Handbook_on_Material_and_Energy_Balance_Calculat1

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Relative Atomic Mass Values and Atomic Numbers

Atomic mass values apply to naturally-occurring isotopes based on the atomic mass of12C=12

Element Symbol Atomic Atomic Element Atomic Atomic
Number Mass Symbol Number Mass
Actinium Ac Lead
Aluminum Al 89 [227] Lithium РЬ 82 207.21
Americium Am 13 26.9815 Lutetium Li 3 6.9412
Antimony Sb 95 [243] Platinum Lu 71 174.9671
Argon Ar 51 121.7601 Plutonium Pt 78 195.0782
Arsenic 18 39.9481 Pu 94 [244]
Astatine As 33 74.9216 Polonium Po 84 [209]
Barium At 85 [210] Potassium К 19 39.0983
Beryllium Ba 56 137.3277 Praseodymium Pr 59 140.9077
Bismuth Be 4 Promethium Pm 61 [145]
Boron Bi 83 9.0122 Protactinium Pa 91 231.0359
Bromine В 5 208.9804 Radium Ra 88 [226]
Cadmium Br 35 Radon Rn 86 [222]
Calcium 48 10.812 Rhenium Re 75 186.2071
Carbon Cd 20 79.9041 Rhodium Rh 45 102.9055
Cerium Ca 6 Rubidium Rb 37 85.4678
Cesium С 58 112.4118 Ruthenium Ru 44 101.072
Chlorine Ce 55 40.0784 Samarium Sm 62 150.363
Chromium Cs 17 12.0108 Scandium Sc 21 44.9559
Cobalt CI 24 140.1161 Selenium Se 34 78.963
Copper Cr 27 132.9055 Silicon Si 14 28.0855
Dysprosium Co 29 35.4532 Silver Ag 47 107.8682
Erbium Cu 66 51.9962 Na 11 22.9898
Europium Dy 68 58.9332 Sodium Sr 38 87.621
Fluorine Er 63 63.5463 Strontium S 16 32.0655
Francium Eu 9 162.5001 Sulfur Та 73 180.9479
Gadolinium F 87 167.2593 Tantalum Тс 43 [98]
Gallium Fr 64 151.9641 Technetium Те 52 127.603
Germanium Gd 18.9984 Tellurium TI 81 204.3833
Gold Ga 31 [223] Thallium Tb 65 158.9253
Hafnium Ge 32 157.253 Terbium Th 90 232.0381
Helium Au 69.7231 Thorium Tm 69 168.9342
Holmium Hf 79 72.641 Thulium Sn 50 118.7107
Hydrogen He 72 196.9666 Tin Ti 22 47.8671
Ho 2 178.492 Titanium 183.841
Indium H 67 4.0026 Tungsten w 74
Iodine 164.9303 и 92 238.0289
Iridium In 1 1.00795 Uranium 50.9415
Iron I Vanadium V 23 131.2936
Krypton Ir 49 114.8183 Xenon Xe 54 173.043
Lanthanum Fe 53 126.9045 Ytterbium Yb 70
Kr 77 192.2173 Yttrium Y 39 88.9059
26 55.8452 Zinc Zn 30 65.4094
La 36 83.7982 Zirconium Zr 40 91.2242
57 138.9055

Summary Descriptions of Nine Quantities that are Quotients of Amount of
Substance, Volume, or Mass
Quantity in numerator

Amount of substance Volume Mass
Symbol: n Symbol: V Symbol: m
SI unit: mol SI unit: m SI unit: kg

Amount of substance amount-of-substance molar volume molar mass
fraction Vm = Win M = т/л
Symbol: n SI unit: m /mol SI unit: kg/mol
xB = nB/n
ои SI unit: mol mass density
SI unit: mol/mol = 1 p = m/V
εfi
amount-of-substance volume fraction SI unit: kg/m
fоi Volume concentration
х в Vm*,B mass fraction
Symbol: V CB = HB/F wB = mB/m
Фв - *
SI unit: m SI unit: mol/m Σ *AVm,A..
fi
fi SI unit: m /m = 1

О Mass molality specific volume
bB = nB/mA v = V/m
Symbol: m

SI unit: kg SI unit: mols/kg SI unit: m /kg SI unit: kg/kg = 1

The volume fraction фв refers to substance В in a mixture of А, В, С, etc., and V* refers to the molar
volume of the pure substance.

The concentration of components is based on mass or amount of substance per unit volume. The
above diagram specifies these as mass density (kg/m3), and amount-of-substance concentration (moles per
m3). The composition of the components is defined as a part-per-part dimension. These are specified as
mass fraction (kilograms of component В divided by the total kilograms), amount-of-substance fraction
(moles of component В divided by the total number of moles), molality (moles of solute В in solution
divided by the mass of the solvent in kg), or volume fraction (m3 of component В divided by the total m3).

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.

Conversion Fact

To convert from to Multiply by

LENGTH m 1.609 344 E+03
mile m 3.048 E-01
foot m 2.54 E-02
inch m 1.0
angstrom (A) E-10

VOLUME m3

cubic yard mm3 7.645 549 E-01
barrel (petroleum, 42 U.S. gal.) m3 1.589 873 E-01
2.831 685 E-02
cubic foot m3 3.785412 E-03
m3 9.463 529 E-04
gallon (U.S.) 2.957 353 E-05
quart (U.S., liquid)
3.155 693 E+07
fluid ounce (U.S.) ™3 3.1536 E+07
TIME 8.64 E+04
m 3.6 E+03

year (tropical) s

year (365 days) s

day s

hour s

MASS kg 1.016 047 E+03
ton (long) kg 9.071 847 E+02
ton (short) kg 1.0 E+03
tonne kg 4.535 924 E-01
pound (avoirdupois) kg 3.110 348 E-02
ounce (troy) kg 2.834 952 E-02
ounce (avoirdupois) kg 6.479 891 E-05
grain
N 8.896 443 E+03
FORCE N 9.806 65 E+00
ton (short)-force N 4.448 222 E+00
kilogram-force N 1.382 550 E-01
pound-force N 1.0 E-05
poundal
dyne

* Exact conversion factors indicated in bold font.

tors and Units to SI*

To convert from to Multiply by

PRESSURE 1.013 25 E+05
atmosphere (standard) 1.0 E+05
bar
psi 6.894 757 E+03
inch of Hg (conventional) 3.386 389 E+03
foot of water (conventional) 2.989 067 E+03
inch of water (conventional) 2.490 889 E+02
mm Hg (torr) (conventional) 1.333 224 E+02
micron of Hg (conventional) 1.333 224 E-01

TEMPERATURE T,K = t, °C + 273.15
degree Celsius T, К = (t, °F + 459.67/1.8
degree Fahrenheit
degree Rankine T, K == (T, R)/1.8

ENERGY AND WORK 1.054 804 E+08
therm (U.S.) 3.6 E+06
kilowatt hour 4.184 E+03
kilocalorieth
British thermal unitth 1.054 350 E+03
horsepower hour (electric) 2.6856 E+06
liter atmosphere 1.013 250 E+02
foot pound-force
erg 1.355 818 E+00
1.0 E-07

POWER wwww 7.46 E+02

horsepower (electric) 1.355 818 E+00
foot pound-force/sec 1.162 222 E+00
kilocalorieth/h 2.928 750 E-01
British thermal unitth/hr

THE GAS CONSTANT R
8.314 47 J/(mol · K); m3 · Pa/(mol · K)
1.987 21 cal(mol · K); BTU/(lb-mol · °R)
0.730 241 ft3 · atm/(lb-mol · °R)
0.0820 575 liter · atm/(mol · K)

HANDBOOK ON
MATERIAL AND ENERGY
BALANCE CALCULATIONS IN
MATERIALS PROCESSING

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HANDBOOK ON
MATERIAL AND ENERGY
BALANCE CALCULATIONS IN
MATERIALS PROCESSING

THIRD EDITION

Arthur E. Morris
Gordon Geiger
H. Alan Fine

TIMS

)WILEY

A JOHN WILEY & SONS, INC., PUBLICATION

Copyright © 2011 by The Minerals, Metals, & Materials Society. All rights reserved

Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada.

