) of alloy, the required amounts are 106 g of Ni, 188 g of Cr, and 706 g of Fe.

Assignment. A sample of cast iron was analyzed and found to contain wC = 3.81 % and wSi = 0.90 %. Calculate the xC and the xSi, in % units.

EXAMPLE 1.15 — Concentration Conversions.

Aqueous NiSO4 solutions are being prepared. Calculate the molality, grams NiSO4/L of solution, and amount of substance fraction as a function of the mass fraction NiSO4.

Data. The density of NiSO4/water solutions vs. wNiSO4 at 25 °C in units of kg/m3 solution is.

|% NiSO4 |1.00 |2.00 |4.00 |8.00 |12.00 |16.00 |18.00 |

|density |1009 |1020 |1042 |1085 |1133 |1183 |1209 |

Solution. A convenient basis for the calculations is 100 g of solution with the density expressed in units of g/L, which has the same value as kg/m3. Owing to the many calculations involved in solving this example, a spreadsheet solution will be used. A sample calculation for the 1% solution is given below.

The molality is expressed as moles of NiSO4/kg of water. The molar mass of NiSO4 is 58.69 + 32.07 + 4(16.00) = 154.76. The molality (expressed here as b) is then:

Molality of NiSO4 = (1/154.76)(1000/99) = 0.06527 b

The amount-of-substance fraction is given by:

[pic]

The concentration in grams NiSO4/liter of solution is obtained from the dimension table as follows, where 1 liter = 1009 g as given by the density.

|1 g NiSO4 |1009 g solution |

|100 g solution |1 liter |

= 10.09 g/L.

The results for all concentrations are shown in the following table.

|grams NiSO4 |1.00 |2.00 |4.00 |8.00 |12.00 |16.00 |18.00 |

|grams H2O |99.00 |98.00 |96.00 |92.00 |88.00 |84.00 |82.00 |

|moles NiSO4 |0.00646 |0.01292 |0.02585 |0.05169 |0.07754 |0.1034 |0.1163 |

|moles H2O |5.495 |5.440 |5.329 |5.107 |4.885 |4.663 |4.552 |

|bNiSO4 |0.06527 |0.1319 |0.2692 |0.5619 |0.8812 |1.231 |1.418 |

|xNiSO4 |0.001175 |0.002370 |0.004827 |0.01002 |0.01563 |0.02169 |0.02492 |

|g NiSO4/L |10.09 |20.40 |41.68 |86.80 |135.96 |189.28 |217.62 |

Assignment. 100 grams of water and 50 grams of nickel sulfate heptahydrate (NiSO4•7H2O) are mixed to prepare an aqueous NiSO4 solution. Calculate the composition and concentration of Ni in the solution. Estimate the density of the solution from a plot of the density data given in the first part of this Example.

1.6.3 Composition of Gases

The composition of gases is most frequently given as amount-of-substance fraction or volume fraction. However, as will be seen in Chapter 2, the volume fraction is virtually identical to the amount-of-substance fraction at most materials processing temperatures and pressures. Therefore, unless otherwise stated:

xB = VB/(ΣVA, VB, VC . . .) [1.14]

The molar volume of a gas has no meaning relative to its mass unless the temperature and pressure are specified. A common convention for standard temperature and pressure (STP) is 1 atm (101 325 Pa) and 0 °C (273.15 K), although other conventions exist (Wikipedia, 2006). This Handbook uses 1 atm and 0 °C for STP. The molar volume at STP of any ideal gas is 22.414 L, or 359.04 ft3 for a lb-mole.* The STP molar volume is often rounded off to 22.4 L and 359 ft3.

When dealing repeatedly with a certain mixture of gases, the mixture’s molar mass is a useful number. For example, to a very good approximation, CO2-free dry air* has the following composition (where φ refers to volume fraction):

φN2 = 0.7812; φO2 = 0.2096; φAr = 0.0092

The molar mass M of air is then given by:

M for dry air = 28.01(0.7812) + 32.00(0.2096) + 39.95(0.0092 = 28.96

The molar mass of dry air is thus 28.96 g/g-mol (and 28.96 lb/lb-mol). The mass density of dry air at STP is 1.292 kg/m3, or 0.08066 lb/ft3 (see Equation [1.11] for non-STP conditions). One kg of air occupies 774.0 L at STP, while one lbm of air occupies 12.40 ft3. Where the above accuracy is not required, the volumetric composition of air is often approximated as 79 % N2, 21 % O2, and M is approximated as 29. In that case, one mole of air can be given an empirical formula N1.58O0.42.

