SECOND EDITION
PROCESS ENGINEERING and
DESIGN USING VISUAL BASIC®
ARUN DATTA
© 2010 Taylor & Francis Group, LLC
SECOND EDITION
PROCESS ENGINEERING and
DESIGN USING VISUAL BASIC®
© 2010 Taylor & Francis Group, LLC
SECOND EDITION
PROCESS ENGINEERING and
DESIGN USING VISUAL BASIC®
ARUN DATTA
Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
© 2010 Taylor & Francis Group, LLC
CRC Press
Taylor & Francis Group
6000 Broken Sound Parkway NW, Suite 300
Boca Raton, FL 33487-2742
© 2014 by Taylor & Francis Group, LLC
CRC Press is an imprint of Taylor & Francis Group, an Informa business
No claim to original U.S. Government works
Version Date: 20130715
International Standard Book Number-13: 978-1-4398-6281-0 (eBook - PDF)
This book contains information obtained from authentic and highly regarded sources. Reasonable efforts
have been made to publish reliable data and information, but the author and publisher cannot assume
responsibility for the validity of all materials or the consequences of their use. The authors and publishers
have attempted to trace the copyright holders of all material reproduced in this publication and apologize to
copyright holders if permission to publish in this form has not been obtained. If any copyright material has
not been acknowledged please write and let us know so we may rectify in any future reprint.
Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented,
including photocopying, microfilming, and recording, or in any information storage or retrieval system,
without written permission from the publishers.
For permission to photocopy or use material electronically from this work, please access www.copyright.
com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood
Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and
registration for a variety of users. For organizations that have been granted a photocopy license by the CCC,
a separate system of payment has been arranged.
Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used
only for identification and explanation without intent to infringe.
Visit the Taylor & Francis Web site at
http://www.taylorandfrancis.com
and the CRC Press Web site at
http://www.crcpress.com
To my late mother Smt. Narayani Datta
© 2010 Taylor & Francis Group, LLC
Contents
Preface...............................................................................................................xxv
Acknowledgments....................................................................................... xxvii
Author..............................................................................................................xxix
Chapter 1 Basic mathematics........................................................................ 1
Introduction......................................................................................................... 1
Physical constants............................................................................................... 1
SI prefixes........................................................................................................ 1
Mensuration......................................................................................................... 1
Triangles.......................................................................................................... 1
Rectangles........................................................................................................ 2
Parallelogram (opposite sides parallel)....................................................... 2
Rhombus (equilateral parallelogram)......................................................... 3
Trapezoid (four sides, two parallel)............................................................. 3
Quadrilateral (four sided)............................................................................. 4
Regular polygon of n sides............................................................................ 4
Circle................................................................................................................ 4
Ellipse............................................................................................................... 6
Parabola........................................................................................................... 6
Prism................................................................................................................ 6
Pyramid........................................................................................................... 7
Right circular cylinder................................................................................... 7
Sphere............................................................................................................... 7
Right circular cone......................................................................................... 8
Dished end...................................................................................................... 8
Irregular shape............................................................................................... 8
Trapezoidal rule.............................................................................................. 8
Simpson’s rule................................................................................................. 8
Irregular volume............................................................................................ 9
Algebra................................................................................................................. 9
Factoring.......................................................................................................... 9
Arithmetic progression................................................................................. 9
Geometric progression................................................................................ 10
© 2010 Taylor & Francis Group, LLC
vii
viii
Contents
Infinite series (in GP)................................................................................... 10
Best-fit straight line (least squares method)............................................. 10
Binomial equation........................................................................................ 11
Polynomial equation.................................................................................... 11
Maxima/minima.......................................................................................... 12
Cubic equation.............................................................................................. 13
General procedure.............................................................................. 13
Matrix............................................................................................................. 16
Addition and multiplication of matrices......................................... 16
Addition of matrices........................................................................... 16
Multiplication of matrices................................................................. 16
Matrix properties involving addition.............................................. 17
Matrix properties involving multiplication.................................... 17
Matrix properties involving addition and
multiplication...................................................................................... 18
Transpose............................................................................................. 18
Symmetric matrix............................................................................... 18
Diagonal matrix.................................................................................. 19
Determinants................................................................................................ 19
Properties of determinants............................................................... 19
Cofactor................................................................................................ 21
Determinant and inverses................................................................. 21
Adjoint.................................................................................................. 21
Cramer’s rule................................................................................................. 22
Trigonometry..................................................................................................... 24
Functions of circular trigonometry........................................................... 24
Periodic functions........................................................................................ 25
Magic identity............................................................................................... 25
Addition formulas........................................................................................ 25
Double angle and half angle formulas...................................................... 26
Product and sum formulas......................................................................... 27
Relations between angles and sides of triangles..................................... 28
Law of sines................................................................................................... 28
Law of tangents............................................................................................ 28
Law of cosines............................................................................................... 28
Other relations.............................................................................................. 29
Inverse trigonometric functions................................................................. 29
Hyperbolic functions................................................................................... 30
Other hyperbolic functions......................................................................... 31
Inverse hyperbolic functions...................................................................... 31
Analytical geometry......................................................................................... 32
Straight line................................................................................................... 32
Straight line through two points...................................................... 32
© 2010 Taylor & Francis Group, LLC
Contents
ix
Three points on one line.................................................................... 32
Circle.............................................................................................................. 33
Tangent................................................................................................. 33
Normal................................................................................................. 33
Four points on a circle........................................................................ 34
Circle through three points............................................................... 34
Conic section................................................................................................. 35
Focus..................................................................................................... 35
Eccentricity.......................................................................................... 35
Directrix............................................................................................... 35
Partial derivatives............................................................................... 35
Parabola................................................................................................ 36
Tangent line with a given slope, m................................................... 38
Ellipse................................................................................................... 38
Hyperbola............................................................................................ 40
Calculus.............................................................................................................. 42
Differential calculus..................................................................................... 42
Understanding the derivatives................................................................... 43
Standard derivatives.................................................................................... 44
Integral calculus........................................................................................... 45
Volume of horizontal dished end..................................................... 45
Volume of vertical dished end.......................................................... 47
Standard integrals........................................................................................ 48
Differential equations....................................................................................... 49
First-order differential equations............................................................... 49
Separation of variables....................................................................... 50
Second-order differential equations.......................................................... 50
Bessel function.................................................................................... 51
Partial differential equations.......................................................................... 52
Laplace transform........................................................................................ 59
Standard Laplace transforms..................................................................... 60
Fourier half-range expansions.................................................................... 61
Fourier half-range cosine series................................................................. 61
Fourier half-range sine series..................................................................... 61
Numerical analysis........................................................................................... 63
Solving linear equations (Newton’s method)........................................... 63
Newton’s method in two variables............................................................ 64
Numerical methods in linear algebra....................................................... 66
Gauss elimination............................................................................... 66
Cholesky method................................................................................ 67
Numerical integration................................................................................. 69
Trapezoidal rule.................................................................................. 69
Simpson’s rule..................................................................................... 70
© 2010 Taylor & Francis Group, LLC
x
Contents
Double integration using Simpson’s rule................................................. 72
Numerical solution of first-order differential equations........................ 73
Euler’s method.................................................................................... 73
Improved Euler’s method........................................................................... 73
Runge–Kutta method......................................................................... 74
Second-order differential equations.......................................................... 76
Runge–Kutta–Nystrom method....................................................... 76
Partial differential equations...................................................................... 77
Heat conduction problem.................................................................. 78
Numerical solution....................................................................................... 79
Alternating direction implicit method............................................ 81
Unit conversions................................................................................................ 86
Programming.................................................................................................... 86
General notes for all programs.................................................................. 86
Vessel.............................................................................................................. 86
Program limitations..................................................................................... 94
Horizontal..................................................................................................... 95
Data entry...................................................................................................... 95
Inclined................................................................................................ 97
Vertical................................................................................................. 99
Conversion................................................................................................... 100
Program limitations................................................................................... 101
Procedure........................................................................................... 101
References........................................................................................................ 102
Chapter 2 Thermodynamics..................................................................... 103
Introduction..................................................................................................... 103
Heat, work, and energy.................................................................................. 103
Force............................................................................................................. 103
Kinetic and potential energy.................................................................... 104
First law of thermodynamics........................................................................ 104
Phase rule......................................................................................................... 105
Reversible process........................................................................................... 105
Heat content or enthalpy................................................................................ 106
Heat capacity at constant volume and constant pressure ........................ 106
Isothermal process.......................................................................................... 107
Adiabatic process............................................................................................ 107
Equation of state.............................................................................................. 108
Boyle’s law and Charles’s law................................................................... 108
Equation of state for real gas.....................................................................110
Comparison between PR and SRK EOSs.................................................110
Acentric factor..............................................................................................110
Vapor pressure of pure components....................................................... 112
Vapor pressure of water.............................................................................115
© 2010 Taylor & Francis Group, LLC
Contents
xi
Vapor pressure calculation using EOSs...................................................116
Second law of thermodynamics.....................................................................118
Carnot’s cycle...............................................................................................119
Entropy........................................................................................................ 120
Sensible heat................................................................................................ 121
Thermodynamic properties........................................................................... 122
Isobaric specific heat of hydrocarbon ideal gases................................. 122
Isobaric specific heat of hydrocarbon real gases................................... 124
Isobaric specific heat of hydrocarbon gas mixtures.............................. 126
Joule–Thomson coefficient........................................................................ 128
Isobaric specific heat of ideal liquids....................................................... 131
Isobaric specific heat of real liquids........................................................ 131
Enthalpy of gases....................................................................................... 132
Enthalpy of gas mixtures.......................................................................... 134
Entropy of ideal gases................................................................................ 137
Entropy of real gases................................................................................. 138
Fugacity correction..................................................................................... 139
Entropy of hydrocarbon gas mixtures.................................................... 140
Viscosities of ideal liquids..........................................................................141
Viscosity of water....................................................................................... 142
Viscosity of ideal hydrocarbon vapors.................................................... 142
Liquid viscosity of defined mixtures at low pressure.......................... 142
Vapor viscosity of defined mixtures at low pressure........................... 143
Thermal conductivity of pure hydrocarbon liquids at low
pressure....................................................................................................145
Thermal conductivity of pure hydrocarbon vapors at low
pressure....................................................................................................147
Flash calculation.............................................................................................. 148
Vapor–liquid equilibrium......................................................................... 148
Programming.................................................................................................. 150
Calculation of JT effect due to drop in pressure.................................... 150
Nomenclature.................................................................................................. 151
Greek characters......................................................................................... 152
References........................................................................................................ 152
Chapter 3 Fluid mechanics....................................................................... 153
Introduction..................................................................................................... 153
Bernoulli’s theorem......................................................................................... 153
Velocity heads............................................................................................. 155
Flow measurements........................................................................................ 156
Orifice/Venturi meter................................................................................ 156
Thermal expansion factor (Fa)................................................................... 158
Coefficient of discharge (CD)..................................................................... 158
Orifice meter...................................................................................... 158
© 2010 Taylor & Francis Group, LLC
xii
Contents
Venturi meter.............................................................................................. 159
Expansion factor (Y)................................................................................... 159
Orifices............................................................................................... 160
Nozzles and Venturi......................................................................... 160
Nonrecoverable pressure drop................................................................. 160
Orifices............................................................................................... 160
Venturi with 15° divergent angle................................................... 160
Venturi with 7° divergent angle..................................................... 160
Critical flow..................................................................................................161
Thickness of flow element..........................................................................162
Thickness of restriction orifice..................................................................162
Area meter: Rotameters............................................................................. 165
Flow through an open channel................................................................ 165
V notch......................................................................................................... 166
Rectangular notch...................................................................................... 166
Frictional pressure drop..................................................................................167
Darcy equation............................................................................................167
Flow in open channel................................................................................ 168
Estimation of friction factor...................................................................... 168
Friction factor: Laminar flow.......................................................... 168
Friction factor: Turbulent flow........................................................ 169
Two-K method................................................................................................. 169
K for reducer/expander............................................................................. 171
Reducer............................................................................................... 171
Expander............................................................................................ 171
Pipe entrance............................................................................................... 172
Pipe exit....................................................................................................... 172
Split flow...................................................................................................... 172
Split 1,3............................................................................................... 172
Split 1,2............................................................................................... 172
Split 3,1............................................................................................... 172
Split 1,2,3............................................................................................ 173
Split 1,3,2............................................................................................ 173
Split 3,1,2............................................................................................. 173
Hydraulics: General guidelines.....................................................................174
Roughness of pipe wall..............................................................................174
Control valve CV.........................................................................................174
Line sizing criteria for liquid lines.......................................................... 175
Line sizing for gravity flow lines..............................................................176
Downpipe sizing.........................................................................................176
Line sizing criteria for vapor lines........................................................... 177
Relief valve inlet line sizing...................................................................... 178
Relief valve outlet line sizing................................................................... 178
Line sizing criteria for two-phase flow................................................... 178
© 2010 Taylor & Francis Group, LLC
Contents
xiii
Hydraulics: Compressible fluids................................................................... 179
Adiabatic flow in a pipe............................................................................ 179
Isothermal flow in a pipe.......................................................................... 181
Heat loss........................................................................................................... 182
Types of cross-country buried pipelines................................................ 183
Yellow jacket............................................................................................... 183
Coating thickness............................................................................. 183
Fusion-bonded epoxy coating.................................................................. 183
Rate of heat transfer................................................................................... 184
Film resistance (Rfilm)........................................................................ 184
Resistance of pipe (Rpipe)................................................................... 186
Resistance of coatings (Rcoating)......................................................... 186
Resistance of environment (Renv).................................................... 187
Viscosity of water....................................................................................... 189
Thermal conductivity of water................................................................. 190
Viscosity of air............................................................................................ 190
Thermal conductivity of air...................................................................... 190
Choked flow................................................................................................ 190
Limiting differential pressure.................................................................. 191
Limiting expansion factor (Y)................................................................... 191
Hydraulics: Two-phase flow.......................................................................... 193
Beggs and Brill correlations...................................................................... 195
Step 1: Estimation of flow regime................................................... 195
Step 2: Estimation of horizontal holdup........................................ 195
Step 3: Estimation of uphill holdup............................................... 196
Step 4: Estimation of downhill holdup.......................................... 197
Step 5: Estimation of friction factor................................................ 197
Step 6: Estimation of pressure drop............................................... 197
Mukherjee and Brill correlations............................................................. 198
Step 1: Estimation of flow regime................................................... 198
Step 2: Estimation of holdup........................................................... 199
Step 3: Estimation of hydrostatic head.......................................... 200
Step 4: Estimation of acceleration head......................................... 200
Step 5: Estimation of friction factor................................................ 201
Step 6: Estimation of frictional pressure drop.............................. 201
CO2 corrosion.............................................................................................. 203
CO2 corrosion mechanism........................................................................ 203
NACE requirements................................................................................... 204
Rate of corrosion......................................................................................... 204
NORSOK model................................................................................ 204
Corrosion 93 model.................................................................................... 210
Corrosion 95 model.....................................................................................211
Programming.................................................................................................. 213
Program for flow elements........................................................................ 213
© 2010 Taylor & Francis Group, LLC
xiv
Contents
General overview.............................................................................. 213
Project details.....................................................................................214
Calculation form................................................................................214
Program limitations and notes................................................................. 215
Program for hydraulic calculations......................................................... 219
General overview.............................................................................. 219
Project details.................................................................................... 220
Program limitations and notes................................................................. 221
Form incompressible fluid............................................................... 221
Form compressible fluid.................................................................. 224
Pressure drop comparison.............................................................. 227
Form for two-phase flow................................................................. 227
Program for corrosion calculations......................................................... 230
General............................................................................................... 231
NORSOK model................................................................................ 232
Calculation of pH: NORSOK model.............................................. 232
Calculation of shear stress: NORSOK model............................... 233
Corrosion 93/95 model..................................................................... 233
Nomenclature.................................................................................................. 236
Greek characters......................................................................................... 239
References........................................................................................................ 240
Chapter 4 Heat transfer............................................................................. 243
Introduction..................................................................................................... 243
Conductive heat transfer................................................................................ 243
Heat conduction through a composite wall................................................ 244
Heat conduction through multiple cylindrical walls................................ 245
Heat conduction through the wall of a sphere........................................... 247
Multidimensional steady-state heat conduction......................................... 248
Rectangular coordinates........................................................................... 248
Cylindrical coordinates............................................................................. 248
Spherical coordinates................................................................................. 248
Conduction shape factors.......................................................................... 248
One-dimensional unsteady heat conduction.............................................. 250
Rectangular coordinates........................................................................... 250
Cylindrical coordinates............................................................................. 250
Spherical coordinates................................................................................. 251
Thermal conductivity of various materials/components......................... 256
Thermal conductivities of hydrocarbon liquids......................................... 256
Thermal conductivity of water (0–100°C)............................................... 264
Convective heat transfer................................................................................ 264
Free or natural convection........................................................................ 264
Free convection outside pipes and immersed body.................... 264
Free convection to air....................................................................... 265
© 2010 Taylor & Francis Group, LLC
Contents
xv
Heat-transfer coefficient for immersed bodies...................................... 266
Gas quenching.................................................................................. 266
Forced convection....................................................................................... 269
Forced convection inside the tube........................................................... 269
Forced convection outside the tube......................................................... 276
Shell-side cross flow area................................................................. 276
Estimation of pressure drop.......................................................................... 280
Shell-side pressure drop............................................................................ 280
Estimation of friction factor............................................................ 281
Tube-side pressure drop............................................................................ 281
Log mean temperature difference................................................................ 283
Overall heat-transfer coefficient.................................................................... 284
Fouling resistance...................................................................................... 285
Extended surface........................................................................................ 285
Fin efficiency..................................................................................... 285
Longitudinal fins.............................................................................. 290
Impact of heat-transfer coefficient on fin efficiency.................... 291
Circular fins....................................................................................... 291
Rectangular fins................................................................................ 292
Film coefficient for finned tube................................................................ 292
Radiation heat transfer................................................................................... 293
Emissivity and absorptivity..................................................................... 293
Blackbody radiation................................................................................... 294
Emissivity of commonly used materials................................................ 294
Radiation shape factor............................................................................... 294
Parallel, equal rectangle................................................................... 294
Parallel, equal, coaxial disks........................................................... 296
Perpendicular rectangles with a common edge........................... 297
Finite, coaxial cylinders................................................................... 297
Parallel, coaxial disks....................................................................... 298
Radiation shield for large surface area................................................... 299
Double-pipe heat exchanger.......................................................................... 303
Heat exchanger nomenclature.................................................................. 307
Standard tube pattern................................................................................ 308
Tube dimensions........................................................................................ 308
Minimum unsupported tube span...........................................................310
Heat exchanger specification......................................................................... 315
Batch heating and cooling............................................................................. 315
Batch cooling, internal coil........................................................................318
Batch heating, internal coil........................................................................318
Batch cooling, counterflow external heat exchanger.............................318
Batch heating, counterflow external heat exchanger............................ 319
Batch cooling, 1−2 multipass external heat exchanger.......................... 319
Batch heating, 1–2 multipass external heat exchanger......................... 320
© 2010 Taylor & Francis Group, LLC
xvi
Contents
Heat transfer in agitated vessels................................................................... 320
Viscosity correction.................................................................................... 321
Film coefficient inside the coil.................................................................. 322
Minimum metal temperature during depressuring operation................ 325
Programming.................................................................................................. 329
Program for double pipe heat exchanger................................................ 329
Program limitations and notes................................................................. 329
Checking Example 4.10 (double-pipe exchanger)........................ 329
Checking Example 4.11 (double-pipe finned exchanger)............ 329
Checking Example 4.13 (batch heating)........................................ 332
Checking Example 4.14 (batch cooling)......................................... 332
Nomenclature.................................................................................................. 334
Subscripts.................................................................................................... 335
Greek............................................................................................................ 335
References........................................................................................................ 336
Chapter 5 Distillation................................................................................ 337
Introduction..................................................................................................... 337
Relative volatility............................................................................................ 337
Vapor–liquid equilibrium.............................................................................. 338
Raoult’s law: Ideal solutions.......................................................................... 339
Material balance for two-component systems............................................ 340
Operating lines........................................................................................... 341
Reflux ratio.................................................................................................. 342
Minimum reflux ratio................................................................................ 342
Feed plate..................................................................................................... 343
McCabe–Thiele method................................................................................. 344
Smoker equations............................................................................................ 346
Approximate column sizing.......................................................................... 348
Sieve tray..................................................................................................... 348
Active hole area................................................................................. 350
Packed column.................................................................................. 350
Tray efficiency.................................................................................................. 352
Murphree tray efficiencies........................................................................ 353
Overall column efficiency......................................................................... 353
Prediction of efficiency.............................................................................. 355
Number of gas-phase transfer units........................................................ 356
Number of liquid-phase transfer units................................................... 356
Mixing factor............................................................................................... 356
Prediction of vapor diffusivity................................................................. 358
Prediction of liquid diffusivity................................................................ 358
Column hydraulics and design..................................................................... 362
Tray pressure drop.......................................................................................... 362
Sieve tray..................................................................................................... 362
© 2010 Taylor & Francis Group, LLC
Contents
xvii
Dry pressure drop............................................................................ 363
Orifice coefficient (Co)....................................................................... 363
Weir liquid crest (how)........................................................................ 363
Residual head (hr)....................................................................................... 364
Valve tray..................................................................................................... 365
Dry pressure drop............................................................................ 365
Downcomer design.................................................................................... 367
Downcomer backup......................................................................... 367
Downcomer residence time............................................................. 368
Flow regimes............................................................................................... 369
Spray regime...................................................................................... 369
Froth regime...................................................................................... 369