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Library of Congress Cataloging-in-Publication Data:

Morris, Arthur E., 1935-

Handbook on material and energy balance calculations in material processing / Arthur E. Morris, Gordon Geiger, H. Alan Fine. — 3rd ed.

p. cm.

Rev. ed. of: Handbook on material and energy balance calculations in metallurgical processes. 1979.

Includes bibliographical references and index.

ISBN 978-1-118-06565-5 (hardback)

1. Chemical processes—Mathematical models—Handbooks, manuals, etc. 2. Manufacturing processes—Mathematical models—Handbooks, manuals, etc.

3. Chemical processes—Mathematical models—Handbooks, manuals, etc. 4. Materials—Handbooks, manuals, etc. 5. Phase rule and equilibrium—

Handbooks, manuals, etc. 6. Heat balance (Engineering)—Mathematics—Handbooks, manuals, etc. 7. Conservation laws (Physics)—Mathematics—

Handbooks, manuals, etc. I. Fine, H. Alan. II. Geiger, Gordon Harold, 1937- III. Fine, H. Alan. Handbook on material and energy balance calculations in

metallurgical processes. IV. Title.

TP155.7.M66 2011

660'.28—dc22 2011010947

Printed in Singapore.

10 9 8 7 6 5 4 3 2 1

Contents

Preface to the First Edition xiv

Preface to the Third Edition xvii

Acknowledgements xix

Chapter 1. Dimensions, Units, and Conversion Factors 1

1.1 The SI System of Units 1 4
1.1.1 Derived Units 2
1.1.2 Units Outside the SI 3
1.1.3 Comments on Some Quantities and Their Units

1.2 The American Engineering System (AES) of Units 4

1.3 Conversion of Units 6 8
1.3.1 Conversion Factor Tables 6
1.3.2 The Dimension Table 7
1.3.3 Conversion Equations — Temperature and Pressure

1.4 Unit Conversions Using the U-Converter Program 11

1.5 Amount of Substance — the Mole Unit 11

1.6 Density and Concentration 13 16
1.6.1 Density 13
1.6.2 Composition and Concentration

1.6.3 Composition of Gases 18

1.7 Electrical Units 20

1.8 Calculation Guidelines 21

1.9 Summary 22

References and Further Reading 23

Exercises 23

Chapter 2. Thermophysical and Related Properties of Materials 26

2.1 State of a System and Properties of a Substance 26

2.2 The Gibbs Phase Rule 27

2.2.1 Consequences of the Phase Rule for Non-Reactive and Reactive Systems
2.2.2 Application of the Phase Rule to One-Phase Non-Reactive Systems 28
2.2.3 Application of the Phase Rule to Multi-Phase Non-Reactive Systems 28
2.2.4 Application of the Phase Rule to Reactive Systems 29

2.3 The Gas Phase 30
2.3.1 The Ideal Gas Law 30
2.3.2 Non-Ideal Gas Behavior 32

2.4 Condensed Phases 34

2.5 Vapor-Liquid Equilibrium (VLE) 35
2.5.1 Mixtures of Condensable and Non-Condensable Gases 3 8
2.5.2 Software for Making Dew Point and Humidity Calculations 39

2.6 Effect of Pressure on Phase Transformation Temperatures 42

2.7 Steam and Air Property Calculators 44

v

vi Contents

2.8 Properties of Solutions 44
2.8.1 Ideal Solutions — Raoult's Law 44
2.8.2 Non-Ideal Solutions — Activity Coefficients 46
2.8.3 Solutions of Gases in Condensed Phases 47
2.8.4 The Solubility Limit 48

2.8.5 The Solubility of Ionic Species in Water; the Solubility Product 49

2.9 Summary 50

References and Further Reading 51

Exercises 52

Chapter 3. Statistical Concepts Applied to Measurement and Sampling 54

3.1 Basic Statistical Concepts and Descriptive Tools 55
3.1.1 Histograms and Frequency Distributions 56
3.1.2 Mean, Standard Deviation, and Variance 59
3.1.3 Median, Percentile and Quantile 60

3.2 Distributions of Random Variables 62
3.2.1 The Uniform Distribution 62
3.2.2 The Normal Distribution 70

3.3 Basic Applications of Inferential Statistics to Measurement 75 78
3.3.1 Sampling Distributions of the Mean and the Central Limit Theorem
3.3.2 Confidence Intervals 82
3.3.3 Treatment of Errors 86
3.3.4 Error Propagation 92

3.4 Curve Fitting 95
3.4.1 Simple Linear Regression and Excel's Trendline Tool 96
3.4.2 Using Solver to Develop Single-Variable Regression Models 101
3.4.3 Multiple Linear and Non-linear Regression 102
3.4.4 Using Solver and Excel's SSD Tool to Find Equation Coefficients 105
3.4.5 Choosing Among Models 106
3.4.6 Polynomial vs. Rational Function Models 116
3.4.7 Outliers 117
3.4.8 Warnings 119

3.5 Experimental Design 119
3.5.1 Factorial Design 120

3.5.2 Fractional Factorial Design 130

3.6 Summary 137

References and Further Reading 137

Exercises 138

Chapter 4. Fundamentals of Material Balances with Applications to Non-Reacting Systems 144

4.1 System Characteristics 144

4.2 Process Classifications 145

4.3 Flowsheets 146

4.4 The General Balance Equation 150

4.5 Material Balances on Simple Non-Reactive Systems 151

4.6 Strategy for Making Material Balance Calculations 154

4.6.1 Guidelines for Setting up a Materials Balance 155
4.6.2 Guidelines for Resolving a Set of Equations 156
4.6.3 Objectives of a Material Balance 157

Contents vii

4.7 Degree-of-Freedom Analysis 158 160 266
4.7.1 DOF Concepts 159
4.7.2 DOF Calculation Strategy for a Single Non-Reactive Device
4.7.3 A Washing Process Having Zero Degrees of Freedom 162
4.7.4 A Washing Process Having a DOF = +1 168
4.7.5 A Leaching Process Having a DOF = - 1 171

4.8 Using Excel-based Calculational Tools to Solve Equations 174
4.8.1 Goal Seek and Solver as Calculational Aids 174
4.8.2 Software for Conversion of Stream Units: MMV-C 178