Often the sampling or analytical procedure has difficulty in dealing with the water vapor content of gases. During sampling, some of the water vapor may condense as the gas cools. Even if the sample is kept warm to prevent condensation, the analytical procedure may not be set up to analyze for water vapor. In such cases, the gas is usually dried before analysis, so that the composition is given on a dry basis. It is very important to know the sampling and analytical basis if a correct materials and heat balance is to be made.

The moisture content of a gas is determined in different ways. Some of these are:

• Moisture is prevented from condensing, and either all species or all but inert species are analyzed.

• The gas is dried and the amount of moisture removed is determined. The rest of the gas analysis is on a dry basis.

• The gas is dried but no attempt is made to determine moisture content. The dry gas is analyzed.

Some of the units used to express the moisture content include the grains of moisture per ft3, grams per m3, and volume fraction, expressed as %. Always take care to determine if the moisture content is expressed on a dry or wet basis (i.e., mass of moisture per volume of dry or wet gas, or mass of moisture per mass of dry or wet gas).

EXAMPLE 1.16 — Composition of a Gas on a Wet and Dry Basis.

A gas sample was extracted from an incinerator duct and analyzed without allowing any moisture to condense. The gas analysis was:

φN2 = 70.0 %; φCO2 = 10.0 %; φCO = 5.5 %; φH2 = 4.5 %; φH2O = 10.0 %

Calculate the volume fraction composition of the gas on a) a dry basis; b) the grains of moisture per ft3 (STP) of wet and dry gas; and c) the mass fraction of each gas on a wet basis.

Solution. a) Since we are free to pick any basis, 100 moles of gas is convenient because the % values are equivalent to the number of moles of each gas. If the gas is dried, it will lose 10 moles of water vapor, and the dry gas will consist of 90 moles. The composition of the dry gas will be:

φN2 = 70.0/90 = 77.8 %; φCO2 = 10.0/90 = 11.1 %; φCO = 5.5/90 = 6.1 %; φH2 = 4.5/90 = 5.0 %.

b) The removal of 10 moles of H2O means the removal of 180.16 grams of water. At STP, the molar volume is 22.4 L, hence the dry volume of the gas is:

V = Vm(n) = 22.4(90) = 2016 L

Conversion of the mass of water removed to grains gives 180.16/0.0648 = 2780 grains of water removed. Conversion of the original and dry gas amounts to ft3 gives 2240/28.32 = 79.1 ft3 wet gas, and 2016/28.32 = 71.2 ft3 dry gas. The moisture content is then:

Wet gas = 2780/79.1 = 35.1 grains/ft3

Dry gas = 2780/71.2 = 39.0 grains/ft3

c) The mass of each component of the wet gas is given by application of Equation [1.13]. For N2, the relationship is:

[pic]

Each of the other gases is treated in the same way. The results are:

wCO2 = 0.160; wCO = 0.056; wH2 = 0.0033; wH2O = 0.0656

Assignment. A gas has the following analysis:

φN2 = 80.0 %; φO2 = 15.0 %; φH2O = 5.0 %

A volume of 10 L (STP) was taken for sampling. During sampling, the gas cooled, and some of the water vapor condensed. A gas analysis for N2 and O2 gave the following results:

φN2 = 82.5 %; φO2 = 15.5 %

Are these results consistent with the original analysis? If so, how many grams of water condensed during sampling?

1.7 Electrical Units

The basic SI electrical unit is the ampere (symbol A), which is defined as follows:

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between those conductors a force equal to 2 ( 10-7 newton per meter of length.

Current can be passed by the flow of either electrons or ions. In solids, current almost always consists of electron flow. In electrolyte solutions, most of the current flows by migration of ionic species (e.g., Cu+, OH–, etc.). The SI unit of charge is the coulomb (symbol C), which is defined as a current of 1 ampere flowing for 1 second. The SI unit of electrical potential is the volt (symbol V), which can be defined in various ways (for example, 1 V = 1 N ( m/C). Another term for potential is electromotive force, sometimes abbreviated emf. Note that current only flows across a potential difference.

The SI unit of resistance is the ohm (symbol Ω), which is defined as the resistance that permits the flow of 1 ampere under a potential difference of 1 V. Electrical energy is defined by the joule, and power (the rate of energy produced or consumed) by the watt (1 W = 1 J/s). Electrical energy is often expressed in watt-hours (W · h) or kilowatt-hours (kW · h), which is commonly abbreviated as kWh. Please see Table 1.3 for a review of these units.