Emulsion regime............................................................................... 369
Bubble regime................................................................................... 370
Pressure drop through packing............................................................... 370
Estimation of pressure drop........................................................... 370
Packing factor.................................................................................... 371
Height equivalent to theoretical plate.......................................................... 373
Entrainment......................................................................................................374
Weeping and dumping.................................................................................. 376
Programming.................................................................................................. 379
Program for Smoker equations................................................................ 379
References........................................................................................................ 381
Chapter 6 Separators.................................................................................. 383
Introduction..................................................................................................... 383
General principles of separation................................................................... 383
Droplet in a vertical vessel........................................................................ 383
Droplet in a horizontal vessel................................................................... 386
Gravity settling: Limiting conditions...................................................... 386
Newton’s law............................................................................................... 387
Stokes’ law................................................................................................... 387
Intermediate law........................................................................................ 387
Critical particle diameter.......................................................................... 387
Vertical vs. horizontal separators................................................................. 388
Advantages of the horizontal separator.................................................. 389
Disadvantages of the horizontal separator............................................. 389
Advantages of the vertical separator....................................................... 389
Disadvantages of the vertical separator.................................................. 389
Design of a gas–liquid separator.................................................................. 389
Critical settling velocity............................................................................ 389
Design constant, KD.................................................................................... 390
API 521 method.......................................................................................... 391
Design of liquid–liquid separators............................................................... 392
© 2010 Taylor & Francis Group, LLC
xviii
Contents
Mist eliminator................................................................................................ 394
Wire mesh mist eliminator....................................................................... 394
Efficiency of the mist eliminator.................................................... 394
Inertial parameter (K)....................................................................... 395
Maximum gas velocity..................................................................... 395
Corrected pad-specific surface area (SO)...................................... 395
Impaction efficiency factor (E)........................................................ 396
Pressure drop of the mist eliminator...................................................... 397
Vane-type mist eliminator........................................................................ 397
Efficiency of vane pack.............................................................................. 398
Terminal centrifugal velocity................................................................... 398
Pressure drop through the vane pack..................................................... 398
General dimensions and setting of levels................................................... 399
The horizontal separator........................................................................... 399
Boot..................................................................................................... 402
Vertical separator....................................................................................... 403
Separator internals.......................................................................................... 404
Inlet nozzle.................................................................................................. 404
Vortex breaker............................................................................................. 405
Separator control............................................................................................. 406
Pressure and flow control......................................................................... 406
Light liquid-level control........................................................................... 407
Heavy phase liquid-level and slug control............................................. 407
High-performance separator......................................................................... 407
Salient features of GLCC........................................................................... 408
Design parameters.......................................................................................... 409
Flow rates.................................................................................................... 409
Slug length .................................................................................................. 409
Density..........................................................................................................411
Viscosity........................................................................................................411
Oil in gas droplet size................................................................................ 413
Oil in water droplet size............................................................................ 413
Water in oil droplet size............................................................................ 413
Inlet nozzle velocity................................................................................... 413
Gas outlet nozzle velocity..........................................................................414
Liquid outlet velocity..................................................................................414
Separator program...........................................................................................414
Program limitations/notes.........................................................................414
Horizontal separators................................................................................ 415
Three-phase flooded weir............................................................... 415
Three-phase nonflooded-weir separator....................................... 415
Three phase with boot separator.................................................... 415
Two-phase vapor–liquid separator................................................ 415
Two-phase liquid–liquid separator................................................ 415
© 2010 Taylor & Francis Group, LLC
Contents
xix
Vertical separators.......................................................................................416
Two-phase vapor–liquid separator.................................................416
Two-phase liquid–liquid separator.................................................416
General overview of the separator.exe program....................................416
Design.......................................................................................................... 421
Slug volume................................................................................................. 421
Further checking and analysis................................................................. 422
Design.......................................................................................................... 423
Analysis....................................................................................................... 424
Nomenclature.................................................................................................. 425
Greek characters......................................................................................... 426
References........................................................................................................ 427
Chapter 7 Overpressure protection........................................................ 429
Introduction..................................................................................................... 429
Impact on plant design.............................................................................. 429
Impact on individual design.................................................................... 429
Definition......................................................................................................... 430
Accumulation.............................................................................................. 430
Atmospheric discharge............................................................................. 431
Built-up back pressure............................................................................... 431
General back pressure............................................................................... 431
Superimposed back pressure................................................................... 432
Balanced-bellows PRV............................................................................... 432
Blowdown.................................................................................................... 432
Closed discharge system........................................................................... 432
Cold differential test pressure.................................................................. 432
Conventional PRV...................................................................................... 432
Design capacity........................................................................................... 432
Design pressure.......................................................................................... 433
Maximum allowable accumulated pressure.......................................... 433
Maximum allowable working pressure.................................................. 433
Operating pressure.................................................................................... 433
Overpressure.............................................................................................. 433
Pilot-operated PRV..................................................................................... 434
Pressure relief valve................................................................................... 434
Pressure safety valve................................................................................. 434
Rated relieving capacity............................................................................ 434
Relief valve.................................................................................................. 434
Relieving conditions.................................................................................. 434
Rupture disk............................................................................................... 435
Safety relief valve....................................................................................... 435
Safety valve................................................................................................. 435
Set pressure................................................................................................. 435
© 2010 Taylor & Francis Group, LLC
xx
Contents
Vapor depressuring system...................................................................... 436
Vent stack..................................................................................................... 436
Types of pressure relief valves...................................................................... 436
Conventional pressure relief valve (vapor service)............................... 436
Conventional pressure relief valve (liquid service)............................... 438
Balanced-bellows pressure relief valve................................................... 439
Pilot-operated pressure relief valve......................................................... 440
Rupture disk............................................................................................... 443
Selection of pressure relief valves................................................................ 445
Conventional pressure relief valve.......................................................... 445
Balanced-bellows pressure relief valve................................................... 445
Pilot-operated pressure relief valve......................................................... 446
Rupture disk............................................................................................... 446
PRV installation and line sizing................................................................... 447
Compressors and pumps.......................................................................... 447
Fired heaters................................................................................................ 448
Heat exchangers......................................................................................... 448
Piping........................................................................................................... 448
Pressure vessels.......................................................................................... 449
PRV isolation valves................................................................................... 449
Inlet piping to PRVs................................................................................... 451
Discharge piping from PRVs.................................................................... 453
Contingency quantification........................................................................... 454
General......................................................................................................... 454
Power failure............................................................................................... 455
Local power failure.......................................................................... 456
Failure of a distribution center....................................................... 456
Total power failure........................................................................... 457
Cooling water failure................................................................................. 457
Instrument air failure................................................................................ 458
Steam failure............................................................................................... 460
Total steam failure............................................................................ 460
Loss of steam to specific equipment.............................................. 460
Partial steam failure......................................................................... 460
Check valve failure.................................................................................... 460
Blocked outlet............................................................................................. 461
Pump or compressor discharge...................................................... 461
Multiple outlet................................................................................... 462
Block valve downstream of control valve..................................... 462
Control valve failure.................................................................................. 462
Vapor breakthrough................................................................................... 463
Maximum flow........................................................................................... 465
Thermal relief............................................................................................. 465
Modulus of elasticity of pipe material (E)..................................... 468
© 2010 Taylor & Francis Group, LLC
Contents
xxi
Coefficient of linear thermal expansion (α).................................. 468
Valve leakage rate (q)........................................................................ 469
Compressibility of liquid (Z)........................................................... 469
Coefficient of cubic expansion of liquids (β)................................. 469
Installation of thermal relief valve................................................. 471
Fire exposure.............................................................................................. 471
General guidelines........................................................................... 472
Estimation of wetted surface area.................................................. 472
Fire circle.............................................................................................474
Estimation of latent heat and physical properties........................474
Liquid wet vessel.............................................................................. 475
Vessels with only gas....................................................................... 477
Two liquid phases............................................................................. 479
Heat exchanger tube rupture................................................................... 480
Contingency calculation.................................................................. 482
Reflux failure and overhead system........................................................ 484
Loss of reboiler heat................................................................................... 485
Venting of storage tanks............................................................................ 485
Venting due to liquid movements.................................................. 486
Thermal venting............................................................................... 486
Fire exposure..................................................................................... 486
Minimum flow area......................................................................... 488
Sizing procedure............................................................................................. 489
Sizing of liquid relief................................................................................. 489
Sizing of vapor relief.................................................................................. 491
Critical flow....................................................................................... 491
Subcritical flow.................................................................................. 492
Conventional and pilot-operated PRV........................................... 492
Balanced-bellows PRV..................................................................... 492
Sizing for steam relief................................................................................ 493
Sizing for two-phase fluids....................................................................... 494
Type 1 (omega method).................................................................... 496
Type 2 (omega method).................................................................... 499
Type 3 (integral method)................................................................. 501
Design of flare stack................................................................................... 503
Minimum distance..................................................................................... 504
Fraction of heat intensity transmitted (τ)................................................ 504
Fraction of heat radiated (F)...................................................................... 505
Heat release (Q)........................................................................................... 505
Sizing of a flare stack: Simple approach................................................. 505
Calculation of stack diameter......................................................... 505
Calculation of flame length............................................................. 506
Flame distortion caused by wind velocity.................................... 506
Sizing of flare stack: Brzustowski and Sommer approach................... 508
© 2010 Taylor & Francis Group, LLC
xxii
Contents
Calculation of flare stack diameter................................................ 508
Location of flame center xc, yc.......................................................... 509
Lower explosive limit of mixtures................................................. 509
Vertical distance (yc)..........................................................................511
Horizontal distance (xc).....................................................................511
SIL analysis.......................................................................................................514
Definitions................................................................................................... 515
Diagnostic coverage......................................................................... 515
Final element..................................................................................... 515
MooN.................................................................................................. 515
Programmable electronics............................................................... 515
Programmable electronic system................................................... 515
Protection layer................................................................................. 515
Safety-instrumented function......................................................... 515
Safety-instrumented systems...........................................................516
Safety integrity...................................................................................516
Safety integrity level.........................................................................516
Safety life cycle...................................................................................516
Matrix for SIL determination....................................................................516
Probability of failure on demand............................................................. 517
ALARP model............................................................................................. 519
Determination of SIL........................................................................ 519
Financial...................................................................................................... 521
Health and safety....................................................................................... 522
Environment and asset.............................................................................. 522
Programming.................................................................................................. 524
Program for pressure relief valve............................................................ 524
Program limitations and notes....................................................... 525
General overview.............................................................................. 525
Project details.................................................................................... 525
File save.............................................................................................. 526
File open............................................................................................. 526
File print............................................................................................. 527
Exit...................................................................................................... 527
Specific message or warning: back pressure................................ 527
Back-pressure correction factor...................................................... 527
Pilot-operated PRV........................................................................... 527
Liquid................................................................................................. 528
Vapor.................................................................................................. 529
Two-phase type 1 calculation......................................................... 529
Two-phase type 2 calculation......................................................... 529
Two-phase type 3 calculation......................................................... 529
Program for flare stack estimation.......................................................... 531
Program limitations and notes....................................................... 531
© 2010 Taylor & Francis Group, LLC
Contents
xxiii
Specific message/warning.............................................................. 532
Nomenclature.................................................................................................. 533
Greek characters......................................................................................... 533
References........................................................................................................ 534
Chapter 8 Glycol dehydration.................................................................. 537
Introduction..................................................................................................... 537
Basic scheme.................................................................................................... 537
Advantages.................................................................................................. 539
Disadvantages............................................................................................. 539
Pre-TEG coalescer....................................................................................... 539
Contactor..................................................................................................... 539
Flash separator............................................................................................ 540
Filters............................................................................................................ 540
Pumping...................................................................................................... 540
Glycol/glycol exchanger............................................................................ 541
Gas/glycol exchanger................................................................................. 541
Regenerator................................................................................................. 541
Physical properties.......................................................................................... 542
Selection of type of glycol......................................................................... 542
Common properties of glycol................................................................... 543
Densities of aqueous glycol solutions............................................ 543
Solubility of various compounds................................................... 543
Fire hazard information............................................................................ 543
Viscosities of aqueous glycol solutions.......................................... 543
Specific heats of aqueous glycol solutions.................................... 543
Thermal conductivities of aqueous glycol solutions................... 543
Design aspects................................................................................................. 544
Water content in hydrocarbon gas........................................................... 544
Equilibrium dew point.............................................................................. 545
Minimum lean-TEG concentration.......................................................... 548
Number of theoretical stages of the contactor....................................... 550
Design of contactor.................................................................................... 551
Type of internals............................................................................... 554
Liquid distributor............................................................................. 562
Flash separator............................................................................................ 563
Filters............................................................................................................ 564
Particulate filter................................................................................. 564
Carbon filter....................................................................................... 564
Glycol/glycol exchanger............................................................................ 564
Gas/glycol exchanger................................................................................. 565
Regenerator................................................................................................. 566
Still column........................................................................................ 567
Reboiler........................................................................................................ 569
© 2010 Taylor & Francis Group, LLC
xxiv
Contents
Fire tube heat density....................................................................... 569
Fire tube heat flux............................................................................. 570
Lean-glycol storage.................................................................................... 570
Energy exchange pump............................................................................. 571
Burner management.................................................................................. 573
Specifications.............................................................................................. 578
Programming.................................................................................................. 578
Program limitations................................................................................... 582
General overview....................................................................................... 582
File menu............................................................................................ 582
Unit menu.......................................................................................... 583
Project details.................................................................................... 584
Data entry.......................................................................................... 584
References........................................................................................................ 585
© 2010 Taylor & Francis Group, LLC
Preface
After publishing the first edition in 2007, I received many suggestions
from professionals who encouraged me to bring out the second edition. Considering the suggestions received from end users and to make
the book more useful, I have added three more chapters to the book:
Thermodynamics, Heat Transfer, and Distillation. I have also developed
the following additional programs in Visual Basic®.
Calculation of JT effect due to drop in pressure
Double-pipe heat exchanger design
Batch heating and cooling calculation
Metal temperature calculation
Design of distillation column using Smoker equations
With an additional 5 programs, the total number of programs has
increased to 14.
Most of the existing Visual Basic® programs have been modified to
make them more user-friendly; however, the possibilities of program bugs
cannot be totally eliminated.
Arun Datta
Brisbane, Australia
© 2010 Taylor & Francis Group, LLC
xxv
Acknowledgments
I wish to express my gratitude to all the professionals who encouraged
me to write the second edition of this book with additional chapters.
Special thanks to my wife, Dr. Nivedita Datta, and my daughter, Raka
Datta, whose support and encouragement were the key factors in the
­completion of this work.
I am also grateful to Anthony Buckley and Malcolm Rough of
WorleyParsons for reviewing some chapters of the book and p
­ roviding
useful suggestions. I also appreciate the contribution I received from
Dr. Pramod Mathur of Technip who reviewed some chapters and provided valuable feedback. I extend my gratitude to Haruo Kikkawa of
WorleyParsons for reviewing and checking the work-out examples of the
new chapters.
I also acknowledge the help received from the Tubular Exchanger
Manufacturers Association and Pergamon Press for permitting me to
publish relevant figures. Special thanks are also due to the editorial and
production staff of CRC Press and Techset Composition, particularly Syed
Mohamad Shajahan, Florence Kizza, and Allison Shatkin, for publishing
this work with outstanding quality.
I will thankfully acknowledge any suggestion to further improve
future editions.
Arun Datta
Brisbane, Australia
© 2010 Taylor & Francis Group, LLC
xxvii
Author
Arun Datta a lead facilities engineer working with Santos, has more
than 30 years experience in the field of process engineering and design.
He holds a master’s degree in chemical engineering from the Indian
Institute of Technology, Delhi, and has worked with several process consultancy organizations both in India and Australia.
Dr. Datta has been a consultant for a large number of process engineering organizations, including refineries, oil and gas industries, fine
chemicals, and pharmaceuticals. His fields of expertise include heat and
mass transfer, process simulations, exchanger design, pressure vessel
design, design of safety systems, and design of control systems. Some
of his clients include BP Refinery, Caltex Refinery, Santos, ExxonMobil,
ONGC, Indian Oil, Incitec, and Oil Search.
He is a chartered professional engineer in Australia (Queensland
chapter) as well as a member of the Institute of Engineers, Australia.
© 2010 Taylor & Francis Group, LLC
xxix
chapter one
Basic mathematics
Introduction
Understanding mathematics is the most fundamental requirement in
understanding engineering. Some fields in chemical engineering require
the solution of complex mathematical equations. The purpose of this
chapter is to get some idea in the field of mathematics commonly used for
chemical engineering design and includes the following:
• Physical constants
• Mensuration
• Algebra
• Trigonometry
• Analytical geometry
• Calculus
• Differential equations
• Partial differential equations
• Numerical analysis
• Equation of states
• Unit conversions
• Programming
Physical constants
The commonly used physical constants are presented in Table 1.1.
SI prefixes
The International System of Units (SI) prefixes are presented in Table 1.2.
Mensuration
Triangles
Area = 1/2 bh
(1.1)
where b = base and h = altitude.
© 2010 Taylor & Francis Group, LLC
1
2
Process engineering and design using visual basic®
Table 1.1 Commonly Used Physical Constants
Name
Symbol
Unit
Value
Speed of light
Planck’s constant
c
h
hbar
G
G
e
me
mp
mn
mH
amu
NA
k
eV
a
sigma
cm/s
erg⋅s
erg⋅s
cm3/(g⋅s2)
Nm2/kg2
ESU
g
g
g
g
g
2.99792458E10
6.6260755E−27
1.05457266E−27
6.67259E−8
6.67259E−11
4.8032068E−10
9.1093897E−28
1.6726231E−24
1.6749286E−24
1.6733E−24
1.6605402E−24
6.0221367E23
1.380658E−16
1.6021772E−12
7.5646E−15
5.67051E−5
2.1798741E−11
8.31439
1.98719
0.0820567
3.1415926536
2.7182818285
Gravitational constant
Electron charge
Mass of electron
Mass of proton
Mass of neutron
Mass of hydrogen
Atomic mass unit
Avogadro’s number
Boltzmann constant
Electron volt
Radiation density constant
Stefan–Boltzmann constant
Rydberg constant
Gas constant
Pi
Napierian (natural)
logarithm base
Euler’s constant
Logarithm conversion
Logarithm conversion
Radian
Degree
Minute
Second
R
R
R
π
e
erg/k
erg
erg/(cm3⋅K4)
erg/(cm2⋅K4⋅s)
erg
J/(mol⋅K)
cal/(mol⋅K)
l.atm/(mol⋅K)
Y
log x
ln x
0.5772156649
0.4342944819 ln x
2.302585093 log x
57.2957795131°
0.0174532925 rad
0.0002908882 rad
0.0000048481 rad
Rectangles
Area = ab
(1.2)
where a and b are the lengths of the sides.
Parallelogram (opposite sides parallel)
Area = ah = ab sin α
© 2010 Taylor & Francis Group, LLC
(1.3)
Chapter one: Basic mathematics
3
Table 1.2 SI Prefixes
Multiplication factor
1 000 000 000 000 000 000
1 000 000 000 000 000
1 000 000 000 000
1 000 000 000
1 000 000
1 000
100
10
0.1
0.01
0.001
0.000 001
0.000 000 001
0.000 000 000 001
0.000 000 000 000 001
0.000 000 000 000 000 001
b
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
10
1015
1012
109
106
103
102
101
10−1
10−2
10−3
10−6
10−9
10−12
10−15
10−18
18
Prefix
Symbol
exa
peta
tera
giga
mega
kilo
hecto
deka
deci
centi
milli
micro
nano
pico
femto
atto
E
P
T
G
M
k
h
da
d
c
m
μ
n
p
f
a
h
α
a
Figure 1.1 Parallelogram.
where a and b are the lengths of the sides, h the height, and α the angle
between the sides. See Figure 1.1.
Rhombus (equilateral parallelogram)
Area = 1/2 ab
(1.4)
where a and b are the lengths of the diagonals.
Trapezoid (four sides, two parallel)
Area = 1/2(a + b)h
(1.5)
where a and b are the lengths of the parallel sides, and h is the height.
© 2010 Taylor & Francis Group, LLC
4
Process engineering and design using visual basic®
Quadrilateral (four sided)
Area = 1/2 ab sin θ
(1.6)
where a and b are the lengths of the diagonals, and θ is the acute angle
between them.
Regular polygon of n sides (refer to Figure 1.2)
Area = 1/4 nL2 cot(180°/n)
(1.7a)
R = L/2 csc(180°/n)
(1.7b)
r = L/2 cot(180°/n)
(1.7c)
β = 360°/n
(1.7d)
θ = (n − 2)180°/n
(1.7e)
L = 2r tan(β/2) = 2R sin(β/2)
(1.7f)
Circle (refer to Figure 1.3)
Let
C = circumference
r = radius
D = diameter
A = area
S = arc length subtended by θ
L = chord length subtended by θ
H = maximum rise of arc above chord, r − H = d
θ = central angle (rad) subtended by arc S
R
r
β
θ
Figure 1.2 Regular polygon (n = 5).
© 2010 Taylor & Francis Group, LLC
L
Chapter one: Basic mathematics
5
S
H
θ
r
d
D
Figure 1.3 Circle.
Then
C = 2πr = πD
(1.8a)
S = rθ = 1/2 Dθ
(1.8b)
L = 2 r 2 − d 2 = 2r sin(θ/2) = 2d tan(θ/2)
(1.8c)
d = 1/2 4r 2 − L2 = 1/2L cot(θ/2)
(1.8d)
θ = S/r = 2cos−1(d/r) = 2sin−1(L/D)
(1.8e)
A(circle) = πr2 = 1/4 πD2
(1.8f)
A(sector) = 1/2 rS = 1/2 r2θ
(1.8g)
A(segment) = A(sector) − A(triangle) = 1/2 r2 (θ − sin θ)
(1.8h)
(r − H )
− (r − H )(2rH − H 2 )0.5
r
(1.8i)
A(segment ) (φ/360°)(π/4) − ( h − h 2 )(0.5 − h)
=
A(total)
π/4
(1.8j)
= r 2cos −1
where
h = H/D, h ≤ 0.5
ϕ = 2{90 − sin−1(1 − 2h)}
© 2010 Taylor & Francis Group, LLC
6
Process engineering and design using visual basic®
b
a
Figure 1.4 Ellipse.
F
y
E
G
x
Figure 1.5 Parabola.
Ellipse (refer to Figure 1.4)
Area = πab
(1.9a)
Circumference = 2π{(a2 + b2)/2}0.5 (approximately)
(1.9b)
Parabola (refer to Figure 1.5)
Length of arc EFG =
(4x 2 + y 2 ) +
{(
y2
ln 2x +
2x
Area of section EFG = 4/3 xy
4x2 + y 2
) y} (1.10a)
(1.10b)
Prism
Lateral surface area = (perimeter of right section) * (lateral edge) (1.11a)
Volume = (area of base) * (altitude)
© 2010 Taylor & Francis Group, LLC
(1.11b)
Chapter one: Basic mathematics
7
Pyramid
Lateral area of a regular pyramid = 1/2 (perimeter of base) * (slant height)
(1.12a)
= 1/2 (number of sides) * (length of one side) * (slant height)
Volume = 1/3 (area of base) * (altitude)
(1.12b)
(1.12c)
Right circular cylinder
Lateral surface area = 2π (radius) * (altitude)
(1.13a)
Volume = π (radius)2 * (altitude)
(1.13b)
Sphere (refer to Figure 1.6)
Area (sphere) = 4πR 2 = πD2
(1.14a)
Area (zone) = 2πRh = πDh
(1.14b)
Volume (sphere) = 4/3πR3 = 1/6πD3
(1.14c)
Volume (spherical sector) = 2/3πR2 h = 1/6πh1(3r22 + h12 )
(1.14d)
Volume (spherical segment of one base) = 1/6πh1 (3r22 + h12 )
(1.14e)
Volume (spherical segment of two bases) = 1/6πh (3r12 + 3r22 + h22 ) (1.14f)
r2
r1
h1
h2
R
h
D
Figure 1.6 Sphere.
© 2010 Taylor & Francis Group, LLC
8
Process engineering and design using visual basic®
Right circular cone
Curved surface area = πr(r2 + h2)0.5
(1.15a)
Volume = 1/3 πr2h
(1.15b)
Surface area (general) = πa2 + π/2(b2/e)ln{(1 + e)/(1 − e)}
(1.16a)
where
r = radius
h = height of the cone
Dished end
Surface area (ellipsoidal, b = a/2) = πa2 + (πa2/8e)ln{(1 + e)/(1 − e)} (1.16b)
= 4.336 a2
Surface area (hemispherical, b = a) = 2πa2
(1.16c)
Volume (general) = 2/3 πa2b
(1.16d)
Volume (ellipsoidal) = 1/3 πa3
(1.16e)
Volume (hemispherical) = 2/3 πa3
(1.16f)
where
a = semimajor axis
b = semiminor axis
e = eccentricity = (1 − b2/a2)0.5
Irregular shape
Let y0, y1, y2, . . ., yn be the lengths of a series of equally spaced parallel
chords and h be the distance between them. The approximate area of the
figure is given by using the trapezoidal rule or by Simpson’s rule.
Trapezoidal rule
Area = h/2 {(y0 + yn) + 2(y1 + y2 + y3 + ⋯ + yn−1)}
(1.17)
Simpson’s rule
Area = h/3 {(y0 + yn) + 4(y1 + y3 + y5 + ⋯ + yn−1)
+ 2(y2 + y4 + ⋯ + yn−2)}
(1.18)
where n is even.
The greater the value of n, the greater the accuracy of approximation.
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
9
Irregular volume
To find the irregular volume, replace y’s by cross-sectional area Aj, and use
the results in the preceding equation.
Algebra
Factoring
a2 − b2 = (a + b)(a − b)
(1.19a)
a3 − b3 = (a − b)(a2 + ab + b2)
(1.19b)
a3 + b3 = (a + b)(a2 − ab + b2)
(1.19c)
an − bn = (a − b)(an−1 + an−2b + an−3b2 + ⋯ + bn−1)
(1.19d)
(a + b)2 = a2 + 2ab + b2
(1.19e)
(a − b)2 = a2 − 2ab + b2
(1.19f)
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
(1.19g)
(a − b − c)2 = a2 + b2 + c2 − 2ab − 2ac + 2bc
(1.19h)
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(1.19i)
(a − b)3 = a3 − 3a2b + 3ab2 − b3
(1.19j)
(a + b)n = an + nan−1b + {n(n − 1)/2!}an−2b2
+ {n(n − 1)(n − 2)/3!}an−3b3 + ⋯ + bn
(1.19k)
Arithmetic progression
A series is said to be in arithmetic progression (AP) if the algebraic difference between any two successive terms is the same throughout the series.
The following series is in AP:
a, a + b, a + 2b, a + 3b, . . ., nth term
where
a = first term
b = common difference
The nth term can be defined as
tn = a + (n − 1)b
© 2010 Taylor & Francis Group, LLC
(1.20a)
10
Process engineering and design using visual basic®
Sum of the series
S = (n/2){2a + (n − 1)b}
(1.20b)
Geometric progression
A series is said to be in geometric progression (GP) if the ratio of any
term to the preceding one is the same throughout the series. The following series is in GP:
a, ar, ar2, ar3, . . ., nth term
where
a = first term
r = common ratio
The nth term can be defined as
tn = arn−1
(1.21a)
Sum of the series
S=
a(1 − r n )
(1 − r )
(1.21b)
Infinite series (in GP)
When r < 1, the sum of the series = a/(1 − r)
(1.21c)
Example
Series 1 + 1/3 + 1/9 + 1/27 + ⋯ + 1/∞
The sum = a/(1 − r) = 3/2
Best-fit straight line (least squares method)
The straight line y = a + bx should be fitted through the given points (x1,y1),
(x2,y2), . . ., (xn,yn) so that the sum of the squares of the distances of those
points from the straight line is minimum, where the distances are measured in the vertical direction (y-direction).
The values of a and b are calculated from the following equations:
an + bΣxj = Σyj
© 2010 Taylor & Francis Group, LLC
(1.22a)
Chapter one: Basic mathematics
11
and
aΣx j + bΣx 2j = Σx j y j
(1.22b)
where n is the number of points.
Example 1.1
Let the four points be (−1,1), (−0.1,1.099), (0.2,0.808), and (1,1).
n=4
Σx j = 0.1
Σx 2j = 2.05
Σy j = 3.907
Σx j y j = 0.0517
4 a + 0.1b = 3.907
0.1a + 2.05b = 0.0517
The solution is
a = 0.9773
b = −0.0224
Therefore, the equation of the best-fit straight line is
y = 0.9773 − 0.0224x.
Binomial equation
The general equation is
ax2 + bx + c = 0
(1.23a)
General solution of Equation 1.23a
x=
−b ± b 2 − 4 ac
2a
(1.23b)
Polynomial equation
General solution of the following type of polynomial equation:
y = ax2 + bx + c
© 2010 Taylor & Francis Group, LLC
(1.24a)
12
Process engineering and design using visual basic®
If three values (x1,y1), (x2,y2), and (x3,y3) are known, then
a=
( y1 − y 2 )( x2 − x3 ) − ( y 2 − y 3 )( x1 − x2 )
( x12 − x22 )( x2 − x3 ) − ( x22 − x32 )( x1 − x2 )
(1.24b)
{( y − y ) − a(x − x )}
b=
(1.24c)
c = y1 − ax12 − bx1
(1.24d)
1
2
2
1
2
2
( x1 − x2 )
Example 1.2
Solve the equation y = ax2 + bx + c for three sets of values (1,10), (3,32),
and (5,70).
SOLUTION
a=
(10 − 32)(3 − 5) − (32 − 70)(1 − 3)
=2
(1 − 9)(3 − 5) − (9 − 25)(1 − 3)
b=
{(10 − 32) − 2(1 − 9)} = 3
−2
c = 10 − 2 − 3 = 5
Therefore, the general solution will be y = 2x2 + 3x + 5.
Maxima/minima
Let a function be
y = f(x)
(1.25a)
If a solution for x at y′= 0 exists, then there will be either a maximum
value or a minimum value for the function.
At a value of x, where y′ = 0
If y″ is negative, then there will be a maximum value for the function.
If y″ is positive, then there will be a minimum value for the function.
Example 1.3A
Let the equation be
y = ax + b
y′ = a = 0
© 2010 Taylor & Francis Group, LLC
(1.25b)
Chapter one: Basic mathematics
13
No solution exists at y′ = 0; this indicates that there will be no
maxima or minima for the function.
This is the equation of a straight line.
Example 1.3B
Let the equation be
y = x3 + x − 1
(1.25c)
2
y′ = 3x + 1 = 0
(1.25d)
or x2 = −1/3
(1.25e)
No general solution exists for Equation 1.25e, meaning that Equation
1.25c will have no general maxima or minima.
Example 1.3C
Let the equation be y = 4x3 − 3x + 2
(1.25f)
y′ = 12x − 3 = 0
2
x = ±1/2
At x = 1/2, y″ = 12 and at x = −1/2, y″ = −12.
Therefore, at x = 1/2, the value of y will be minimum. The value
of y = 1.
At x = −1/2, the value of y will be maximum. The value of y = 3.
The nature of this graph is such that for x values less than −1/2,
the value of y will decrease continuously, and for x values more than
1/2, the value of y will increase continuously.
Cubic equation
An accurate solution of a cubic equation is very important in process engineering calculations. The commonly used Newton’s method (discussed
later) may not be adequate to establish all the roots of a cubic equation.
One root is always real and the other two roots can be real or imaginary
depending on the equation.
General procedure
Let the equation be
ax3 + bx2 + cx + d = 0
(1.26)
The following factors are calculated as
 3c  b  2 
p =  −   3
 a  a  
© 2010 Taylor & Francis Group, LLC
(1.27a)
14
Process engineering and design using visual basic®
9bc 27 d 
  b 
q = 2   − 2 +
 27