4.9 Balances on Systems with Multiple Devices 179

4.10 Extension of Excel's Calculational Tools for Repetitive Solving 194
4.10.1 SuperGS 194
4.10.2 SuperSolver 196

4.11 Special Multiple-Device Configurations I — Recycle and Bypass 197

4.12 Special Multiple-Device Configurations II — Counter-Current Flow 205

4.13 Using FlowBal for Material Balance Calculations 216 219
221
4.13.1 FlowBal Example #1: Mixer/Splitter 217
4.13.2 FlowBal Example #2: Evaporation/Condensation Process
4.13.3 FlowBal Example #3: Systems with Multi-Phase Streams

4.14 Continuous-Mixing Devices 223
4.14.1 Steady-State Processes 223
4.14.2 Unsteady-State Systems 226

4.14.3 Inert Gas Flushing 229

4.15 Graphical Representation of Material Balances 232

4.16 Measures of Performance 232

4.17 Controllers 234

4.18 Summary 239

References and Further Reading 240

Exercises 241

Chapter 5. Stoichiometry and the Chemical Equation 248

5.1 Atomic and Molecular Mass 248

5.2 Composition of Compounds and the Gravimetric Factor 249

5.3 Writing and Balancing Chemical Equations 251
5.3.1 Chemical Reaction Concepts 252

5.3.2 Writing and Balancing Chemical Reactions for Simple Processes 253

5.4 Calculations Involving Excess and Limiting Reactants 256

5.5 Progress of a Reaction 258

5.5.1 Extent of Species Reaction and Rate of Reaction Terminology 258

5.5.2 Chemical Reaction Kinetics 259 265
5.5.3 Reaction Progress and Кщ 263
5.5.4 Кщ Values from FREED 264
5.5.5 Guidelines for Using Кщ to Determine Maximum Reaction Extent

5.5.6 Application of Equilibrium Limitations for Gas-Condensed Phase Reactions

5.5.7 Application of Equilibrium Limitations to Gas-Phase Reactions 268

5.6 Practical Indicators of the Progress of Reactions and Processes 269

5.7 Parallel, Sequential and Mixed Reactions 273

5.8 Independence of Chemical Reactions 274

viii Contents

5.9 Practical Examples of Reaction Writing and Stoichiometry 274
5.9.1 Calculations in Gas-Condensed Phase Processes 275
5.9.2 Calculations in Gas-Phase Processes 277

5.10 Use of Chemical Reactions in FlowBal 279
5.10.1 FlowBal's Extent of Reaction Tool 280

5.10.2 FlowBal's Insert Equation Tool 282

5.11 Balancing Aqueous (Ionic) Reactions 283

5.12 Summary 286

References and Further Reading 288

Exercises 288

Chapter 6. Reactive Material Balances 294

6.1 The General Material Balance Procedure for a Reactive System 294
6.1.1 Independent Chemical Reactions, Independent Species, and Independent Elements 295
6.1.2 Molecular Species Material Balance Method 296
6.1.3 Atomic Species Method 300
6.1.4 Atomic and Molecular Species Balance Examples 302

6.2 The Use of Excel-based Computational Tools in Reactive System Balances 306
6.2.1 Application of SuperSolver 306
6.2.2 Reactive System Material Balances Using FlowBal 309

6.3 Combustion Material Balances 321 329
6.3.1 Material Balance for the Combustion of a Gaseous Fuel 322
6.3.2 Combustion of Liquid Fuels 327
6.3.3 Combustion of Solid Fuels 327
6.3.4 Use of Feed-Forward and Stack Gas Analysis for Combustion Control
6.3.5 Use of FlowBal for Combustion Calculations 333
6.3.6 Trace Combustion Products 335

6.4 The Production of a Reducing Gas 340

6.5 Gas-Solid Oxidation-Reduction Processes 345
6.5.1 Oxidation-Reduction During Calcination 345
6.5.2 The Reduction of Iron Ore Concentrate 348
6.5.3 The Chemistry of Fluidized Bed Reduction of Iron Ore by Hydrogen 349
6.5.4 Excel Simulation of the Fluidized Bed Reduction of Hematite 351
6.5.5 FlowBal Simulation of the Fluidized Bed Reduction of Hematite 355
6.5.6 Shaft Furnace Reduction of Iron Ore Concentrate 359
6.5.7 The Roasting of a Sulfide Concentrate 366

6.6 The Production of Gases with Controlled Oxygen and Carbon Potential 370

6.7 Processes Controlled by Chemical Reaction Kinetics 371

6.8 The Reconciliation of an Existing Materials Balance 372

6.9 The Use of Distribution Coefficients in Material Balance Calculations 377
6.9.1 Use of Tabulated Distribution Coefficients 377

6.9.2 Thermodynamic Databases as a Source of Distribution Coefficient Data 381

6.10 Time-Varying Processes 384

6.11 Systems Containing Aqueous Electrolytes 389

6.11.1 The Stability of Ions 390
6.11.2 Aqueous Processes 392

6.11.3 The Solubility of Ionizable Gases in Water 398

6.12 Summary 402

References and Further Reading 404

Exercises 405

Contents

Chapter 7. Energy and the First Law of Thermodynamics 410

7.1 Principles and Definitions 410

7.2 General Statement of the First Law of Thermodynamics 413

7.3 First Law for an Open System 415

7.4 Enthalpy, Heat Capacity, and Heat Content 416

7.5 Enthalpy Change of Phase Transformations 418

7.6 Enthalpy Change of Chemical Reactions 420

7.7 Thermodynamic Databases for Pure Substances 421

7.8 Effect of Temperature on Heat of Reaction 426

7.8.1 Application of Kirchhoff s Equation to Chemical Reactions 426
7.8.2 Heat of Transformation for Non-Standard and Non-Physical States 428

7.9 The Properties of Steam and Compressed Air 431 433
7.9.1 Properties of Steam 431
7.9.2 Properties of Compressed Air 432
7.9.3 Temperature Change for Free Expansion of a Gas
7.9.4 Cooling by Steam Venting 435
7.9.5 Enthalpy of Psychrometry 43 6

7.10 The Use of FREED in Making Heat Balances 437

7.11 Heat of Solution 441
7.11.1 Formation of Non-ideal Metallic Solutions 441
7.11.2 Polymeric Solutions 442

7.11.3 Aqueous Solutions 444

7.12 Summary 445

References and Further Reading 446

Exercises 447

Chapter 8. Enthalpy Balances in Non-Reactive Systems 450

8.1 Combined Material and Heat (System) Balances 450

8.2 Heat Balances for Adiabatic Processes 458

8.3 Psychrometric Calculations 462

8.4 Energy Efficiency 468

8.5 Recovery and Recycling of Heat 469 474

8.5.1 Heat Exchange Between Fluids 469
8.5.2 Heat Exchange between Solids and Fluids 474
8.5.3 Application of Heat Recovery Techniques to Aluminum Melting