Resistance is related to resistivity (specific resistance) by the relation that resistivity is resistance multiplied by the area of the conductor, and divided by the length of the conductor. The dimension of resistivity is ohm · meter (not to be confused with an ohmmeter, which is a device for measuring resistance).

Some basic equations of electrical flow are:

V = Ω ( A [1.15]

W = A2 ( Ω [1.16]

J = A ( V ( s [1.17]

An electrical quantity that is of importance in electrometallurgy is the faraday, which is the magnitude of electric charge per mole of electrons (i.e., NA of electrons). Recall that we defined the mole in terms of the mass (0.012 kg) of one mole of 12C. One faraday is 96 485 coulombs. One faraday will discharge 1 mole of univalent ions (or ½ mole of divalent ions, etc.).

EXAMPLE 1.17 — Electrical Flow in a Wire.

A potential difference of 2.0 V is applied along a circular bar of 0.10 m radius and 2.0 m long. The bar material has a resistivity of 1.7 ( 10–6 Ω ( m. Calculate the current flow.

Solution. The bar area is π(0.1)2 = 0.0314 m2. Therefore, the resistance is

2(1.7 ( 10–6)/0.0314 = 1.08 ( 10–4 ohms.

The current is given by application of Equation [1.15]:

A = V/Ω = 2/1.08 ( 10–4 = 18 500 amperes.

Assignment. Calculate the power involved in the process.

EXAMPLE 1.18 — Electrical Energy for Metal Deposition.

Calculate the mass of Al+3 ions discharged in 1.0 minute by the current in the above example, if no current is lost to leaks or other processes (i.e., 100% current efficiency).

Solution. In one minute, a current of 18 500 amperes will carry 60(18 500) = 111 000 coulombs. The number of faradays is given by:

111 000/96 485 = 1.15

The aluminum ion is trivalent, so one faraday will discharge ⅓ mole of aluminum. The mass is then 1.15(26.98)/3 = 10.3 grams of aluminum.

Assignment. Calculate how many kWh are required to produce 1.00 kg of aluminum from Al+3, assuming 100% current efficiency.

1.8 Calculation Guidelines

The calculations so far have been relatively simple, but later sections of this Handbook will deal with processes that are more complex. It’s a good idea to develop ways to check if the results are reasonable as you go along, to avoid carrying errors through the problem. There are three ways you can do this.

• Substitute your answer back into the equations to see if you have solved them correctly.

• Replace the numbers you used with integers of the same order of magnitude, and solve the equations “in your head”. This identifies gross errors involving wrong exponents, etc.

• Check your answer for reasonableness. With experience, you will develop a feeling for the magnitude of certain dimensions, such as volume, pressure, temperature, enthalpy, etc. If the volume of a substance produced comes out as millions of m3, or the degrees K is below 200 or above 3000, you should be suspicious. Enormous errors can be caused by using the number instead of its logarithm, dividing by a tiny number instead of multiplying it, or switching signs.

Another important guideline involves tracking the significant figures in your calculation, and expressing the answer properly. The significant figures of a number are the numbers from the first non-zero digit on the left to either the last digit on the right (zero or non-zero) if there is a decimal point or the last non-zero digit if there is no decimal point. For example, the following four numbers all have four significant figures:

|1425 |142 500 |1420 |1.420 × 103 |

The number of significant figures is an indication of the precision with which the quantity is known. Your calculator or computer will carry and may display many more significant figures than the final answer justifies, so it’s important to express your answer with the correct number of figures. There are two general rules to follow:

1. In multiplication and division, round off your final answer to the lowest number of significant figures of the multiplicands or divisors.

2. In addition and subtraction, one should compare the last significant figures of each number relative to the decimal point. Of these positions, the one farthest to the left is the position of the last permissible significant figure.

The above rules don’t apply in two cases. First, when conversion factors are exact (like the 1.8 ratio between kelvin and Rankin degrees). Second, when an exact count (instead of a measurement) of something is stated, like 10 ingots of silicon. In each case, there are an infinite number of significant figures.

The above discussion is relevant to rounding off significant figures in your final answer. In carrying out successive calculations, it’s a good idea to avoid rounding off intermediate answers. This logic may justify keeping an extra significant figure in the “final” answer if the value may be used in subsequent calculations. These simple rules don’t cover all cases; specific guidance is available (Taylor 2008; Thompson 2008). Rounding off calculated results from uncertain data will be discussed in more detail in Chapter 3.*

EXAMPLE 1.19 — Number of Significant Figures.