a
a 
 a
3
3
 p
 q
∆ =  + 
 3
 2
2
(1.27b)
(1.27c)
If Δ < 0, then all three roots will be real.
Otherwise, one root will be real and the other roots will be imaginary.
When Δ < 0, all roots are real
|p|3 
v=

 27 
0.5
 −q 
φ = cos −1  
 2v 
|p|
w= 
3
0.5
(1.28)
(1.29)
(1.30)
y1 = 2w cos(ϕ/3)
(1.31a)
 φ + π
y 2 = −2w cos 
 3 
(1.31b)
 φ − π
y 3 = −2w cos 
 3 
(1.31c)
The final roots will be
x1 = y1 −
b
3a
(1.32a)
x2 = y 2 −
b
3a
(1.32b)
x3 = y 3 −
b
3a
(1.32c)
q
+ ∆ 0.5
2
(1.33a)
When Δ >= 0, only one root is real
u1 = −
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
15
If u1 < 0, then
u = − u1
1/3
(1.33b)
Else
u = u11/3
k1 = −
(1.33c)
q
− ∆ 0.5
2
(1.34a)
1/3
(1.34b)
If k1 < 0, then
k = − k1
Else
k = k11/3
(1.34c)
y1 = u + k
(1.35)
b
3a
(1.36)
The final root will be
x1 = y1 −
Example 1.4A
Solve the equation
x3 − x2 + 0.1x − 0.002 = 0
SOLUTION
p = −0.2333
q = −0.04274
Δ = −0.000013
Since the value is negative, there will be three real roots
v = 0.02169
ϕ = 0.1722
w = 0.27889
y1 = 0.55685
y2 = −0.250715
y3 = −0.30614
The final roots will be
x1 = 0.89019
x2 = 0.08262
x3 = 0.02719
© 2010 Taylor & Francis Group, LLC
16
Process engineering and design using visual basic®
Example 1.4B
Solve the equation
3x3 − x2 + 0.1x − 0.002 = 0
SOLUTION
p = −0.0037
q = 0.000294
Δ = 1.9E−8
Since the value is positive, there will be one real root
u1 = −0.00000656
u = −0.01872
k1 = −0.000287
k = −0.06596
y1 = −0.08468 The final real root will be
x1 = 0.02643
Matrix
Addition and multiplication of matrices
• If two m by n matrices A and B are given, the sum A + B can be
defined as the m by n matrix, adding corresponding elements, that is
(A + B) [i,j] = A[i,j] + B[i,j]
(1.37a)
• If a matrix A and a number c are given, the multiplication of the
matrix can be defined as
cA = cA[i,j]
(1.37b)
Addition of matrices
a
i. 
d
b
e
c  u
+
f   x
v
y
w a + u
=
z   d + x
a
ii. 
d
b
e
c  a
+
f   d
b
e
c   2a
=
f   2d
b+v
e+y
2b
2e
Multiplication of matrices
a
iii. 
c
b  e
*
d   g
f   ae + bg
=
h   ce + dg
© 2010 Taylor & Francis Group, LLC
af + bh 
cf + dh 
c + w
f + z 
2c 
a
= 2