8.5.4 Heat Exchange Accompanied by Material Transfer 478

8.6 Multiple-Device System Balances 483

8.7 Use ofFlowBal for System Balances 488

8.8 Heat Balances Involving Solution Phases 494

8.9 Enthalpy Change During Dissolution of an Electrolyte 496

8.10 Graphical Representation of a Heat Balance 499

8.11 Summary 500

References and Further Reading 501

Exercises 502

X Contents

Chapter 9. System Balances on Reactive Processes 505

9.1 Thermal Constraints on a Material Balance 505
9.1.1 Uncoupled System Balances 506
9.1.2 Strategy for Coupled System Balance 508

9.2 Combustion of Fuels 511
9.2.1 Heat of Combustion Calculations 512
9.2.2 Use of Wobbe Index for Combustion Burner Control 514
9.2.3 Combustion of Fuels of Uncertain Molecular Composition 516

9.3 Adiabatic Processes 517
9.3.1 System Balances Involving Combustion Reactions 518
9.3.2 ART for Condensed-Phase Reaction Processes 531

9.4 System Balances Using FlowBal 533

9.5 Quality of Heat and Thermal Efficiency 542

9.6 System Balances with Heat Exchangers 548

9.7 Aqueous Processes 565

9.8 Electrolytic Processes 571
9.8.1 Energy Requirement for Electrorefming 571

9.8.2 Energy Requirement for Electrowinning 572

9.9 Summary 573

References and Further Reading 574

Exercises 574

Chapter 10. Case Studies 577

10.1 Material Balance for an H-Iron Reduction Process with Gas Tempering and Recycle 577

10.2 Mass and Heat Balance Simulation for the Use of DRI in EAF Steelmaking 581

10.3 Natural Gas Combustion Control and the Wobbe Index 588

10.3.1 The Stoichiometry of NG Combustion with Excess Air 588
10.3.2 The Wobbe Index as a Natural Gas Combustion Control Parameter 591

10.4 Reduction of Hematite to Magnetite 592
10.4.1 Preliminary Calculations — Single Reactor 593
10.4.2 Simulation of Hematite Reduction by a Multi-Stage Process 595

10.5 Conversion of Quartz to Cristobalite in a Fluidized Bed 598
10.5.1 Process Characteristics 598
10.5.2 Device Sizing and Heat Loss Calculation 598
10.5.3 Material and Heat Balance Calculations 599

Exercise 600

Appendix. Computational Tools for Making Material and Heat Balance Calculations 601
A.l U-Converter 601
A.2 Thermophysical Properties of Steam and Air 602
A.3 Stream Units Conversion Calculator (MMV-C) 602
A.4 Extension of Excel Tools for Repeat Calculation 603
A.5 Thermodynamic Database Programs 604
A.6 Flowsheet Simulation and System Balancing 604

General References 605

Index 606

Contents

List of Examples

Chapter 1. Dimensions, Units, and Conversion Factors

1.1 Mass and Weight of Aluminum 5 19
1.2 Kinetic Energy 6
1.3 Energy of Lifting 6
1.4 Units of Energy 7
1.5 Dimensions for Flowrate 8
1.6 Conversion of Temperature 9
1.7 Conversion Formula 9
1.8 Conversion of Pressure - 1 10
1.9 Conversion of Pressure - II 10
1.10 Pressure in a Liquid 11
1.11 The SI and AES Mole 13
1.12 The Density of a Slurry 14
1.13 Bulk Density of a Solid-I 16
1.14 Bulk Density of a Solid - II 17
1.15 Concentration Conversion 17
1.16 Composition of a Gas on a Wet and Dry Basis
1.17 Electrical Flow in a Wire 20
1.18 Electrical Energy for Metal Deposition 20
1.19 Number of Significant Figures 22

Chapter 2. Thermophysical and Related Properties of Material
2.1 Removal of Air by a Vacuum Pump 31
2.2 Gas Volume and Flowrate 31
2.3 Compressibility of Steam 32
2.4 Thermal Expansion of Titanium 34
2.5 Evaporation of Water in a Closed Vessel 36
2.6 Humidity and Dew Point 38
2.7 Moisture Content of Clay Dryer Streams 40
2.8 Effect of Pressure on the Freezing Point of Water. 42
2.9 Effect of Pressure on the Vapor Pressure of Water 43
2.10 Vapor and Liquid Phase Composition for the Cu - Ni System 45
2.11 Evaporation from Liquid Cd - Mg Alloys at 700 °C 47
2.12 Volumetric Solubility of C02 in Water 48
2.13 The Solubility of CaF2 in Water 49

Chapter 3. Statistical Concepts Applied to Measurement and Sampling
3.1 A Histogram of Ceramic Strength Measurements 57
3.2 Percentiles of the %Cu Data using Excel 61
3.3 Uniformity of Vermiculite Particles 66
3.4 Evaluation of the Normal Distribution for Ceramic Strength Data 74
3.5 Finding a 90% Confidence Interval 84
3.6 Relationship Between Sample Size and Interval Width Using the Si2ON2 Example

xii Contents

3.7 Heat Capacity Systematic Error 88

3.8 Ore Assay 89
3.9 Improving Measurement Precision 91

3.10 The Professor Tries Again 92 94
3.11 Linear Random Error Propagation 93
3.12 Multiplicative Random Error Propagation

3.13 Other Random Error Propagation 94
3.14 The Difference Between Propagation of Random and Systematic Errors 95

3.15 Modeling the Heat Capacity of TiOx 99
3.16 Non-linear Models for the CpTiOx Data 100

3.17 An Asymptotic Model for the Heat Capacity of TiOx 102
3.18 Using the Regression Tool to Find Non-Linear Models 104

3.19 Removing the Thickness Variable from the Galvanized Corrosion Example 107 108
3.20 Hypothesis Testing for the Galvanized Steel Model with Three Independent Variables
3.21 Selecting a Model for the Hydrogen Reduction of NiO 114

3.22 Calculating the Pressure-Catalyst Interaction 124
3.23 Effect of a Fractional Factorial Design: How Much Information is Lost? 135

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

4.1 Distillation ofaCd-Zn Alloy 151 153
4.2 Charge Calculation for Feed to a Brass Melting Furnace
4.3 Vacuum De-Zincing of Lead 153

4.4 Leaching of Salt Cake from Aluminum Recycling 177
4.5 Refining Crude Boric Acid by a Two-Stage Aqueous Process 181

4.6 Recovery of KMn04 by Evaporation 184 187
4.7 Removing Dust and S02 from a Roaster Gas
4.8 Absorption of HC1 197

4.9 Preparation of a Pigment Precursor 202 208
4.10 Removal of CuS04 from a Pollution Control Residue
4.11 Catalyst Reactivation 225