Calculate the answer to the following problems, expressed to the correct number of significant figures. a) A temperature of 356.4 °F increases by 8°. What is the temperature in ºC? b) Calculate the mass of silver that would be deposited from a solution of AgNO3 by the passage of a current of 1.237 A for 22.7 minutes?

Solution. a) The addition of 8° gives 364.4 °F as an intermediate value. This is converted to °C by Equation [1.10]:

°C = (364.4 – 32)/1.8 = 184.7°C.

However, the 8° increase in temperature contains only 1 significant figure before the decimal point, hence the answer should be rounded off to 185 °C.

b) The quantity of electricity is given as:

C = 1.237(60)(22.7) = 1684.8 coulombs

The number of faradays is given by:

1684.8/96 485 = 0.017 462 faradays.

One faraday will discharge 1 mole of silver. The mass of silver discharged is then 107.87(0.017 462) = 1.8836 grams. The time is the unit with the least number of significant figures (three), so the final answer is 1.88 grams of silver deposited.

Assignment. Hydrogen gas is produced by electrolysis of acidified water at a cell voltage of 1.80. Calculate the energy required to produce 1.000 kg of hydrogen.

When using Excel, the value displayed in the cell may appear to have only 4 or 5 significant figures (depending on the width of the cell), but the actual value stored in Excel may have many more than that. Even if you use the “number” function to format the cell contents to specify the number of significant figures to display, Excel still uses the actual value. If you want to round off the number to display (and for Excel to use), you should use Excel’s ROUND function on the values. Please refer to Excel’s on-line help for entering and using functions.

A significant-figure quirk is that a number like 985 has nearly the same number of significant figures as a number like 1025 because an error of 1 in the last digit of both numbers is about the same percent error in the number.

1.9 Summary

The SI provides an internally consistent set of units and dimensions that has been universally adopted by all countries in the world, although the acceptance is not yet complete in engineering work in the US. The SI is based on a well-defined set of fundamental units, with additional or derived units made up of simple combinations of the fundamental units. For practical and historic reasons, some non-SI units are accepted for use in conjunction with the SI. Certain NIST Special Publications are recommended for a complete guide to the SI and for conversion factors between SI and other systems. These should be consulted to not only assure adherence to SI standards, but also for rules and style conventions for spelling and using SI in documents.

This Handbook will emphasize the use of SI. For illustrative purposes, some problems will be stated and worked with less-preferred metric units or AES units. Conversion factors for cases of mixed units are found on the inside cover, as derived from NIST values. A unit conversion program (U-Converter) is on the Handbook CD for making units conversion in Excel, and calculating conversion factors not found in these sources. Where more than one conversion factor is required, the dimension table technique should be used to develop a new conversion factor, or to make the conversion. For temperature and pressure, a conversion equation may be required because of the different size of the units and the different zero reference points.

References and Further Reading

Antoine, Valerie, Guide to the Use of the Metric System [SI Version], U. S. Metric Association, 15th Edition, 2001.

Butcher, Kenneth et. al., The International System of Units (SI) - Conversion Factors for General Use. NIST Special Publication 1038 [Online] Available: . May 2006.

Coursey, J.S., Schwab, D.J., and Dragoset, R.A. Atomic Weights and Isotopic Compositions (version 2.4.1) [Online]. Available: , May 2005.

Thompson, Ambler, and Taylor, Barry N., Guide for the Use of the International System of Units (SI), NIST Special Publication 811, 2008 Edition (version 3.0). [Online] Available: . April 2009.

Taylor, Barry N., and Thompson, Ambler, Eds., The International System of Units (SI), NIST Special Publication 330, [Online] Available: . March 2008.

U. S. Environmental Protection Agency on-line course, Basic Concepts in Environmental Sciences, Module 1: Basic Concepts. . December 2006.

Wikipedia contributors, “International System of Units”, “Conversion of units”, “Units conversion by factor-label”, “Pressure”, “Mole (unit)”, “Standard conditions for temperature and pressure”, “Density”, “Relative density”, “Bulk density”, “Concentration”, “Significance arithmetic”. Wikipedia, The Free Encyclopedia, December 2009.

.