2f
d
b
e
c
f 
Chapter one: Basic mathematics
a
iv. 
d
b
e
17
x
c     ax + by + cz 
*
y =
f     dx + ey + fz 
 z 
The multiplication of two matrices is well defined only if the number
of columns of the first matrix is the same as the number of rows of the second matrix. If A is an m by n matrix (m rows, n columns) and B is an n by
p matrix (n rows, p columns), then their product AB is the m by p matrix
(m rows, p columns) given by (AB)[i,j] = A[i,1] * B[1,j] + A[i,2] * B[2,j] + ⋯ +
A[i,n] * B[n,j] for each pair of i and j.
Multiplication has the following properties:
• (AB)C = A(BC) for all k by m matrices A, m by n matrices B, and n by
p matrices C (associativity).
• (A + B)C = AC + BC for all m by n matrices A and B and k by m matrices C (distributivity).
• In general, commutativity does not hold, meaning generally AB ≠ BA.
Matrix properties involving addition
Let A, B, and C be m by n matrices. Then
a. A + B = B + A
(1.38a)
b. (A + B) + C = A + (B + C)
(1.38b)
c. A + 0 = A
(1.38c)
where 0 is the m by n zero matrix (all its entries are equal to 0).
d. A + B = 0 if and only if B = −A
(1.38d)
Matrix properties involving multiplication
a. Let A, B, and C be three matrices. If the products AB, (AB)C, BC, and
A(BC) then
(AB)C = A(BC)
(1.39a)
b. If α and β are numbers and A is a matrix, then
α(βA) = (αβ)A
(1.39b)
c. If α is a number, and A and B are two matrices such that the product
AB is possible, then
α(AB) = (αA)B = A(αB)
© 2010 Taylor & Francis Group, LLC
(1.39c)
18
Process engineering and design using visual basic®
d. If A is an n by m matrix and 0 is the m by k zero matrix, then
A0 = 0
(1.39d)
Matrix properties involving addition and multiplication
a. Let A, B, and C be three matrices, then
and
(A + B)C = AC + BC
(1.40a)
A(B + C) = AB + AC
(1.40b)
b. If α and β are numbers and A and B are matrices, then
and
α(A + B) = αA + αB
(1.40c)
(α + β)A = αA + βA
(1.40d)
Transpose
The transpose of a matrix is another matrix, produced by turning rows
into columns and vice versa. The transpose of an n by m matrix A is the n
by m matrix AT defined by
AT[i,j] = A[j,i]
(1.41)
For the matrix
a
d
A=
g

j
c
f 
i

l
b
e
h
k
the transpose
a

A = b
 c
T
d
e
g
h
f
i
j

k
l 
Symmetric matrix
A symmetric matrix is a matrix equal to its transpose. A symmetric matrix
must be a square matrix, for example
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
a
b

19
a
b

and b
c 
 c
b
d
e
c

e
f 
Diagonal matrix
A diagonal matrix is a symmetric matrix with all of its entries equal to
zero, except possibly the values of the diagonal, for example
a
0

a
0

and 0

b
0
0
0
0
0

0
b 
Determinants
Determinants are important in both calculus and algebra.
The determinant of a 1 × 1 matrix is the element itself: det[a] = a.
The 2 × 2 matrix
a
A=
c
b
d 
has the determinant
det(A) = ad − bc
(1.42)
The 3 × 3 matrix
a

A = d
 g
b
e
h
c

f
i 
has the determinant
det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
(1.43)
Properties of determinants
1. Any matrix A and its transpose have the same determinant, meaning det A = det AT.
2. If the elements of one row (or column) of a determinant are all zero,
the value of the determinant is zero.
© 2010 Taylor & Francis Group, LLC
20
Process engineering and design using visual basic®
3. The determinant of a triangular matrix is the product of the entries
on the diagonal, that is
a
0

b  a
=
d  b
0
= ad
d 
4. If one determinant is obtained from another by interchanging any
two rows (or columns), the value of either is the negative of the value
of the other.
a
c

b
c
= −

d
a
d
b 
5. If the elements of one row (or column) of a determinant are multiplied by the same constant factor, the value of the determinant is
multiplied by this factor.
 na
c

nb 
a
= n
d 
c
b  a
=
d   nc
b
nd 
6. If two rows (or columns) of a determinant are identical, the value of
the determinant is zero.
7. If two determinants are identical except for one row (or column),
the sum of their values is given by a single determinant obtained by
adding corresponding elements of dissimilar rows (or columns) and
leaving the remaining elements unchanged.
3
1

2  4
+
5  7
2  7
=
5   8
2
= 19
5 
8. The value of a determinant is not changed if a constant multiple of
the corresponding elements of any other row (or column) is added to
the elements of any row (or column).
 a + nc
 c

b + nd   a
=
d   c
b  a
=
d  c + na


d + nb 
b
9. If all the elements but one in a row (or column) are zero, the value of
the determinant is the product of that element and its cofactor.
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
21
Cofactor
Let A be an n × n matrix. The ij-th cofactor of A, denoted by Aij, will be
Aij = (−1)i+j|Mij|
(1.44)
Example 1.5
Find the cofactors A32 and A24 of a 4 × 4 determinant A
−3
4
5
0
1
2
A=
1

 4
5
0
9
2
6
3 
−2 

7 
SOLUTION
1

A32 = (−1)3 + 2 M32 = −  2
 4
A24 = (−1)
2+ 4
1

1
 4
5
0
2
−3
5
0
6

3  = −8
7 
5

9  = −192
2 
Determinant and inverses
Theorem 1
If A is invertible, then det A ≠ 0 and
det A−1 = 1/det A
(1.45)
Adjoint
Let a determinant be
 A11
A
21
B=
 
 An1
A12
A22
An 2
...
...
...
A1n 
A2 n 


Ann 
(1.46)
Let A be an n × n matrix and let B, given in Equation 1.46, denote the
matrix of its cofactors. Then the adjoint of A, written as adjA, is the transpose of the n × n matrix B.
© 2010 Taylor & Francis Group, LLC
22
Process engineering and design using visual basic®
 A11
A
12
adjA = BT = 
 
 A1n
A21
A22
A2 n
...
...
...
An1 
An 2 


Ann 
(1.47)
0
0
detA
0
...
...
...
0
0 
0 
0  = (detA)I


detA 
(1.48)
Theorem 2
Let A be an n × n matrix. Then
 detA
 0

( A)( adjA) =  0

  0

0
detA
0
0
Theorem 3
Let A be an n × n matrix. Then A is invertible if and only if detA ≠ 0.
If detA ≠ 0, then
A −1 =
1
adjA
detA
(1.49)
Cramer’s rule
Consider a system of n equations and n unknowns.
a11x1 + a12x2 + ⋯ + a1nxn = b1
a21x1 + a22x2 + ⋯ + a2nxn = b2
⋮
⋮
⋮
(1.50a)
⋮
an1x1 + an2x2 + ⋯ + annxn = bn
which can be written in the form
Ax = b
(1.50b)
If detA ≠ 0, then Equation 1.50b will have a unique solution given by
x = b/A
© 2010 Taylor & Francis Group, LLC
(1.50c)
Chapter one: Basic mathematics
23
Let D be detA. The other matrix can be defined as
 b1
b
2
A1 = 


bn
a1n 
 a11

a
a2 n 
21
, A2 = 
 


ann 
 an1
...
...
...
a12
a22
an 2
b1
b2
bn
...
...
...
a1n 
a2 n 
, etc.