4.12 Dissolution of ZnCl2 228
4.13 Removal of Hydrogen from Steel 230

4.14 Vacuum Refining of a Cd-Zn Alloy 230
4.15 Control Strategy for Upgrading Spent Reducing Gas 235

Chapter 5. Stoichiometry and the Chemical Equation

5.1 Use of the Gravimetric Factor for Silicon 250

5.2 Mineralogical Constituents of a Concentrate 250

5.3 Reduction ofWustite by CO 253
5.4 Production of Molybdenum Carbide 255

5.5 Production of Titanium by the Kroll Process 257

5.6 The Reaction Between Oxygen and Carbon 272
5.7 Reduction of Molybdenum Oxide with Hydrogen 275

5.8 Carbothermic Reduction of Zinc Oxide 276
5.9 Steam Reforming of Methane 278

Contents Xlll

5.10 Controlled Oxidation of Pyrite 284
5.11 Dissolution of Gold in Cyanide Solution 285

Chapter 6. Reactive Material Balances

6.1 Production of Sulfur by Reduction of Sulfur Dioxide 302
6.2 Chlorination of Silicon 304

6.3 Application of FlowBal to Stack Gas Desulfurization 314

6.4 Combustion of Natural Gas with XSA 325

6.5 Effect of Oxygen Enrichment on the Oxidant Required for Complete Combustion 326
6.6 Stack Gas Composition and Dew Point for Coal Combustion with Dry Air 328
6.7 Calculation of % Excess Air from Stack Gas Analysis 332

6.8 Calculation of CO, H2 and NO content in Hot Stack Gas 338
6.9 Calculation of Reformer Gas Composition 342
6.10 Calculation of dpt from Gas Analysis 344

6.11 Calcination of Wet Pickling Cake 347

6.12 Simulation of a Pre-Reduction Fluidized Bed Process 357
6.13 Material Balance on Shaft Furnace Reduction of Hematite 363

6.14 Roasting a Zinc Sulfide Concentrate 369
6.15 Material Balance for BOF Steelmaking 378
6.16 Leaching of Scrubber Dust 395

6.17 The Optimum Precipitation of CaC03 by C02 398

Chapter 7. Energy and the First Law of Thermodynamics

7.1 Work and Heat During the Compression of an Ideal Gas 415

7.2 Heat Capacity and Enthalpy for a Flux 417
7.3 Heat of Fusion of Lead 419

7.4 Standard Heat of the Water-Gas Shift Reaction from 800 to 1500 К 427

7.5 Supercooling Liquid Tin 430
7.6 Combustion of CO with Preheated Air 437

7.7 Adiabatic Compression of Steam 438

7.8 Enthalpy Change During Reduction of NiO with С 439
7.9 Temperature Change of an Adiabatic Reaction 440

Chapter 8. Enthalpy Balances in Non-Reactive Systems

8.1 Heat Balance for Melting Aluminum 454
8.2 Heat Balance for Spray Cooling of Hot Air 455

8.3 Fog Cooling of Ceramic Parts 457 463
8.4 Atomization of a Molten Metal 459
8.5 Dehumidifying Spent Gas from an Iron Ore Reducing Furnace
8.6 Using Stack Gas to Dry Cadmium Powder 466

8.7 Heat Exchange in a Waste Heat Boiler 471

8.8 Preheating HC1 in a Pebble-Bed Vertical Shaft Heat Exchanger 474
8.9 Lowering the Water Temperature from a Crystallizer 481

8.10 Condensation of Zinc Vapor from a Gas 486
8.11 Production of Distilled Water 492

xiv Contents 513

Chapter 9. System Balances on Reactive Processes
9.1 Heat of Combustion of a Spent Gas from a Reduction Process
9.2 Effect of Preheating Combustion Air on AFT 519
9.3 Production of a Reducing Gas 521
9.4 Adiabatic Reforming of Fuel Oil 523
9.5 Oxidation of S02 to S03 for Sulfuric Acid Production 525
9.6 Metallothermic Reduction of Uranium Tetrafluoride 531
9.7 Lime-Assisted Reduction of Magnetite 539
9.8 Calcination of Magnesium Carbonate 552
9.9 Formation of Nickel Ferrite by Spray Roasting 561

CD Contents

1 CD Content Descriptions
2 Air
3 Atmospheres
4 Charts
5 Combustion Documents
6 Copper Smelting
7 FlowBal and MMV-C
8 Material and Heat Balance Notes
9 NG Combust & Wobbe Index
10 Statistics
11 Steel
12 SuperGoalSeek
13 SuperSolver
14 Thermodynamic Database
15 Unit Conversions

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 have 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
December, 1979

XV

This page intentionally left blank

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 25 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. However, 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.
Finally, on-line versions of two commercial programs for steam table and psychrometric
calculations were identified and incorporated in the text examples. These programs simplify the
rather difficult calculations commonly required in making material and heat balances. 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. The Appendix 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 (CD Content
Descriptions.doc).

Many 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 that covers the main principles of
thermochemistry, or at least to have available a chemistry textbook typical of those used in the first
year of a materials engineering program.

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

xvii

xviii Prefaces

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. You are also encouraged to
improve any of its articles that are in your area of expertise. The web pages cited in the General
References section and as Chapter references may disappear or change after publication, and other
sites may appear.

A number of useful and interesting public domain articles were found during revision of the
Handbook. These articles have been collected to the Handbook CD in various folders. Many of
these articles give background information on processes that were used as Handbook examples.
Some of the documents contain articles on processes described in the FlowBal User's Guide.

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 Handbook users to alert the authors to useful information and
to submit material for posting on the page.

http ://thermart.net/

Arthur E. Morris
Thermart Software
San Diego, California

Gordon H. Geiger
University of Arizona
Tucson, Arizona

December, 2010

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.
Professor Geiger organized this effort with DOE, and made many valuable suggestions during the
preparation of the manuscript. In particular, he prepared an outline for a greatly revised chapter on
statistics, and contributed advice on the treatment of psychrometry and controllers.

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 made their contributions while they were candidates for a PhD. in
statistics, and made what was a very rough draft into a polished product. Second, Mr. Knut
Lindqvist wrote the code for Super Goal Seek and Super Solver, and Robert Baron wrote U-
Converter. These three programs are on the Handbook CD.

Dr. Semih Perdahcioglu was a prime contributor to the Handbook by his development of
FlowBal and MMV-C. FlowBal, in particular, became a mainstay of the computational tools used
throughout the Handbook. As a result of his dedicated work, both students and process engineers
now have a flowsheet simulation tool capable of dealing with complex multi-device flowsheets.
Semih also revised FREED to include a reaction tool. He is now a post-doc research assistant at
University of Twente, Netherlands.

We are obligated to three faculty members for miscellaneous advice with various topics.
Professor Eric Grimsey (WASM Kalgoorlie) was an inspiration from the beginning, and provided
valuable suggestions, encouragement, and help throughout, especially with the application of the
degree-of freedom concept. Professor Grimsey also contributed a set of his course notes ("Basic
Material and Heat Balances for Steady State Flowsheets") for inclusion on the Handbook CD.
These notes provide a shorter (and somewhat different) approach to the construction of system
balances, and are recommended without reservation as an adjunct to the Handbook approach.

Professor David Robertson (Missouri University of Science and Technology) cleared up a
number of points regarding the material on continuously mixed and unsteady-state processes, and
pointed out a number of text errors in Chapter 3. Some of his graduate course examples were
adapted for use in FlowBal, and he provided suggestions for FlowBal changes to help the user.
Professor Mark Schlesinger (Missouri University of Science and Technology) permitted use of
several of his examples and exercises.

Finally, no text is generated in isolation. Items from the General References Section (page
605, just before the Index) provided background information and material data that was used for
working out examples and exercises. This Section also cites 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 more computationally intensive approach adopted in this text.