Exercises

1.1 Convert 400 in3/d to cm3/min

1.2 A bucket contains 2 lb of NaOH.

a) How many atoms does it contain?

b) What is the quantity of NaOH in units of kmol and lb-mol?

1.3 The precise value of the gas constant R in SI is 8.314 471 J/mol ( K. Use U-Converter to obtain R in L ( atm/(mol ( K) and kcal/(mol ( K).

1.4 An alloy is prepared by melting together 10 lb of Cu and 15 lb of Zn. What is the mass fraction and amount-of-substance fraction, expressed in %?

1.5 Calculate the mass fraction of H2S if 1.00 mole of H2S is added to 16.6 mole of dry air.

1.6 Gasoline contaminated by water is pumped out of a reservoir into a holding tank. The tank was found to have a layer of water on the bottom 0.15 m thick, and a gasoline layer 0.75 m thick. Assume water and gasoline are immiscible, and the gasoline has a specific gravity of 0.75. Calculate the mass fraction of water in the mixed fluids.

1.7 Limestone flux is added to a wastewater neutralization system in 55-gallon drums. The operation uses a guideline of 1 ton = 1 drum. The density of CaCO3 is 2900 kg/m3. What is the assumed void space in the bulk limestone?

1.8 Convert the absolute pressure 1.00 ( 10–8 atm to pressure on the mm Hg and torr scale. Check your answer with U-Converter.

1.9 Calculate the kinetic energy of a kilogram of water moving at 66 kilometers/hour in units of joules, watt-seconds, and liter-atmospheres.

1.10 The emissive power of a black body depends on the fourth power of temperature, and is given by:

W = AT4

where W is the emissive power in Btu/(ft2 ( h), A is the Stefan-Boltzmann constant, = 0.1714 ( 10–8 Btu/(ft2 ( h ( R4). Calculate the value of A in:

a) cal ( sec–1 ( cm–2 ( ºC–4.

b) cal ( sec–1 ( cm–2 ( K–4.

c) J ( sec–1 ( m–2 ( K–4.

1.11 For the first flight of the day, an airplane is normally fueled with 20 000 gallons of jet fuel. The plane consumes 4000 gallons per hour in flight. By error, one day the plane is mistakenly filled with 20 000 kg of fuel. How long can the plane fly before running out of fuel? The specific gravity of jet fuel is 0.95.

1.12 The density of various fluids is measured by inserting a glass tube containing some lead shot. The tube is 10 cm long, and in water, sinks to a point where 2.0 cm of tube is above the surface. In a pool of jet fuel, 2.2 cm extends above the surface. Calculate the specific gravity of the jet fuel. Make a sketch of the tube to scale, and mark on it the indicator marks of specific gravity at 0.1 unit intervals.

1.13 Stainless steel is made by melting 2000 lb of iron, 506 lb of chromium, 225 lb of nickel, and 85 lb of molybdenum. The iron was not pure, but contained 0.5 %C. Calculate the mass fraction (%) and amount-of-substance fraction of carbon in the stainless steel.

1.14 Convert 150 ºC to degrees kelvin, Fahrenheit, and Rankin.

1.15 Fill in the following table.

| |Mass, kg |Mass fraction, % |Vol. fraction, % |Amount-of-substance fraction |

|CO2 |12.0 | | | |

|O2 |8.0 | | | |

|N2 |75.0 | | | |

|H2O |5.0 | | | |

1.16 Fill in the following table.

| |Mass fraction |Amount-of-substance fraction |Vol. fraction, % |

|CaCO3 |22.0 % | | |

|MgCO3 |18.0 % | | |

|CaO |23.0 % | | |

|NaCl |30.0 % | | |

|K2SO4 |7.0 % | | |

1.17 The thermal conductivity of aluminum at 32 ºF is 117 Btu ( h–1 ( ft–1 ( ºF–1. Use U-Converter to find the equivalent value in terms of cal ( sec–1 ( cm–1 ( ºC–1.

1.18 A column of water 30 cm high and 20 mm in diameter is in a cylinder supported by a piston. Calculate the force in N required to prevent the piston from moving, and calculate the pressure on the piston face in mm Hg and psi.

1.19 A government surplus auction is being held to dispose of some scrap titanium. A prospective bidder examined the scrap, which was in the form of a conical pile of broken pieces. The circumference of the pile was estimated to be 18 m, and the angle of the pile surface was 30º from the horizontal. The bulk density was estimated to be 60 %. Estimate the mass of the titanium.