ann 
Let D1 = detA1, D2 = detA2 , . . ., Dn = detAn, then
x1 =
D1
D
D
, x2 = 2 , … , xn = n
D
D
D
Example 1.6
Solve using Cramer’s rule, the system
2x1 + 4x2 + 6x3 = 18
4x1 + 5x2 + 6x3 = 24
3x1 + x2 − 2x3 = 4
SOLUTION
The determinant is calculated first as
2

D = 4
 3
4
5
1
6

6  = 2(−10 − 6) − 4 ( −8 − 18 ) + 6 ( 4 − 15)
−2 
= −32 + 104 − 66 = 6
Because the value of D ≠ 0, the system will have a unique solution.
18

D1 =  24
 4
4
5
1
6

6  = 18(−10 − 6) − 4(−48 − 24)
−2 
+ 6 ( 24 − 20 ) = 24
2

D2 =  4
 3
18
24
4
6

6  = 2(−48 − 24) − 18(−8 − 18)
−2 
+ 6 (16 − 72) = −12
© 2010 Taylor & Francis Group, LLC
(1.51)
24
Process engineering and design using visual basic®
2

D3 =  4
 3
4
5
1
18 

24  = 2 ( 20 − 24 ) − 4 (16 − 72)
4 
+ 18 ( 4 − 15) = 18
Therefore
x1 = D1/D = 24/6 = 4
x2 = D2/D = −12/6 = −2
x3 = D3/D = 18/6 = 3
Trigonometry
Functions of circular trigonometry
Trigonometric functions are the ratios of various sides of the reference
angles presented in Figure 1.7.
The trigonometric relationships are:
Sine of α = sin α = p/h; secant of α = sec α = h/b
Cosine of α = cos α = b/h; cosecant of α = csc α = h/p
Tangent of α = tan α = p/b; cotangent of α = cot α = b/p
The magnitude and sign of trigonometric functions are presented in
Table 1.3.
h
p
b
h
p
α
b
b
p
α
α
b
h
h
Figure 1.7 Triangles.
Table 1.3 Magnitude of Trigonometric Functions
Function
sin α
csc α
cos α
sec α
tan α
cot α
0–90°
90–180°
180–270°
270–360°
+0 to +1
+∞ to +1
+1 to 0
+1 to +∞
+0 to +∞
+∞ to +0
+1 to +0
+1 to +∞
−0 to −1
−∞ to −1
−∞ to −0
−0 to –∞
0 to –1
−∞ to −1
−1 to −0
−1 to −∞
+0 to +∞
+∞ to +0
−1 to −0
−1 to −∞
+0 to +1
+∞ to +1
−∞ to −0
−0 to −∞
© 2010 Taylor & Francis Group, LLC
p
Chapter one: Basic mathematics
25
Periodic functions
The following periodic functions are commonly used in trigonometry:
cos(α + 2π) = cos α
(1.52a)
sin(α + 2π) = sin α
(1.52b)
cos(α + 2nπ) = cos α,
n = 0, ±1, ±2, . . .
(1.52c)
sin(α + 2nπ) = sin α,
n = 0, ±1, ±2, . . .
(1.52d)
Magic identity
The following identities are very basic to the analysis of trigonometric
expressions:
sin2 α + cos2 α = 1
(1.53a)
1 + tan2 α = sec2 α
(1.53b)
1 + cot2 α = csc2 α
(1.53c)
Addition formulas
The following addition formulas are very important for the analysis of
trigonometric expressions:
cos (a + b) = cos a cos b − sin a sin b
(1.54a)
cos(a − b) = cos a cos b + sin a sin b
(1.54b)
sin(a + b) = sin a cos b + cos a sin b
(1.54c)
sin(a − b) = sin a cos b − cos a sin b
(1.54d)
tan( a + b) =
tan a + tan b
1 − tan a tan b
(1.54e)
tan( a − b) =
tan a − tan b
1 + tan a tan b
(1.54f)
© 2010 Taylor & Francis Group, LLC
26
Process engineering and design using visual basic®
Example 1.7
Find the exact value of cos 165°, given that cos 120° = −1/2,
sin 120° = 3 /2, and 45° = sin 45° = 2 /2 .
SOLUTION
cos 165° = cos (120° + 45°)
= cos 120° cos 45° − sin 120° sin 45°
=
−1
2
3
2
−
*
*
2
2
2
2
=−
6+ 2
4
Double angle and half angle formulas
Double- and half-angle formulas are very useful in trigonometry. The following equations are widely used:
cos(2a) = cos2 a − sin2 a = 2 cos2 a − 1 = 1 − 2sin2 a
(1.55a)
sin(2a) = 2sin a cos a
(1.55b)
2 tan a
1 − tan 2 a
(1.55c)
tan(2a) =
1
 a
cos 2   = (1 + cos a)
 2
2
(1.55d)
1
 a
sin 2   = (1 − cos a)
 2
2
(1.55e)
sin a
 a  1 − cos a
tan   =
=
 2
sin a
1 + cos a
(1.55f)
Example 1.8
Use the half-angle formulas to find cos(π/8) and sin(π/8).
SOLUTION
Let a = π/4, then π/8 = a/2.
Now, from the half-angle formula, we have
1
π
1
2
 π
cos 2   =  1 + cos  =  1 +
 8


2
4
2
2 
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
27
Because 0 < π/8 < π/2, the value of cos(π/8) will be positive.
Therefore
 π
cos   =
 8
2+ 2
2
 π
sin   =
 8
2− 2
2
Similarly
Product and sum formulas
The following product and sum formulas are commonly used in
trigonometry:
cos a cos b =
1
{cos( a + b) + cos(a − b)}
2
(1.56a)
sin a sin b =
1
{cos( a − b) − cos(a + b)}
2
(1.56b)
sin a cos b =
1
{sin( a + b) + sin(a − b)}
2
(1.56c)
cos a sin b =
1
{sin( a + b) − sin(a − b)}
2
(1.56d)
 a + b
 a − b
cos a + cos b = 2 cos 
cos 
 2 
 2 
(1.56e)
 a + b
 a − b
cos a − cos b = −2 sin 
sin 
 2 
 2 
(1.56f)
 a + b
 a − b
sin a + sin b = 2 sin 
cos 
 2 
 2 
(1.56g)
 a + b
 a − b
sin a − sin b = 2 cos 
sin 
 2 
 2 
(1.56h)
© 2010 Taylor & Francis Group, LLC
28
Process engineering and design using visual basic®
β
c
r
h
a
γ
α
b
Figure 1.8 Triangle.
Relations between angles and sides of triangles
Refer to Figure 1.8.
Let a, b, and c be the sides of a triangle.
Let α, β, and γ be the angles opposite the sides in the triangle.
2s = a + b + c
A = area
r = radius of the inscribed circle
R = radius of the circumscribed circle
h = altitude
Law of sines
sin α/a = sin β/b = sin γ/c
(1.57)
a + b tan 1/2(α + β)
=
a−b
tan 1/2(α − β)
(1.58a)
b + c tan 1/2(β + γ )
=
b−c
tan 1/2(β − γ )
(1.58b)
a + c tan 1/2(α + γ )
=
a−c
tan 1/2(α − γ )
(1.58c)
a 2 = b 2 + c 2 − 2bc cos α
(1.59a)
b 2 = a 2 + c 2 − 2ac cosβ
(1.59b)
Law of tangents
Law of cosines
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
29
c 2 = a 2 + b 2 − 2ab cos γ
(1.59c)
Other relations
b 2 + c2 − a2
2bc
(1.60a)
a = bcos γ + ccos β
(1.60b)
2
s(s − a)(s − b)(s − c)
bc
(1.60c)
cos α =
sin α =
 α
sin   =
 2
(s − b)(s − c)
bc
 α
cos   =
 2
A=
a 2 sin β sin γ
1
1
bh = ab sin γ =
=
2
2
2 sin α
r=
s(s − a)
bc
(1.60d)
(1.60e)
s(s − a)(s − b)(s − c) = rs (1.60f)
(s − a)(s − b)(s − c)
s
(1.60g)
a
abc
=
2 sin α
4A
(1.60h)
R=
h = c sin α = a sin γ =
2rs
b
(1.60i)
Inverse trigonometric functions
General inverse trigonometric functions with the following restrictions:
−π/2 ≤ sin−1 x ≤ π/2
0 ≤ cos−1 x ≤ π
−π/2 ≤ tan−1x ≤ π/2
© 2010 Taylor & Francis Group, LLC
30
Process engineering and design using visual basic®
are
sin −1 x = cos −1 1 − x 2 = tan −1
= sec −1
1
1− x
2
= csc −1
cos −1 x = sin −1 1 − x 2 = tan −1
= csc −1
1
1− x
2
=
tan −1 x = sin −1
x
1 − x2
= cot −1
1 − x2
x
1
π
= − cos −1 x
x
2
(1.61a)
1 − x2
x
1
= cot −1
= sec −1
2
x
x
1− x
π
− sin −1 x
2
x
1+ x
2
(1.61b)
1
= cos −1
= sec −1 1 + x 2 = csc −1
1+ x
2
1 + x2
x
= cot −1
1
x
(1.61c)
Hyperbolic functions
The hyperbolic functions are similar to the trigonometric functions and
are presented as follows:
cosh x =
ex + e−x
2
(1.62a)
sinh x =
ex − e−x
2
(1.62b)
tanh x =
ex − e−x
ex + e−x
(1.62c)
coth x =
ex + e−x
ex − e−x
(1.62d)
sech x =
2
ex + e−x
(1.62e)
csch x =
2
e − e−x
(1.62f)
© 2010 Taylor & Francis Group, LLC
x
Chapter one: Basic mathematics
31
Other hyperbolic functions
cosh2 x − sinh2 x = 1
(1.63a)
sech2 x + tanh2 x = 1
(1.63b)
coth2 x − csch2 x = 1
(1.63c)
sinh (x ± y) = sinh x cosh y ± cosh x sinh y
(1.63d)
cosh (x ± y) = cosh x cosh y ± sinh x sinh y
(1.63e)
tanh 2x =
2s tanh x
1 + tanh 2 x
(1.63f)
Inverse hyperbolic functions
The inverse of sinh x can be established as follows:
Let y = sinh x
or 2y = ex − e−x
or 2yex = e2x − 1
or (ex)2 − 2y(ex) − 1 = 0
or e x = y ±
y2 + 1
Because ex > 0 for all x, and y 2 + 1 > y for all y, we can have the solution as
ex = y +
y2 + 1
(
y2 + 1
Therefore
x = ln y +
)
and
(
x2 + 1
)
(1.64a)
(
x2 − 1
)
(1.64b)
sinh −1 x = ln x +
Similarly
cosh −1 x = ln x +
tanh −1 x =
© 2010 Taylor & Francis Group, LLC
1 1+ x
ln
2 1− x
(1.64c)
32
Process engineering and design using visual basic®
coth −1 x =
1 x+1
ln
2 x−1
(1.64d)
 1 + 1 − x2 
sech −1x = ln 

x


(1.64e)
 1 + 1 + x2 
csch −1x = ln 

x


(1.64f)
Analytical geometry
Straight line
The general equation of a straight line is
ax + by + c = 0
(1.65)
If b ≠ 0, the slope is −a/b.
The intersection at the y-axis when x = 0 is −c/b.
Straight line through two points
It is possible to draw a straight line through two points. The general equation of a straight line through two points P1 (a1,b1) and P2 (a2,b2) can be
represented as
x

 a1
 a2
y
b1
b2
1

1 = 0
1
(1.66a)
or
x(b1 − b2) − y(a1 − a2) + (a1b2 − a2b1) = 0
(1.66b)
Three points on one line
Although it is possible to draw a straight line through two points, any
three points may not fall on a straight line. Three points P1 (a1,b1), P2 (a2,b2),
and P3 (a3,b3) can be on a straight line if and only if
 a1

 a2
 a3
b1
b2
b3
© 2010 Taylor & Francis Group, LLC
1

1 = 0
1
(1.67)
Chapter one: Basic mathematics
33
Circle
The equation of a circle with coordinates of the center as (a,b) and radius
r is
(x − a)2 + (y − b)2 = r2
(1.68a)
Expanding Equation 1.68a, the general equation of a circle can be
developed as
x2 + y2 + 2mx + 2ny + c = 0
(1.68b)
m 2 + n2 − c > 0
(1.68c)
with
The coordinates of the center of the circle are (−m, −n) and the radius is
r=
m2 + n2 − c
(1.68d)
Tangent
The tangent at any point (h,k) of a circle is defined to be a straight line that
meets the circle at the specified point but, when produced, does not cut it.
This tangent is always perpendicular to the radius drawn from the center
to the point of contact.
The equation of the tangent at a point (h,k) on a circle with center (a,b)
and radius r is given by
y−k =−
a−h
( x − h)
b−k
(1.69)
The slope of the line is −(a − h/b − k).
Normal
The normal at any point of a curve is the straight line that passes through
the point and is perpendicular to the tangent at that point. The general
equation of the normal with the coordinates of the center as (a,b) and that
of the point as (h,k) is
x

a
 h
y
b
k
© 2010 Taylor & Francis Group, LLC
1

1 = 0
1
(1.70)
34
Process engineering and design using visual basic®
Four points on a circle
Four points can be on a circle if the following condition is satisfied. Let
the four points be (a1,b1), (a2,b2), (a3,b3), and (a4,b4). The four points can be
on a circle if
 a12 + b12
 2
2
 a2 + b2
 2
2
 a3 + b3
 2
2
 a 4 + b4
a1
b1
a2
b2
a3
b3
a4
b4
1

1
=0
1

1
(1.71)
Example 1.9
Check if the four points (5,0), (4.33,2.5), (2.5, −4.33), and (−3.536, −3.536)
are on the same circle.
SOLUTION
The condition of the four points on the same circle is given in Equation
1.71. The value of the matrix will be
 25
 25

 25

 25
5
4.33
2.5
−3.536
 25

− 5  25
 25
0
2.5
−4.33
−3.536
2.5
−4.33
−3.536
1
 4.33
1

= 25  2.5
1
 −3.536

1
1
 25


1 − 1  25
 25
1
4.33
2.5
−3.536
2.5
−4.33
−3.536
1

1
1
2.5 

−4.33  = −1067 + 1067 = 0
−3.536 
Therefore, all the four points will be on the same circle.
Circle through three points
The general equation of a circle through the three points (a1,b1), (a2,b2),
and (a3,b3) (provided the three points do not fall on a straight line) can be
given as
x2 + y 2
 2
2
 a1 + b1
 2
2
 a2 + b2
 2
2
 a3 + b3
x
y
a1
b1
a2
b2
a3
b3
© 2010 Taylor & Francis Group, LLC
1

1
=0
1

1
(1.72)
Chapter one: Basic mathematics
35
Conic section
The locus of a point P that moves so that its distance from a fixed point
is always in a constant ratio to its perpendicular distance from a fixed
straight line is called a conic section.
Focus
The previously mentioned fixed point is called the focus and is usually
denoted by S.
Eccentricity
The constant ratio is called eccentricity and is denoted by e.
Directrix
The fixed straight line is called the directrix.
An algebraic curve with an equation of the form
ax2 + bxy + cy2 + dxz + eyz + fz2 = 0
(1.73)
is called a conic section.
The Cartesian equation of the conic section is
ax2 + bxy + cy2 + dx + ey + f = 0
(1.74)
Partial derivatives
The three partial derivatives of the equation of a conic section are
Fx′ (x,y,z) = 2ax + by + dz
(1.75a)
Fy′ (x,y,z) = bx + 2cy + ez
(1.75b)
Fz′ (x,y,z) = dx + ey + 2fz
(1.75c)
The matrix formed by the coefficients of x, y, and z is called Δ.
 2a