One of us (AEM) also wishes to express his appreciation for the steadfast support of his wife
Helen throughout the revision project.

xix

About the Authors

Arthur E. Morris joined the University of Missouri - Rolla (now the Missouri University of Science and Technol-
ogy) in 1965 after receiving his PhD from the Pennsylvania State University in 1965. During his tenure on the fac-
ulty of the Department of Metallurgical Engineering, he taught courses in extractive metallurgy, thermodynamics,
and process simulation, and also carried out research that resulted in theses for several MS and PhD students. Dr.
Morris was a consultant to several industrial corporations in their research laboratories, and was asked by the U.S.
Bureau of Mines to organize a new research group at UMR called the Center for Pyrometallurgy. He was a Princi-
pal Investigator at the Center until his retirement in 1996. While at UMR, Professor Morris published nearly 70 pa-
pers on various aspects of extractive metallurgy and materials processing, and conducted short courses and sym-
posia on the applications of computer modeling to metallurgical processes. He presently develops educational
software and prepares CDs for materials-related textbooks.

Gordon H. Geiger earned his Bachelor of Engineering degree in Metallurgy at Yale University and his M.S. and
Ph.D. degrees in Metallurgy and Materials Science at Northwestern University. He worked in the research depart-
ments of Allis-Chalmers Mfg. Co. and Jones and Laughlin Steel Company before teaching process metallurgy at
the University of Wisconsin, the University of Illinois at Chicago, and the University of Arizona. In addition to his
teaching career, Dr. Geiger worked in industry as a technical officer for a major international bank, a multi-plant
steel company and founded a new steel company. He is now retired and lives in Arizona, where he consults and
where as Academic Director, he assisted the University of Arizona in establishing an Engineering Management de-
gree program.

H.Alan Fine graduated with a PhD degree in Metallurgy from the Massachusetts Institute of Technology in 1974.
He then joined the faculty of the University of Arizona's Metallurgical Engineering Department as an Assistant Pro-
fessor. Dr. Fine remained on the faculty until 1981, when he joined the University of Kentucky as Associate Pro-
fessor in the Department of Metallurgical Engineering and Materials Science. During his time at Kentucky, he also
worked with the Environmental Protection Agency. Dr. Fine co-authored the first two editions of this handbook
with Dr. Geiger, and he is now retired and living with his family in Florida.

CHAPTER 1

Dimensions, Units, and Conversion Factors

Most 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 System Internationale d'Unites
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 emphasizes 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). Some of the Chapter examples require thermophysical data,
which can be obtained from one of the General References (page 605).

1.1 The SI System of Units

In 1960, the General Conference on Weights and Measures (CGPM, Conference General 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.

2 Chapter 1 Dimensions, Units, and Conversion Factors

The quantity of a substance can be expressed in two ways: its mass or its amount. The mass
unit is the kilogram (abbreviation kg). 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. Name Symbol
meter m
Base quantity kilogram
length second kg
mass ampere s
time kelvin A
electric current mole К
thermodynamic temperature candela
amount of substance mol
luminous intensity 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.

Derived quantity SI derived unit Symbol
area Name m2
volume square meter m3
speed, velocity cubic meter
acceleration meter per second m/s
mass density (density) meter per second squared m/s2
specific volume kilogram per cubic meter kg/m3
current density cubic meter per kilogram m3/kg
amount-of-substance concentration (concentration) amperes per square meter A/m2
mole per cubic meter mol/m3

Chapter 1 Dimensions, Units, and Conversion Factors 3

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

SI derived unit

Derived quantity Special name Special Expression in Expression in
symbol terms of other terms of SI base

Hz SI units unit
N s-1
frequency hertz Pa
force newton J N/m2 m · kg · s~2
pressure, stress pascal W rrf ' · kg · s~2
energy, work, heat joule С
power watt V N-m m2 · kg · s~2
quantity of electricity coulomb Ω J/s m2 · kg · s~3
electric potential, emf volt °C
electric resistance ohm W/A s-A
Celsius temperature degree Celsius m3 · kg · s~3 · A-1

V/A m2 · kg · s"3 · A"2

К

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

Derived quantity Name Symbol SI derived unit

heat flux density watt per square meter W/m2 Expression in terms
heat capacity, entropy joule per kelvin J/K of SI base units
specific heat capacity, entropy joule per kilogram kelvin J/(kg · K)
molar entropy, heat capacity joule per mole kelvin J/(mol · K) kg-s-3
thermal conductivity watt per meter kelvin W/(m · K) m2 · kg · s-2 · K"1
m2 · s"2 · K"1
m2-kg-s-2-K-1-mor11
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.

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 composition term % is acceptable in place of 0.01. For example, it is
preferable to state, "the mass fraction of В is 0.02", or "wB = 0.02", but acceptable to state, "the
mass fraction of В 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 (cai, in
various dimensions) as a unit of energy. Similarly, composition terms such as ppm or ppb (parts
per million, parts per billion) are unacceptable unless required by law. As mentioned earlier, some
thermodynamic data may be available only in units of cai and atm, in which case, they will be used
in this Handbook with no further explanation.

4 Chapter 1 Dimensions, Units, and Conversion Factors

1.1.3 Comments on Some Quantities and Their Units

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

T-To [1.1]

where T0 = 273.15 К by definition, and is exactly 0.01 К 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 (К), 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 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 1.6, 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 1.6.2 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 x 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 x 103 or 2.400 x 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 even 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 mass unit (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
produces an acceleration in 1 lbm of 1 ft/s2. This unit is the poundal, with 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

Chapter 1 Dimensions, Units, and Conversion Factors 5

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, engineers have selected 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.

The weight of a body is 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 SI weight is:
W— m · g

Weight (SI) - (405.3 kg)(9.8066 m/s2) - 3975 kg · m/s2 = 3975 newton.

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 x IO-5 m2. The
AES area of the wire is π(0.09845)2 = 0.03043 in2. The stress σ is calculated as:

Q = FIA

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

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

6 Chapter 1 Dimensions, Units, and Conversion Factors

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 0 2 is 0.0320 kg (0.07055 lbm). The kinetic energy (Ek) is Vi the
product of the mass times the velocity squared.
Solution. Ek (SI) = x/2(0.0320 kg)(101 m/s)2 = 163 J

Ek (AES) = 1/2(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 NIST
and other handbooks (Butcher 2006, Thompson 2008) list many more conversion factors. Section
1.4 and the Appendix describe the Excel unit conversion program U-Converter (on the Handbook
CD).

Two types of conversions are required. In the first, multiplication or division converts one
unit into another 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, and expressed as factors. Tables of conversion factors are
readily available in various handbooks. The conversion factors used in this Handbook are based
on 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.

Chapter 1 Dimensions, Units, and Conversion Factors J

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 two or more 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.

m _ km 3600 _ km
s " 1000 h " ' h

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 C02 from
calth/mol to kWh/kg.

Solution. The molar mass of C02 = (12.01 + 32) = 44.01 g. The dimension table is:

cai mol 1000 g 4.184 J W - s h kW

mol 44.01 g kg cai J 3600 s 1000 W

The conversion factor is then 2.6408 x 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 C02 from caWmol to
Btuth/lbm.