1.20 The Reynolds number is a dimensionless number defined for a fluid flowing in a pipe as

Re = Duρ/µ

where D is pipe diameter, u is fluid velocity, ρ is fluid density, and µ is fluid viscosity. When the value of Re is less than about 2100, the flow is laminar, and if above 2100, the flow is turbulent.

Sulfuric acid flows through a pipe with an inner diameter of 3.067 inches, and an average velocity of 0.52 ft/s. At the fluid temperature of 25 ºC, the density of H2SO4 is 1826 kg/m3, the viscosity is 19 centipoises [1 cP = 1 ( 10–3 kg/(m ( s)]. Without using your calculator (relying on integers and orders-of-magnitude), determine if the flow is laminar or turbulent.

1.21 An instrument has been developed to measure the mass flow rate of a gas, and is calibrated in kg/min. Another instrument has been developed to measure the volumetric flow rate, and display it in L/s. These two instruments are to be used to measure the moisture content of flowing air, in terms of mg/L. Assume that dry air has a volume fraction of 78 % N2, 21 % O2, and 1 % Ar. Set up a spreadsheet solution to calculate the moisture content from a number derived from the readings of these instruments, and plot a graph from 0 to 100 mg/L. If the desired precision of the moisture content reading is 2 significant figures, how many significant figures are required from the instrument readings?

1.22 A balloon of volume 1.00 m3 has a mass of 0.50 g, and is filled with helium. Helium has a specific volume of 6.10 m3/kg, and air has a specific volume of 0.861 m3/kg. Calculate the mass capable of being lifted by the balloon.

1.23 Calculate a conversion factor to convert the heat of combustion of a gaseous fuel from units of calories per mole to Btu per standard cubic foot.

1.24 The composition of dry air was listed in a recent reference handbook as follows:

|gas: |N2 |O2 |Ar |CO2 |

|φ, %: |78.10 |20.95 |0.92 |0.03 |

Calculate the mass fraction of each constituent, and the mass fraction of each element. Calculate the molar mass of dry air, and compare to the value given in Section 1.6.3.

* In temperature work of extreme precision, the International Practical Temperature Scale must be recognized because almost all thermodynamic data is based upon it. The IPT scale is based on certain “defining fixed points” that are used in calibration of instruments.

* The symbol ρ will be used for den楳祴愠摮猠数楣楦⁣牧癡瑩⹹†桔⁥敭湡湩⁧楷汬戠⁥慭敤挠敬牡戠⁹灳捥晩楹杮琠敨甠楮獴椠⁦敤獮瑩⁹獩洠慥瑮മ‪慎畴慲汬⁹捯畣牲湩⁧業敮慲獬愠敲猠汥潤異敲‬湡⁤敨据⁥桴⁥敤獮瑩⁹景洠条敮楴整挠畯摬瘠牡⁹牦浯㔠ㄮ琠⸵′⽧浣⸲⨍䤠䥓‬桴⁥浡畯瑮漭ⵦ畳獢慴据⁥牦捡楴湯琠牥敲汰捡獥琠敨洠汯⁥牦捡楴湯琠牥⹭†楓据⁥潭敬映慲瑣潩獩猠楴汬椠潣浭湯甠敳‬桴獩䠠湡扤潯楷汬甠敳戠瑯⁨整浲⹳⨍吠敨挠湯散瑰愠摮愠灰楬慣楴湯漠sity and specific gravity. The meaning will be made clear by specifying the units if density is meant.

* Naturally occurring minerals are seldom pure, and hence the density of magnetite could vary from 5.1 to 5.2 g/cm2.

* In SI, the amount-of-substance fraction term replaces the mole fraction term. Since mole fraction is still in common use, this Handbook will use both terms.

* The concept and application of the ideal gas law will be discussed in detail in Chapter 2.

* Air also contains φCO2 ( 0.0003.

* In text examples, an additional significant figure is often displayed in case you want to duplicate the solution.

-----------------------

P = 1.98 × 105 Pa

= 24.3 psia

= 98.31 kPa

= 2.44 Pa

Header should have centered text here that says: Acknowledgements. Even page numbers are always left-aligned, which is correct here.

Pí[1]mí[2]ní[3]rí[4]wí[5]ñ2ñ$

$[pic]¤x$[pic]If[pic]a$[pic]gdžkd?y$[pic]$[pic]If[pic]T[pic][6]–sÖ[pic][pic][pic][pic] |”¦þÖr?ÿ÷[pic]ø·

'

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download