∆ = b
 d
b
2c
e
d

e 
2 f 
Table 1.4 characterizes the curve represented by the equation.
Example 1.10
Establish the type of curve for the equation x2 + 2xy + y2 + 3x −
4y + 6 = 0.
© 2010 Taylor & Francis Group, LLC
(1.76)
36
Process engineering and design using visual basic®
Table 1.4 Conic Section
b2 − 4ac < 0
Δ≠0
Δ=0
b2 − 4ac = 0
aΔ < 0
a ≠ c, an ellipse
a = c, a circle
aΔ > 0, no locus
Point
b2 − 4ac > 0
Parabola
Hyperbola
Two parallel lines if
Q = d2 + e2 − 4(a + c)f > 0;
one straight line if Q = 0;
no locus if Q < 0
Two intersecting
straightlines
SOLUTION
2

∆ = 2
 3
2
2
−4
3

−4  = −98 ≠ 0
12 
b2 − 4ac = 0
Therefore, the curve will be a parabola.
Parabola
As in Figure 1.9, let AA′ be a straight line and e be a point (e is not on line
AA′). The locus of all points B such that the distance of B from line AA′ is
equal to the distance of B from point e will form a parabola.
The point e is the focus and the line AA′ is the directrix.
The general equation of a parabola is
y2 = 4ax
(1.77)
A y
B
e
A′
Figure 1.9 Parabola.
© 2010 Taylor & Francis Group, LLC
x
Chapter one: Basic mathematics
37
Parametric equations of a parabola Let a and b be two lines in a plane
with equations
x = 4at2
(1.78a)
y = 4at
(1.78b)
respectively.
The real number t is a parameter.
Now, to obtain an equation of the curve, we eliminate the parameter
t from the two equations. Eliminating t from Equations 1.78a and 1.78b,
we get
 y
x = 4a  
 4a 
2
or
y2 = 4ax
which is the equation of the parabola.
Equation of tangent of a parabola
obtained as
The slope of a tangent line can be
2yy′ = 4a or slope y′ = 2a/y
The slope of the tangent line at a point (x0, y0) = 2a/y0.
The equation of the tangent line will be
2a
( x − x0 )
y0
(1.79a)
yy0 − 4ax0 = 2ax − 2ax0
(1.79b)
yy0 = 2a(x + x0)
(1.79c)
y − y0 =
or
or
Equation 1.79c is the equation of the tangent of the parabola y2 = 4ax
at point (x0, y0).
© 2010 Taylor & Francis Group, LLC
38
Process engineering and design using visual basic®
Tangent line with a given slope, m
For a parabola with equation y2 = 4ax, the equation of the tangent line with
a constant slope m can be presented by
y = mx +
a
m
(1.80)
Ellipse (refer to Figure 1.10)
Let there be two points F and F′ such that the distance F – F′ is always less
than a fixed positive value 2a.
The locus of all points A such that distance AF + AF′ = 2a is an ellipse.
Two points F and F′ are called the foci of the ellipse.
The general equation of an ellipse is
x2 y2
+ 2 =1
a2
b
(1.81)
where
a = semimajor axis
b = semiminor axis
Parametric equations of an ellipse Take two lines l and m in a plane
with the respective equations
x = a cos t
(1.82a)
y = b sin t
(1.82b)
The real number t is the parameter.
y
A
F
Figure 1.10 Ellipse.
© 2010 Taylor & Francis Group, LLC
F′
x
Chapter one: Basic mathematics
39
Now, to obtain an equation of the curve, we eliminate the parameter
t from the two equations. Eliminating t from Equations 1.82a and 1.82b,
we get
x/a = cos t
y/b = sin t
or
x2 y2
+ 2 = cos 2 t + sin 2 t = 1
a2
b
is the general equation of the ellipse.
General properties of an ellipse
Eccentricity The eccentricity of an ellipse is defined as
e=
1−
b2
, e<1
a2
(1.83)
Focus The coordinates of the focus are (±ae, 0).
ellipse is
Tangent to a Point P of an Ellipse
The general equation of an
x2 y2
+ 2 =1
a2
b
The slope of the tangent line can be obtained by differentiating
implicitly
2x 2 yy ′
+ 2 =0
a2
b
(1.84a)
or
y′ = −
b2x
a2 y
(1.84b)
The equation of the tangent line at point (x0, y0) is
y − y0 = −
b 2 x0
( x − x0 )
a2 y0
(1.84c)
or
x0 x y 0 y
+ 2 =1
a2
b
© 2010 Taylor & Francis Group, LLC
(1.84d)
40
Process engineering and design using visual basic®
y
F
P
F′
x
Figure 1.11 Hyperbola.
Normal at a point P
(x0, y0) can be given as
The equation of a normal at a point P
( y − y o )b 2
( x − x o )a 2
=
xo
yo
(1.85)
Hyperbola (refer to Figure 1.11)
Let there be two points F and F′ and a strictly positive value 2a. The locus
of all points P such that the absolute difference between PF′ and PF is 2a
will be a hyperbola.
The general equation of a hyperbola is
x2 y2
−
=1
a2 b 2
(1.86)
Parametric equations of a hyperbola Take two lines l and m in a plane
with the respective equations
x = a sec t
(1.87a)
y = b tan t
(1.87b)
The real number t is the parameter.
Now, to obtain an equation of the curve, we eliminate the parameter
t from the two equations. Eliminating t from Equations 1.87a and 1.87b,
we get
x/a = sec t
y/b = tan t
or
x2 y2
−
= sec 2 t − tan 2 t = 1
a2 b 2
is the general equation of the hyperbola.
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
41
General properties of a hyperbola
Eccentricity The eccentricity of a hyperbola is defined as
e=
1+
b2
, e>1
a2
(1.88)
Focus The coordinates of the focus are (±ae, 0).
Tangent to a point P of a hyperbola The general equation of a
hyperbola is
x2 y2
−
=1
a2 b 2
The slope of the tangent line can be obtained by differentiating
implicitly
2x 2 yy ’
− 2 =0
a2
b
(1.89a)
or
y’ =
b2x
a2 y
(1.89b)
The equation of the tangent line at point (x0, y0) is
y − y0 =
b 2 x0
( x − x0 )
a2 y0
(1.89c)
or
x0 x y 0 y
− 2 =1
a2
b
(1.89d)
Normal at a point P The equation of a normal at a point P (x0, y0) can
be given as
( y − y0 )b 2
( x − x 0 )a 2
=−
x0
y0
© 2010 Taylor & Francis Group, LLC
(1.90)
42
Process engineering and design using visual basic®
Calculus
Differential calculus
A function f is said to be differentiable if the following limit exists:
lim
h→ 0
f ( x + h) − f ( x )
h
(1.91)
If this limit exists, it is called the derivative of f at x and is denoted
by f′(x).
Example 1.11
Estimate the derivative of 1/x.
SOLUTION
The derivative can be established based on Equation 1.91 as
1
1
−
f ( x + h) − f ( x )
−1
x
+
h
x
=
=
h
h
x( x + h)
Therefore
f ’( x) = lim
h→0
f ( x + h) − f ( x )
−1
−1
= lim
= 2
h → 0 x( x + h)
h
x
provided x ≠ 0.
Example 1.12
If f(x) = 5 − x3, establish the value of f′(2).
SOLUTION
f ′ ( 2) = lim
h→0
f (2 + h) − f (2)
h
= lim
{5 − (2 + h) } − {5 − 2 }
= lim
{5 − (8 + 12h + 6h + h )} + 3
3
h→0
3
h
2
h→0
h
= lim (−12 − 6 h − h 2 )
h→ 0
= −12
© 2010 Taylor & Francis Group, LLC
3
Chapter one: Basic mathematics
43
Understanding the derivatives
The purpose of derivatives can be explained by using the following
example.
Example 1.13
Estimate the derivative of f(x) = 2x at x = 0.
SOLUTION (SEE FIGURE 1.12)
The slope of line BA = (20 − 2−1)/1 = 1/2 and the slope of line
AC = (21 − 20)/1 = 1. Therefore, the derivative at x = 0 will be in
between 1/2 and 1.
Now, for small values of h, Table 1.5 presents the value of 2h with
the values of the difference quotients.
From Table 1.5, it is clear that difference quotients calculated for
negative h values are less than calculated for positive h values, and the
derivative is in between 0.693123 and 0.693171. If we use a value up to
three decimal places, f′(0) = 0.693.
2
C
1.5
A
B
1
Tangent line
slope = f ′(0)
0.5
–1
0
1
Figure 1.12 Graph y = 2x.
Table 1.5 Solution of Example 1.13
h
−0.003
−0.002
−0.001
0
0.001
0.002
0.003
2h
Difference quotient: (2h − 1)/h
0.99979
0.99986
0.99993
1
1.00007
1.00014
1.00021
0.693075
0.693099
0.693123
© 2010 Taylor & Francis Group, LLC
0.693171
0.693195
0.693219
44
Process engineering and design using visual basic®
Therefore, the equation of the tangent at point A will be
y = 0.693x + 1 (because the intercept is at 1)
Standard derivatives
d(xn) = n xn−1 dx
d(ex) = ex dx
d(ax) = ax ln a dx
d(ln x) = (1/x) dx
d(log x) = {log (e/x)} dx
d(sin x) = cos x dx
d(cos x) = −sin x dx
d(tan x) = sec2 x dx
d(cot x) = −csc2 x dx
d(sec x) = tan x sec x dx
d(csc x) = −csc x cot x dx
d(sin−1 x) = (1 − x2)−1/2 dx
d(cos−1 x) = −(1 − x2)−1/2 dx
d(tan−1 x) = (1 + x2)−1 dx
d(cot−1 x) = −(1 + x2)−1 dx
d(sec−1 x) = x−1 (x2 − 1)−1/2 dx
d(csc−1 x) = −x−1 (x2 − 1)−1/2 dx
d(sinh x) = cosh x dx
d(cosh x) = sinh x dx
d(tanh x) = sech2 x dx
d(coth x) = −csch2 x dx
d(sech x) = −sech x coth x dx
d(sinh−1 x) = (x2 + 1)−1/2 dx
d(cosh−1 x) = (x2 − 1)−1/2 dx
d(tanh−1 x) = (1 − x2)−1 dx
d(coth−1 x) = −(x2 − 1)−1 dx
d(sech−1 x) = x(1 − x2)−1/2 dx
d(csch−1 x) = −x−1 (x2 + 1)−1/2 dx
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
45
Integral calculus
Suppose f is continuous for a ≤ t ≤ b. The definite integral of f from a to b,
written as
b
∫ f (t) dt
(1.92)
a
is the limit of the left- or right-hand sums with n subdivisions of [a, b], as
n gets arbitrarily large. In other words
b

 n −1
f (t) dt = lim(left-hand sum) = lim 
f (ti )∆t
n →∞
n →∞

 i=0
a
(1.93a)

 n
f (t) dt = lim(right-hand sum) = lim 
f (ti )∆t
n →∞
n →∞

 i =1
a
(1.93b)
∑
∫
and
b
∑
∫
Each of these sums is called a Riemann sum, f is called the integrand,
and a and b are called the limits of integration.
Volume of horizontal dished end (refer to Figure 1.13)
The volume of the strip is
dV = π/2 * y * d/2 * dx
(1.94)
b
d
2a
dV
dx
x
(0,0)
(x – a)2/a2 + y2/b2 = 1
Figure 1.13 Volume of horizontal dished end.
© 2010 Taylor & Francis Group, LLC
a–x
a
x
46
Process engineering and design using visual basic®
The volume up to a height of x is
V =
π
2
x=x
1
∫ 2 dydx
(1.95)
x=0
(d/2)2 = a2 − (a − x)2 = 2ax − x2
(1.96a)
y2/b2 = 1 − (x − a)2/a2 = (2ax − x2)/a2
(1.96b)
and
Therefore, Equation 1.94 will become
πb
V =
2a
=
x=x
∫ (2ax − x ) dx
2
x=0
πb  2 x 3 
ax − 
2a 
3
(1.97)
Equation 1.97 is the general equation of the horizontal dished end
volume.
Total volume (x = 2a)
The total volume of the dished end
V T = 2/3 πba2
Elliptical dished end (b = a/2)
(1.98)
The general equation of volume
V = π/4 (ax2 − x3/3)
(1.99a)
The total volume (x = 2a)
V T = π/3 a3
Hemispherical dished end (b = a)
(1.99b)
The general equation of volume
V = π/2 (ax2 − x3/3)
(1.100a)
The total volume (x = 2a)
V T = 2/3 πa3
© 2010 Taylor & Francis Group, LLC
(1.100b)
Chapter one: Basic mathematics
47
Volume of vertical dished end (refer to Figure 1.14)
The equation of the curve with a shifted coordinate is
x 2 ( y − b )2
+
=1
a2
b2
(1.101)
The volume up to a distance y from the bottom will be
y=y
V =
π
∫ 4 4x dy
2
(1.102)
y =0
Integrating Equation 1.102 and using the value of x2 from Equation
1.101
V =
πa 2  2 y 3 
by −

b2 
3 
(1.103)
Equation 1.103 is the general equation of the volume of the vertical
dished end up to a height of y from the bottom.
Total volume (y = b) The total volume of the dished end
V T = 2/3 πba2
Elliptical dished end (b = a/2)
(1.104)
The general equation of volume
V = 4π (ay2/2 − y3/3)
(1.105a)
The total volume (x = 2a)
V T = π/3 a3
(1.105b)
y
2a
2x
dy
y
(0,0)
Figure 1.14 Volume of vertical dished end.
© 2010 Taylor & Francis Group, LLC
b
x
48
Process engineering and design using visual basic®
Hemispherical dished end (b = a)
The general equation of volume
V = π (ay2 − y3/3)
(1.106a)
The total volume (x = 2a)
V T = 2/3 π a3
Standard integrals
∫ (du + dv + dw) = ∫ du + ∫ dv + ∫ dw
∫ adv = a∫ dv
∫
v n +1
+ c(n ≠ −1)
n+1
v n dv =
dv
∫ v = ln v + c
∫
av dv =
av
+c
ln a
∫ e dv = e + c
v
v
∫ sin v dv = − cos v + c
∫ cos v dv = sin v + c
∫ sec v dv = tan v + c
2
∫ csc v dv = − cot v + c
2
∫ sec v tan v dv = sec v + c
∫ csc v cot v dv = − csc v + c
1
dv
 v
∫ v + a = a tan  a  + c
2
2
© 2010 Taylor & Francis Group, LLC
−1
(1.106b)
Chapter one: Basic mathematics
49
dv
 v
∫ a − v = sin  a  + c
2
−1
2
1
dv
v − a
∫ v − a = 2a ln  v + a  + c
2
2
∫ v ± a = ln {v + v ± a } + c
dv
2
2
2
2
∫ sec v dv = ln(sec v + tan v) + c
∫ csc v dv = ln(csc v − cot v) + c
Differential equations
First-order differential equations
An equation of the following type:
dy
= 50 − y
dx
(1.107)
is called a first-order differential equation because it involves the first
derivative, but no higher derivative.
Example 1.14
Solve the following first-order differential equation:
dy
= 2x
dx
SOLUTION
Integrating the preceding equation
y =
∫ 2x dx = x + C (C = constant of integration)
2
The solution curves are parabolas.
© 2010 Taylor & Francis Group, LLC
50
Process engineering and design using visual basic®
Separation of variables
Let us use the following equation and solve the equation using the separating variables method:
dy
= ky
dx
Separating the variables
1
dy = k dx
y
Integrating
1
∫ y dy = ∫ k dx
The result is
ln (y) = kx + C
or
y = ekx+C = ekxeC = Aekx
Therefore, the general solution of the equation
dy
= ky
dx
is
y = Aekx
(1.108)
for any value of A.
Second-order differential equations
The concept of a second-order differential equation can be expressed with
the problem of free body movement under gravity. The general equation
of a free-falling body under gravity is
d 2s
= −g
dt 2
(1.109a)
where s is the distance, t is the time, and g is the acceleration due to gravity. Equation 1.98a is a second-order differential equation.
Now, we know that the velocity can be expressed as v = ds/dt.
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
51
Therefore, a solution of Equation 1.109a will be
ds
= − gt + v0
dt
(1.109b)
where v0 is the initial velocity. Integrating Equation 1.109b, we get
s=−
1 2
gt + v0t + s0
2
(1.109c)
where s0 is the initial height.
Example 1.15
Solve the following differential equation:
d 2s
+ w 2s = 0
dt 2
(1.110a)
SOLUTION
We need to find a solution whose second derivative is the negative
of the original function. This can be achieved by using properties
s(t) = cos wt and s(t) = sin wt.
d
d2
(cos t) =
(− sin t) = − cos t
dt 2
dt
and
d
d2
(sin t) =
(cos t) = − sin t
dt 2
dt
Therefore, the general solution of Equation 1.110a is
s(t) = C1 cos wt + C2 sin wt
(1.110b)
where C1 and C2 are arbitrary constants and the values of C1 and C2
depend on the boundary conditions.
Bessel function
Bessel’s differential is one of the most important equations in applied
mathematics, and the standard Bessel functions are used for the solution
of the following types of equations:
x2y″ + xy′ + (x2 − v2)y = 0
© 2010 Taylor & Francis Group, LLC
(1.111a)
52
Process engineering and design using visual basic®
or
y ′′ +