8 Chapter 1 Dimensions, Units, and Conversion Factors

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
I d ton 8.64 x 104s

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 mass of mercury that the plant would emit operating 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 Rankine 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 К 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 (К). 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 Rankine (°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 Rankine:

1.0 ΔΚ = 1.0 Л°С; 1.0 A°R = 1.0 A°F [1.6]

1.0A°C=1.8A°F; 1.0AK=1.8A°R [1.7]

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

7(K) = /(°C) + 273.15 [1.8]

7(°R) - /(°F) + 459.67 [1.9]

/(°F)=1.8/(°C) + 32 [1.10]

Chapter 1 Dimensions, Units, and Conversion Factors 9

Remember 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 К 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:

/(°F)=1.8(K-273.15) + 32
Insert 1337.58 for К to get:

t(°¥) = 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 /(°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 Г refers
to the unit kelvin, and the lower-case t refers to the unit Celsius:

Cp = 22.30 + 0.00720(7) J/(mol · K)

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

Solution. The Tin 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(0 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 χ 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

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.

10 Chapter 1 Dimensions, Units, and Conversion Factors

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 x 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.

f

TGage Pressure f T
t
Gage Pressure

Absolute Atmospheric (Negative) or VacuumAtmospheric
Pressure Pressure Pressure
Absolute
Pressure
i

Figure 1.1 Relations of various pressures.

EXAMPLE 1.8 — Conversion of Pressure - I.
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 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 χ 105 Pa at its base at the standard location. The dimension table for
converting μπι of Hg to Pa is:

18.3 μ η ι ^ Im 1.013 25 x 10s Pa = 2.44 Pa
10>m 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 - II.

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?

Chapter 1 Dimensions, Units, and Conversion Factors 11

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 lm 1.013 25 x 105Pa = 98.31 kPa
Next, the gage pressure: 39.37 in 0.76 m Hg

101bf lin2 4.448 N - 6.894 x 104 N/m2 - 68 94 kPa.
in2 6.4516 x lO^m2 llbf

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

1.6725 x 105Pa psia - 24.3 psia

6895 Pa
Assignment. A manometer gives a reading of 3.5 in H20 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:

pFe = 8300-0.836r(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 P=1.98x 105Pa
m 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 by dividing the pressure in Pa by 1.013 x 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 Unit Conversions folder. 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 the Appendix. 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):

12 Chapter 1 Dimensions, Units, and Conversion Factors

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 7i2 of the mass of an atom of the nuclide 12C; 1 μ = 1.660 540 2 x 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 AT9 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
molecule is Mr (F2) = 2Ar(¥) - 10"3
fluorine = 37.9968 g/mol. The amount 37.9968. The molar mass of F2 is then 37.9968 x 100
kg/mol of substance of 0.100 kg of F2 is then n(Y2)
=

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 x 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 AT 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 AT for 95 elements are listed on the inside cover of the Handbook (Coursey 2005).
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 composition in
an element from one raw material to another. An element extracted from seawater may have a
different isotopie composition than if extracted from bedrock. The higher the variation in isotope
composition, 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).

The preferred source of atomic and molecular mass is the thermodynamic database program
FREED, on the Handbook CD. Please see the description of this program in the Appendix.

Chapter 1 Dimensions, Units, and Conversion Factors 13

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:

Xlbm 453.5924 g 1 g-mol

1 lb-mol Hbm Xg

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:

pair (kg/m3)- [1.11]
air 0.7737 + 0.002833/
LJ
where t = degrees Celsius, and P is in atmospheres.*

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.

The symbol p will be used for density and specific gravity. The meaning will be made clear by specifying
the units if density is meant.

14 Chapter 1 Dimensions, Units, and Conversion Factors

Quantity in numerator

Amount of substance Volume Mass
Symbol: n Symbol: V Symbol: m
SI unit: mol SI unit: m3 SI unit: kg

Amount of substance amount-of-substance molar volume molar mass
fraction M = m/n
SI unit: kg/mol
Symbol: n x B = n B/n Vm=V/n
mass density
ио SI unit: mol SI unit: mol/mol = 1 SI unit: m3/mol p = m/V
с*
с SI unit: kg/m
о Volume amount-of-substance volume fraction
concentration mass fraction

а Symbol: V cB = nB/V *B^m,B
2^ xA^m,A..

SI unit: m SI unit: mol/m5 SI unit: m3/m3 = 1
fi

fi Mass molality specific volume

Symbol: m bB = nB/mA v = V/m wB = m Blm

SI unit: kg SI unit: mols/kg SI unit: m3/kg SI unit: kg/kg = 1

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

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 (Fe304) 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 x 10"4
m3/kg, so 100 kg of magnetite will occupy 1.93 x 10"2 m3. The total volume of slurry will then be
1.0193 m3. The density of the slurry is then 1100/1.0193 - 1079 kg/m3.

Chapter 1 Dimensions, Units, and Conversion Factors 15

The above procedure was used to calculate the density of several slurry compositions. A chart

was prepared showing the density and wFe304 vs. the selected input variable. Table 1.5 and Figure
1.3 show the results.

Table 1.5 Results of Excel calculations on slurry density vs. kg Fe304/Mg water.

kg Fe304 m3 water kg slurry m3 Fe304 m3 slurry kg/m3 slurry wFe304

100 1100 0.0193 1.0193 1079 9.1%

150 1150 0.0290 1.0290 1118 13.0%

200 1200 0.0386 1.0386 1155 16.7%

250 1250 0.0483 1.0483 1192 20.0%

300 1300 0.0579 1.0579 1229 23.1%

350 1350 0.0676 1.0676 1265 25.9%

400 1400 0.0772 1.0772 1300 28.6%

An additional column of data was calculated to represent the wFe304 in the slurry. This was
obtained by dividing the kg of Fe304 (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 wFe304 columns would be
constant. Nevertheless, for many engineering applications, and considering the slightly variable

density of magnetite*, assuming a linear relationship may be adequate.

Slurry Density

1300 1 "~ 1 I - — — " ^^M 30%

<*> 1250 density of slurry ^-"4^.^" 25%
! - ■*- - mass % Fe304
"3) Jm^"" """ 9
■£* 1200
1150 ^»*—'■"i S
α>
r<^ 4 20%

^ χ ^ * ΐ' 15% $
CO
■£σ* 1100 1 ^^^'t* E
1050
10%

1000 1! Г 5%
100 400
150 200 250 300 350
kg Fe304 per tonne water

Figure 1.3 Relationship between input variable of kg Fe304/tonne water, slurry density, and
wFe304 in the slurry.

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.

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:

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

16 Chapter 1 Dimensions, Units, and Conversion Factors

Pbulk - ( 1 - CO) /theoretical

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 -1.

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:

l m 3 150 000kgFe3O4
3108kgFe3O4
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 °C 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 В divided by the total kilograms), amount-of-
substance fraction x (moles of component В divided by the total number of moles),* molality b
(moles of solute В in solution divided by the mass of the solvent in kg), or volume fraction φ (m3
of component В 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 В is 0.10" or "the mass (or volume)
fraction of В 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 В (xB) and mass fraction
of В (wB) can be accomplished by using the following equations:

WA /^A + WB /M# + WQ IMQ + . . . .

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.