1
v2 
y′ +  1 − 2  y = 0
x
x 

(1.111b)
The general solution of Equation 1.111a or Equation 1.111b is denoted
by Jn(x) and is given by
∞
J n ( x) = x n
(−1)m x 2 m
22 m + n m !(n + m)!
m=0
∑
(1.111c)
The solution is called the Bessel function of the first kind of order n.
This series converges for all x, as the ratio test shows, and in fact converges
very rapidly because of the factorials in the denominator.
Bessel functions J0(x) and J1(x)
∞
J 0 ( x) =
(−1)m x 2 m
x2
x4
x6
∑ 2 (m !) = 1 − 2 (1!) + 2 (2 !) − 2 (3 !) + 2m
2
2
2
4
2
6
2
(1.112a)
m=0
This looks similar to a cosine curve.
∞
J1 ( x ) =
(−1)m x 2 m + 1
x
x3
x5
x7
=
−
+
−
+
22 m + 1 m !(m + 1)! 2 231! 2 ! 252 ! 3 ! 27 3 ! 4 !
m=0
∑
(1.112b)
This looks similar to a sine curve.
Standard Bessel functions Standard Bessel functions are very important in applied mathematics. These are presented in Table 1.6 [1].
Partial differential equations
An equation involving one or more partial derivatives of an (unknown)
function of two or more independent variables is called a partial differential
equation. The order of the highest derivative is called the order of the equation. A few important partial differential equations are presented in the
following text:
One-dimensional wave equation:
2
∂ 2u
2 ∂ u
=
c
∂t 2
∂x 2
© 2010 Taylor & Francis Group, LLC
(1.113a)
Chapter one: Basic mathematics
53
Table 1.6 Standard Bessel Functions
x
J0(x)
J1(x)
x
J0(x)
J1(x)
x
J0(x)
J1(x)
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
1.0000
0.9975
0.9900
0.9776
0.9604
0.9385
0.9120
0.8812
0.8463
0.8075
0.7652
0.7196
0.6711
0.6201
0.5669
0.5118
0.4554
0.3980
0.3400
0.2818
0.2239
0.1666
0.1104
0.0555
0.0025
−0.0484
−0.0968
−0.1424
−0.1850
−0.2243
0.0000
0.0499
0.0995
0.1483
0.1960
0.2423
0.2867
0.3290
0.3688
0.4059
0.4401
0.4709
0.4983
0.5220
0.5419
0.5579
0.5699
0.5778
0.5815
0.5812
0.5767
0.5683
0.5560
0.5399
0.5202
0.4971
0.4708
0.4416
0.4097
0.3754
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
−0.2601
−0.2921
−0.3202
−0.3443
−0.3643
−0.3801
−0.3918
−0.3992
−0.4026
−0.4018
−0.3971
−0.3887
−0.3766
−0.3610
−0.3423
−0.3205
−0.2961
−0.2693
−0.2404
−0.2097
−0.1776
−0.1443
−0.1103
−0.0758
−0.0412
−0.0068
0.0270
0.0599
0.0917
0.1220
0.3991
0.3009
0.2613
0.2207
0.1792
0.1374
0.0955
0.0538
0.0128
−0.0272
−0.0660
−0.1033
−0.1386
−0.1719
–0.2028
−0.2311
−0.2566
−0.2791
−0.2985
−0.3147
−0.3276
−0.3371
−0.3432
−0.3460
−0.3453
−0.3414
−0.3343
−0.3241
−0.3110
−0.2951
6.0
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
7.0
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
8.0
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
0.1506
0.1773
0.2017
0.2238
0.2433
0.2601
0.2740
0.2851
0.2931
0.2981
0.3001
0.2991
0.2951
0.2882
0.2786
0.2663
0.2516
0.2346
0.2154
0.1944
0.1717
0.1475
0.1222
0.0960
0.0692
0.0419
0.0146
−0.0125
−0.0392
−0.0653
−0.2767
−0.2559
−0.2329
−0.2081
−0.1816
−0.1538
−0.1250
−0.0953
−0.0652
−0.0349
−0.0047
0.0252
0.0543
0.0826
0.1096
0.1352
0.1592
0.1813
0.2014
0.2192
0.2346
0.2476
0.2580
0.2657
0.2708
0.2731
0.2728
0.2697
0.2641
0.2559
One-dimensional heat equation:
∂u
∂ 2u
= c2 2
∂t
∂x
(1.113b)
Two-dimensional Laplace equation:
∂ 2u ∂ 2u
+
=0
∂x 2 ∂y 2
© 2010 Taylor & Francis Group, LLC
(1.113c)
54
Process engineering and design using visual basic®
Two-dimensional Poisson equation:
∂ 2u ∂ 2u
+
= f (x, y)
∂x 2 ∂y 2
(1.113d)
Three-dimensional Laplace equation:
∂ 2u ∂ 2u ∂ 2u
+
+
=0
∂x 2 ∂y 2 ∂z 2
(1.113e)
Example 1.16
Refer to Figure 1.15 above. A string of length l is stretched between
two fixed points, and the motion is started by drawing aside, through
a small distance b, a point of the string at distance 1/5 from one end.
Establish the displacement y at distance x from the end at time t.
SOLUTION
Let us take the fixed end O of the string as the origin and the equilibrium position OA along the x-axis, where OA = l. OY is taken in the
plane OBA, and it is perpendicular to OX. The point A′ of the string,
which is at a distance l/5 from the origin O, is drawn aside through
a small distance b to the position B, and then released from rest. The
motion starts from the position OB and BA (A is also a fixed point)
of the string, and we measured time from this position of the string.
Boundary conditions:
1. y = 0 when x = 0 and x = l, for all t
2. y = 5bx/l for 0 ≤ x ≤ l/5, when t = 0
3. y = 5(l − x)b/(4l) for l/5 ≤ x ≤ l, when t = 0
4. ∂y/∂t = 0 for 0 ≤ x ≤ l, when t = 0
y
B
0
A
l/5
A′
l
Figure 1.15 Example 1.16.
© 2010 Taylor & Francis Group, LLC
x
Chapter one: Basic mathematics
55
The differential equation for the vibrating string is the wave
equation
2
∂2 y
2 ∂ y
=
c
∂t 2
∂x 2
(1.114a)
Because y = 0 when x = 0, we can assume
y = T sin mx
(1.114b)
where T is a function of t only, and m ≠ 0.
Substituting Equation 1.114b in Equation 1.114a
T″ sin mx = −T m2c2 sin mx
For all values of x, sin mx ≠ 0.
Therefore
T″ + m2c2 T = 0
(1.114c)
The general solution of Equation 1.114c is
T = A cos(mct) + B sin(mct)
or
y = [A cos(mct) + B sin(mct)] sin(mx)
Because
∂y
= 0, when t = 0
∂t
0 = T′ sin mx when t = 0 or T′ = 0 when t = 0 because sin mx ≠ 0.
Now, T′ = mc [−A sin(mct) + B cos(mct)]. Given T′ = 0 when t = 0, or B = 0.
Therefore
y = A cos(mct) sin mx;
m≠0
(1.114d)
Because y = 0 when x = l, from Equation 1.114b
0 = T sin ml,
T≠0
Therefore, sin ml = 0 or ml = nπ, when n is any positive or negative integer. Because the positive and negative values of n will result
in identical sets of solutions of Equation 1.114a, we can assume only
positive values, that is, n = 1, 2, 3, 4, . . ., ∞.
Therefore, the general solution of Equation 1.114a will be
∞
y =
∑ b cos(mct)sin(mx)
n
n =1
∞
=
nπct
nπx
∑ b cos l sin l
n
(1.114e)
n =1
Now, y = 5bx/l, 0 ≤ x ≤ l/5 for t = 0 and y = 5b (l − x)/(4l), l/5 ≤ x ≤ l
for t = 0.
© 2010 Taylor & Francis Group, LLC
56
Process engineering and design using visual basic®
Let us assume that y = V0 when t = 0.
Therefore
∞
V0 =
nπx
∑ b sin l
n
(1.114f)
n =1
The right-hand side of Equation 1.114f represents the Fourier halfrange sine series for the function V0 defined in (0,l); therefore
bn =
=
l
 l/5

2  5bx
5b(l − x)
nπx
nπx 
sin
dx +
sin
dx

4l
l
l
l
l
l/5
0

∫
∫
25b
nπ
sin
2n 2 π 2
5
Therefore, the solution will be
∞
y =
25b
1
nπx
nπ
nπct
sin
cos
sin
2π 2 n = 1 n 2
5
l
l
∑
(1.114g)
Example 1.17
Solve the heat equation
∂u
∂ 2u
= c2 2
∂t
∂x
(1.115a)
for the following boundary conditions:
u = 0 when x = 0 and x = l
u = x when t = 0
SOLUTION
Because u = 0 when x = 0, we can assume that u = T sin mx, and T is
the function of t only for a solution of Equation 1.115a (m ≠ 0).
Substituting in Equation 1.115a
T′ sin mx = −c2Tm2 sin mx
or
T′ = −Tc2m2 (because sin mx ≠ 0)
The solution is
2 2
T = Ae − c m t
Equation 1.115a will be
2 2
u = Ae − c m t sin mx
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
57
Now, u = 0 when x = l.
Therefore
2 2
0 = Ae − c m t sin ml
Therefore, sin ml = 0, or
m = πn/l (n = ±1, ±2, ±3, ±4, . . . , ±∞)
Therefore
2
2
 nπ 
u = Ae − c ( nπ/l ) t sin   x
 l 
(1.115b)
Because the positive and negative values of n will give identical
sets of solutions of Equation 1.115a, we can therefore assume either
the positive or negative values of n to have an independent set of solutions of Equation 1.115a.
The general solution will be
∞
u=
∑b e
n
− c 2 ( nπ / x )2 t
n =1
 nπ 
sin   x
 l 
(m ≠ n ≠ 0)
Now, u = x when t = 0. Therefore
∞
x=
nπ
∑ b sin  l  x
n
n =1
The right-hand side of the above equation is the Fourier halfrange sine series for the function f(x) defined in (0,l).
Therefore
l
2
 nπ 
bn =
x sin   x dx
 l 
l
∫
0
2l
=
(−1)n + 1
nπ
Therefore, the solution will be
∞
u=
2l
(−1)n + 1 − c2 ( nπ/l )2 t
 nπx 
sin 
e
 l 
π n =1
n
∑
(1.115c)
Example 1.18
Solve the following Laplace equation:
∂ 2u ∂ 2u
+
=0
∂x 2 ∂y 2
(1.116a)
subject to the conditions u = 0 when x = 0, x = c, and y = b and u = u0
when y = 0.
© 2010 Taylor & Francis Group, LLC
58
Process engineering and design using visual basic®
SOLUTION
Because u = 0 when x = 0, we assume u = Y sin mx, where m ≠ 0, for a
solution of Equation 1.116a.
Substituting in Equation 1.116a
−Y m2 sin mx + Y″ sin mx = 0
or
Y″ − Y m2 = 0
(1.116b)
(sin mx cannot be zero for all values of x) whose solution is
Y = A cosh my + B sinh my
Therefore
u = (A cosh my + B sinh my) sin mx,
m≠0
(1.116c)
Because u = 0 when x = c, we have 0 = Y(y) sin mc. Y(y) cannot be zero
for all values of y. Therefore, sin mc = 0 or mc = nπ, n = 0 ± 1 ± 2 ± 3 ± 4
± . . . ± ∞. u = 0 when y = b.
0 = (A cosh mb + B sinh mb) sin mx. Because x is arbitrary, therefore, we have 0 = A cosh mb + B sinh mb.
Therefore
cosh mb


u = A cosh my −
sinh my  sin mx
sinh mb


=
A
[cosh my sinh mb − cosh mb sinh my]sin mx
sinh mb
=
A
sinh m(b − y )sin mx
sinh mb
Because the positive and negative values of m will result in
identical sets of solutions of Equation 1.116a, we can use either the
positive or negative values of m to have an independent set of solutions of Equation 1.116a. In this case, we will use the positive values
of m.
Therefore, the general solution of Equation 1.116a will be
∞
u=
∑A
n
n =1
sinh(nπ/c)(b − y )
nπx
sin
sinh( nπb/c)
c
Because u = u0 when y = 0, we have
∞
u0 =
nπx
∑ A sin c
n
n =1
© 2010 Taylor & Francis Group, LLC
(1.116d)
Chapter one: Basic mathematics
59
The right-hand side of Equation 1.116d represents the Fourier
half-range sine series for all functions of u0.
Therefore
c
An =
2
nπx
u0 sin
dx
c
c
∫
0
=
2u0
[1 − (−1)n ]
nπ
When n is even, the value will be zero, and when n is odd, we use
n = 2m − 1 (m = 1, 2, . . . , ∞).
A2 m − 1 =
4u0
(2m − 1)π
(1.116e)
Hence, the general solution of Equation 1.116a is
∞
∑
4u0
u=
2
(
m
− 1)π
m =1
2m − 1
π(b − y )
2m − 1
c
sin
πx
2m − 1
c
sinh
πb
c
sinh
(1.116f)
Laplace transform
The Laplace transform of a function x(t) is the function x (s) defined by
∞
x (s) =
∫ x(t)e dt
− st
0
where s > 0, and is written as L[x(t)].
Therefore, from definition
L [ x(t)] =
∞
∫ x(t)e dt
− st
0
where s > 0, which is a function of x (s).
Example 1.19
Work out the Laplace transform of t.
SOLUTION
x(t) = t
∞
L[t] =
∫ te dt
− st
0
© 2010 Taylor & Francis Group, LLC
(1.117)
60
Process engineering and design using visual basic®
∞
 e − st e − st 
= t
− 2 
s 0
 −s
∞
1
te − st  e − st 
= − lim
− 2 
s t →∞ 1
 s 0
1
1
t
= − lim st − 2 [−1]
s t →∞ e
s
1
1
1
1
= − lim st + 2 = 2
s t →∞ se
s
s
(L′ Hospital ’s rule)
Therefore
L[t] = 1/s2
Standard Laplace transforms
L[1] =
1
s
L[t] =
1
s2
L[t n ] =
n!
sn +1
L[e at ] =
1
s−a
L[e − at ] =
1
s+a
L[sin at] =
a
s + a2
L[cos at] =
s
s + a2
L[sinh at] =
a
s − a2
© 2010 Taylor & Francis Group, LLC
2
2
2
(1.118)
Chapter one: Basic mathematics
61
L[cosh at] =
s
s − a2
2
L[t cos at] =
s2 − a 2
( s 2 + a 2 )2
L[t sin at] =
2sa
( s 2 + a 2 )2
Fourier half-range expansions [1]
In various applications, there is a practical need to use Fourier series
in connection with function f(x) that are given on some intervals only.
We could extend f(x) periodically with period L and then represent the
extended function by a Fourier series, which in general would involve
both cosine and sine terms.
Fourier half-range cosine series
The cosine half-range expansion is
∞
f ( x) = a0 +
nπ
∑ a cos L x
n
(1.119a)
n =1
where
a0 =
2
an =
L
1
L
L
∫ f (x)dx
(1.119b)
0
L
nπx
∫ f (x)cos L dx
(1.119c)
0
Fourier half-range sine series
The sine half-range expansion is
∞
f ( x) =
nπ
∑ b sin L x
n
n =1
© 2010 Taylor & Francis Group, LLC
(1.120a)
62
Process engineering and design using visual basic®
where
bn =
2
L
L
nπx
∫ f (x)sin L dx
(1.120b)
0
Example 1.20 [1]
Find the two half-range expansions of the following function:
f(x) = 2kx/L
if 0 < x < L/2
and
f(x) = 2k(L − x)/L
if L/2 < x < L
SOLUTION
Even periodic extension:
L
 L/ 2