Chapter 1 Dimensions, Units, and Conversion Factors 17

xBxMB [1.13]
wB = xA x M A + x B x MB + xc x Mc + . . . .

where Mis 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 — Bulk Density of a Solid - II.

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 x 58.69 + xCr x 52.00 + xFe x 55.85 = 55.36

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

wNi - (0.1 x 58.69)/55.36 = 0.106
wCr = (0.2 x 52.00)/55.36 = 0.188
wFe - (0.7 x 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
ofFe.

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 NiS04 solutions are being prepared. Calculate the molality, grams N1SO4/L of
solution, and amount of substance fraction as a function of the mass fraction NiS04.

Data. The density of NiS04/water solutions vs. wNiS04 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 NiS04/kg of water. The molar mass of NiS04 is 58.69
+ 32.07 + 4(16.00) = 154.76. The molality is then:

Molality of NiS04 = (l/154.76)(1000/99) = 0.06527 molai.
The amount-of-substance fraction is given by:

xNiS04 = 1/154.76 = 0.0011745
99/18.016 + 1/154.76

The concentration in grams NiS04/liter of solution is obtained from the dimension table
below as follows, where 1 liter = 1009 g as given by the density. The results for all concentrations

are shown in the next page table.

I g N i S O , 1009 g solution = 10.09 g/L

100 g solution 1 liter

18 Chapter 1 Dimensions, Units, and Conversion Factors

grams N1SO4 1.00 2.00 4.00 8.00 12.00 16.00 18.00
grams H20 99.00 98.00 96.00 92.00 88.00 84.00 82.00
0.00646 0.01292 0.02585 0.05169 0.07754 0.1034 0.1163
moles N1SO4 5.495 5.440 5.329 5.107 4.885 4.663 4.552
moles H2O 0.06527 0.1319 0.2692 0.5619 0.8812 1.231 1.418
0.001175 0.002370 0.004827 0.01002 0.01563 0.02169 0.02492
6NiS04 10.09 20.40 41.68 86.80 135.96 189.28 217.62
xNiS04
g N1SO4/L

Assignment. 100 grams of water and 50 grams of nickel sulfate heptahydrate (NiS04#7H20) are
mixed to prepare an aqueous N1SO4 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/(LVA, 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 or 14.696 psi) and 0 °C (273.15 К or 32 °F), although other conventions exist
(Wikipedia 2010). 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-mol.* At STP, 1 kg-mol occupies 791.54 ft3, and 1 lb-
mol occupies 10.167 m3 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, C02-free dry air** has the following
composition (where φ refers to volume fraction):

φΝ2 - 0.7812; φ02 = 0.2096; фАг - 0.0092
The molar mass M of air is then given by:

Mfor 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. The volumetric
composition of air is often approximated as 79 % N2, 21 % 02, and Mis approximated as 29. In
that case, one mole of air can be given an empirical formula Ni.58Oo.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:

* The concept and application of the ideal gas law will be discussed in detail in Chapter 2.
** Air contains <pC02 « 0.0003.

Chapter 1 Dimensions, Units, and Conversion Factors 19

• 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:

φΝ2 = 70.0%; <pCO2=10.0%; <pCO = 5.5%; φΗ2 = 4.5%; <pH2O=10.0%

Calculate the volume fraction composition of the gas on a) a dry basis; b) the concentration
(in terms of 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:

<pN2 = 70.0/90 = 77.8 %; <pC02 = 10.0/90 = 1 1 . 1 % ; <pCO = 5.5/90 = 6.1 %; φΗ2 = 4.5/90 = 5.0 %.

b) The removal of 10 moles of H20 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 concentration is then:

Wet gas = 2780/79.1 = 35.1 grains of water/ft3 of wet gas
Dry gas = 2780/71.2 = 39.0 grains of water/ft3 of dry gas

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

wNi = 7°(28.01) ППЛА
--
= 0.714

70(28.01) +10(44.01) + 5.5(28.01) + 4.5(2.02) +10(18.02)

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

wC02 = 0.160; wCO= 0.056; wU2 = 0.0033; wH20 = 0.0656

Assignment. A gas has the following analysis: <jpN2 = 80.0 %; φ02 = 15.0 %; <pH20 = 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 0 2 gave the following results:

φΝ2 = 82.5%; φ 0 2 = 1 5 . 5 %

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

20 Chapter 1 Dimensions, Units, and Conversion Factors

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 aforce equal to2x IO"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:

ν=Ω·Α [1.15]
Ψ=Α2Ώ [1.16]
[1.17]
J = A-V-s

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 mol of univalent ions (or lA 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 x 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 x Ю^УО.0314 - 1.08 x 10"4 ohms.
The current is given by application of Equation [1.15]:

A=V/Q = 2/1.08 x IO"4 - 18 500 amperes.
Assignment. Calculate the power involved in the process.

EXAMPLE 1.18 — Electrical Energyfor 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.

Chapter 1 Dimensions, Units, and Conversion Factors 21

The aluminum ion is trivalent, so one faraday will discharge Уз 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 К 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 x 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 Rankine 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

22 Chapter 1 Dimensions, Units, and Conversion Factors

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 °C? b)
Calculate the mass of silver that would be deposited from a solution of AgN03 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.

The value displayed in an Excel 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 ExcePs ROUND function on the values. Please
refer to Excel's on-line help for entering and using functions.

A significant-figure quirk is that the first significant figure in a number like 985 has nearly the
same consequence as the second significant figure in a number like 1025 because an error of 1 in
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

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

Chapter 1 Dimensions, Units, and Conversion Factors 23

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: http://ts.nist.gov/WeightsAndMeasures/Metric/mpo_pubs.cfm. May 2010.

Coursey, J.S., Schwab, DJ., and Dragoset, R.A., Atomic Weights and Isotopie Compositions
(version 2.4.1) [Online]. Available: http://www.nist.gov/pml/data/comp.cfm, October 2010.

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:
http://physics.nist.gov/cuu/pdf/sp811 .pdf. April 2009.

Taylor, Barry N., and Thompson, Ambler, Eds., The International System of Units (SI), NIST
Special Publication 330 [Online]. Available: http://physics.nist.gov/Pubs/SP330/sp330.pdf.
March 2008.

U. S. Environmental Protection Agency on-line course, Basic Concepts in Environmental Sciences,
http://www.epa.gov/apti/bces/modulel/index.htm. December 2009.

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 2010. http://en.wikipedia.org/wiki/Main_Page.

Exercises

1.1 Convert 400 in3/d to cmVmin

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.

24 Chapter 1 Dimensions, Units, and Conversion Factors

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 СаСОз is 2900 kg/m3. What is the
assumed void space in the bulk limestone?

1.8 Convert the absolute pressure 1.00 x 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:

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

a) cai - sec"1 - cm"2 · °C^.
b) cai · sec-1 · cm-2 · ΚΓ4.
c) J · sec"1 - nf2. КГ*.

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 ofjet 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 °C to degrees kelvin, Fahrenheit, and Rankine.

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

н2о 5.0

1.16 Fill in the following table.

Mass fraction Amount-of-substance fraction Vol. fraction, %
22.0 %
CaC03 18.0%
MgC03 23.0 %
30.0 %
CaO 7.0 %
NaCl

K2SO4