1  2k
2k
k
a0 =
x dx +
(L − x) dx  =
 2
L L
L
0
L/ 2


∫
an =
∫
L
 L/ 2

2  2k
2k
nπ
nπ
x cos
x dx +
(L − x)cos
x dx 

L L
L
L
L
0
L/ 2


∫
∫
Integrating by parts
L/ 2
∫
x cos
0
nπ
L2
nπ
L2 
nπ

x dx =
sin
+ 2 2  cos
− 1

2nπ
2
2
L
nπ 
and
L
nπ
L2
nπ
L2 
nπ 
∫ (L − x)cos L x dx = − 2nπ sin 2 − n π  cos nπ − cos 2 
2
2
L/ 2
Therefore
an =
4k 
nπ

2 cos
− cos nπ − 1
2 2 


nπ
2
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
63
Thus
a2 = −
16k
16k
, a4 = 0, a6 = − 2 2 , a8 = 0,…
22 π 2
6 π
Therefore, the first half-range expansion of f(x) is
f ( x) =
2π
1
6π
k 16k  1

− 2  2 cos
x + 2 cos
x + 

2
6
π 2
L
L
Odd periodic extension:
Similarly, for the odd periodic extension
bn =
nπ
8k
sin
n2π 2
2
Hence, the second half-range expansion of f(x) is
f ( x) =
8k  1
1
3π
1
5π
π

x + 2 sin
x − 
 sin x − 3 2 sin

5
π 2  12
L
L
L
Numerical analysis
Solving linear equations (Newton’s method)
Given a function f(x) that is continuous and has a continuous derivative
and also, given a starting value of x0:
For n = 1, 2, 3, . . . until termination
Compute f′(xn).
If f′(xn) = 0, signal and stop.
Else, compute
xn + 1 = xn −
f ( xn )
f ’( xn )
(1.121a)
Test for termination.
Example 1.21
Let the function f(x) = x3 + x − 1 = 0.
xn + 1 = xn −
xn3 + xn − 1 2xn3 + 1
=
3 xn2 + 1
3 xn2 + 1
Starting value x0 = 1.
© 2010 Taylor & Francis Group, LLC
(1.121b)
64
Process engineering and design using visual basic®
Therefore,
x1 = 3/4 = 0.75
x2 =
2 * 0.753 + 1
= 0.686047
3 * 0.752 + 1
x3 = 0.682340
x4 = 0.682328
The value of x4 is exact to the 6th decimal. A larger number of
analyses may be required if the function does not converge rapidly.
Therefore, the solution of the function x3 + x − 1 = 0 will be
x = 0.682328.
Newton’s method in two variables [2]
Let (x0, y0) be chosen. We need to generate the sequence of vectors (xn, yn)
recursively as follows:
Let
D(x, y) = fx(x, y)gy(x, y) − f y(x, y)gx(x, y)
Then
xn + 1 = xn −
f ( x n , y n ) g y ( x n , y n ) − f y ( x n , y n ) g( x n , y n )
D( xn , y n )
(1.122a)
yn+1 = yn −
− f ( x n , y n ) g x ( x n , y n ) + f x ( x n , y n ) g( x n , y n )
D( xn , y n )
(1.122b)
Under suitable conditions, the sequence (xn, yn) will converge to a vector (u, v) that satisfies
f(u, v) = g(u, v) = 0
(1.122c)
Newton’s method is most likely to work when the initial vector (x0, y0)
is close to a solution. It is not necessary that the system has only one
solution.
Example 1.22
Use Newton’s method to find a solution to the system
x2 − 2x − y + 1/2 = 0
x2 + 4y2 − 4 = 0
starting at the initial vector (2,0.25).
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
65
SOLUTION
f = x2 − 2x −y + 1/2
fx = 2x − 2
and f y = −1
g = x2 + 4y2 − 4
gx = 2x
and g y = 8y
D(x,y) = fx g y − f y gx = (2x − 2)(8y) − (−1)(2x)
= 2x − 16y + 16xy
fg y − f yg = (x2 − 2x − y + 1/2)(8y) − (−1)(x2 + 4y2 − 4)
= −4 + 4y + x2 − 4y2 − 16xy + 8x2y
−fgx + fxg = −(x2 − 2x − y + 1/2)(2x) + (2x − 2)(x2 + 4y2 − 4)
= 8 − 9x + 2xy + 2x2 − 8y2 + 8xy2
Thus
xn + 1 = xn −
−4 + 4 y n + xn2 − 4 y n2 − 16 xn y n + 8 xn2 y n
2xn − 16 y n + 16 xn y n
yn+1 = yn −
8 − 9xn + 2xn y n + 2xn2 − 8 y n2 + 8 xn y n2
2xn − 16 y n + 16 xn y n
Starting at x0 = 2 and y0 = 0.25
x1 = 2 − 0.75/8 = 1.90625
and
y1 = 0.25 − (−)0.5/8 = 0.3125
The first four iterations are presented in Table 1.7.
The solution is correct up to six decimal places.
The values of the above-mentioned two equations can be calculated using x = 1.900677 and y = 0.311219 from Table 1.7 as
x2 − 2x − y + 1/2 = (1.900677)2 − 2(1.900677) − 0.311219 + 0.5
= 0.000000058
x2 + 4y2 − 4 = (1.900677)2 + 4(0.311219)2 − 4 = 0.000002112
Table 1.7 Solution of Example 1.21
n
xn
yn
xn+1
yn+1
0
1
2
3
2
1.90625
1.900691
1.900677
0.25
0.3125
0.311213
0.311219
1.90625
1.900691
1.900677
1.900677
0.3125
0.311213
0.311219
0.311219
© 2010 Taylor & Francis Group, LLC
66
Process engineering and design using visual basic®
Numerical methods in linear algebra
Gauss elimination [3]
In this method, unknowns are eliminated by combining equations such
that the n equations and n unknowns are reduced to an equivalent upper
triangular system, which is then solved by back substitution. Let us consider the following system:
a11x1
a21x1
a31x1
a12 x2
a22 x2
a32 x2
a13 x3 = b1
a23 x3 = b2
x33 x3 = b3
(1.123)
For the first stage of elimination, multiply the first row of Equation
1.123 by a21/a11 and a31/a11, respectively, and subtract from the second and
third rows.
a22( 2) x2
a32( 2) x2
a23( 2) x3 = b2( 2)
a33( 2) x3 = b3( 2)
(1.124)
where
a22(2) = a22 − (a21/a11)a12
a23(2) = a23 − (a21/a11)a13
a32(2) = a32 − (a31/a11)a12
a33(2) = a33 − (a31/a11)a13
b2(2) = b2 − (a21/a11)b1
b3(2) = b3 − (a31/a11)b1
In the next stage of elimination, multiply the first row of Equation
1.124 by a32(2)/a22(2) and subtract from the second row in Equation 1.124.
a33(3) x3 = b3(3)
where
a33(3) = a33(2) − (a32(2)/a22(2))a23(2)
b3(3) = b3(2) − (a32(2)/a22(2))b2(2)
The first equation from each stage results in
a11x1
a12 x2
a22( 2) x2
a13 x3 = b1
a23( 2) x3 = b2( 2)
a33( 3 ) x3 = b3( 3 )
(1.125)
The elements a11, a22(2), and a33(3), which are assumed to be nonzero, are
called pivot elements and this elimination method is called Gauss elimination.
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
67
One condition of the analysis is that the pivots must be different from
zero. To achieve this, it may be required to change the order of the equations. This is called partial pivoting. In some cases, it may require not only
an interchange of equations, but also an interchange of the position of the
variables. This is called complete pivoting.
Example 1.23
Solve the following equations using Gauss elimination:
x1 + x2 + x 3 = 6
3x1 + 3x2 + 4x3 = 20
2x1 + x2 + 3x3 = 13
SOLUTION
Eliminating x1 from the last two equations
x1 + x2 + x 3 = 6
x3 = 2
−x2 + x3 = 1
In this case, the pivot point in the second equation is zero, and we
need to interchange the preceding second and third equations before
the second step. Therefore
x1 + x2 + x 3 = 6
−x2 + x3 = 1
+ x3 = 2
The solution is x3 = 2, x2 = 1, and x1 = 3.
Cholesky method
This method is also known as the square root method. If the coefficient
matrix A is symmetric and positively definite, the matrix A can be decomposed as A = LLT.
The popular method of solving an equation
Ax = b
(1.126a)
A = LLT
(1.126b)
LLTx = b
(1.126c)
LTx = z
(1.126d)
Lz = b
(1.126e)
is
or
This can be written as
© 2010 Taylor & Francis Group, LLC
68
Process engineering and design using visual basic®
Based on the preceding equations, a matrix can be factorized as
 a11

 a21

 a31
a13   m11
 
a23  =  m21
 
a33   m31
a12
a22
a32
m32
0   m11

0  0

m33   0
m11 =
a11
0
m22
m21
m22
0
m31 

m32 

m33 
(1.126f)
where
(1.126g)
i −1
mii =
aii −
∑m
2
ij
i = 2, … , n
(1.126h)
j =1
ai1
m11
mi1 =
j −1

1 
mij =
m jk mik 
 aij −

m jj 
k =1
∑
i = 2, … , n
(1.126i)
i = j + 1, j + 2, … , n, k ≥ 2
(1.126j)
Example 1.24
Using the Cholesky method, solve the following equations:
4
 −1

0

 0
−1
4
−1
0
0
−1
4
−1
0   x1   1 
0   x2  0 
=
−1  x3  0 
   
4   x 4  0 
The form of factorization
4
 −1

0

 0
−1
4
−1
0
0
−1
4
−1
0   m11
0   m21
=
−1  m31
 
4   m41
0
m22
m32
m42
0
0
m33
m43
0   m11
0   0
0  0

m44   0
Using the preceding equations
m11 = 2
m21 = −1/2 ​ ​
m22 = (15/4)0.5
© 2010 Taylor & Francis Group, LLC
m21
m22
0
0
m31
m32
m33
0
m41 
m42 
m43 

m44 
Chapter one: Basic mathematics
69
m31 = 0 ​ ​m32 = −(4/15)0.5 ​ ​m33 = (56/15)0.5
m41 = 0 ​ ​m42 = 0 ​ ​m43 = −(15/56)0.5 ​ ​
m44 = (209/56)0.5
The solution for z will be
 2

 −1/2

 0
 0

0
0
15/4
0
− 4/15
56/15
0
− 15/56
  z1   1 
   
  z2   0 
  z  = 0 
0
 3  
209/56   z4  0 
0
0
or
z1 =
1
, z2 =
2
1
, z3 =
60
1
, z4 =
840
1
11704
The final solution for x will be
2

0

0
0

−1/2
0
15/4
− 4/15
0
56/15
0
0
  x1   1/2 
  

0
  x2   1/ 60 
=


− 15/56   x3   1/ 840 
 
209/56   x 4  1/ 11704 
0
or
x4 =
1
4
15
56
, x3 =
, x2 =
, x1 =
209
209
209
209
Numerical integration
Two methods are extremely popular in numerical integration:
1. Trapezoidal rule
2. Simpson’s rule
Trapezoidal rule
Under this rule, the interval (a,b) is divided into N subintervals of length
h = (b − a)/N, and if the subintervals are denoted as (x0,x1), (x1,x2), (x2,x3), . . . ,
(xN−1,xN), then the general integration can be written as
x1
b
I =
∫
a
f ( x) dx =
∫
xN
x2
f ( x) dx +
x0
© 2010 Taylor & Francis Group, LLC
∫
x1
f ( x) dx + +
∫ f (x) dx
xN − 1
(1.127)
70
Process engineering and design using visual basic®
Evaluating each integral by the trapezoidal rule, the integration can
be written as
I =
h
 f 0 + 2 ( f1 + f 2 + + f N −1 ) + f N 
2
(1.128)
Simpson’s rule
Under this rule, the interval of integration (a,b) is divided into an even
number of equal subintervals, say into N = 2M subintervals of length
h = (a − b)/2M, with end points x0 = a, x1, x2, . . . , x2M−1, x2M = b. The general
form of the integration will be
b
I =
x2
x4
∫ f (x) dx = ∫ f (x) dx + ∫
a
x0
x2 M
f ( x) dx + … +
x2
∫ f (x) dx
(1.129)
x2 M − 2
Using Simpson’s rule, the final integration will be
I =
h
 f 0 + 4 ( f1 + f 3 + + f 2 M −1 ) + 2 ( f 2 + f 4 + + f 2 M − 2 ) + f 2 M  (1.130)
3
Example 1.25
A slug catcher 60 m long, having a 0.5 m diameter, and at a 2° angle
with the horizontal plane, has a liquid-filled length of 5 m as shown
in Figure 1.16. Calculate the filled volume of the slug catcher.
SOLUTION
The volume can be calculated by integration using Simpson’s rule.
0.5 m
Angle = 2º
60 m
Figure 1.16 Slug catcher volume.
© 2010 Taylor & Francis Group, LLC
h
h
5m
2b
Chapter one: Basic mathematics
71
The distance from the bottom corner of the slug catcher up to the
liquid level (h) can be calculated as h = 5 tan2° = 0.175 m. This is less
than the radius of the pipe. For this condition, the top view of the
surface will always look like a half-ellipse.
For a liquid length of l, if the major axis is 2x and the minor axis
is 2y (the values of x and y depend on the value of l), the surface area
(A) of the half-ellipse will be
A = πxy/2
The volume of the liquid (V) can be estimated as
l=5
πxy
∫ 2 sin 2°dl
(1.131a)
x = l/cos2°
(1.131b)
y = {l tan 2° (2r − l tan2°)}0.5
(1.131c)
V =
l=0
Now
and
where r = radius of the slug catcher = 0.25 m.
Therefore, the volume will be
l=5
V =
πl
∫ 2 cos 2° * {l tan 2°(2r − l tan 2°)} sin 2°dl
0.5
l=0
or
l=5
V = 0.054853
∫ l * {0.03492l * (0.5 − 0.03492l)} dl (1.131d)
0.5
l=0
Now, a = 0 and b = 5.
The number of divisions between limits (l = 0 and l = 5), 2M = 50,
and the value of each division, h = 5/50 = 0.1.
The value of the integrals will be
f0 = 0, f50 = 0.054853 * 5{0.1746 * (0.5 − 0.1746)}0.5 = 0.06537
f1 = 0.054853 * 0.1 * {0.003492 * (0.5 − 0.003492)}0.5 = 0.000228
f 2 = 0.054853 * 0.2 * {0.006984 * (0.5 − 0.006984)}0.5 = 0.000644
and so on.
The sum of the odd integrals will be
( f1 + f3 + f5 + ⋯ + f49) = 0.701109
© 2010 Taylor & Francis Group, LLC
72
Process engineering and design using visual basic®
and the sum of the even integrals (2–48) will be
(f 2 + f4 + f6 + . . . + f48) = 0.668797
Therefore, the volume will be
V =
0.1
{0 + 4 * 0.701109 + 2 * 0.668797 + 0.06537} = 0.14 m 3
3
Double integration using Simpson’s rule [3]
A double integration in the following form:
b

I =  f ( x , y )dx dy


c  a
d
∫∫
(1.132)
can be solved using Simpson’s rule with h = (b − a)/2 and k = (d − c)/2. The
value of the integral will be
I =
hk
[ f ( a, c) + f ( a, d) + f (b , c) + f (b , d) + 4{ f ( a, c + k ) + f ( a + h, c)
9
+ f ( a + h, d) + f (b , c + k )} + 16 f ( a + h, c + k )]
(1.133)
Example 1.26
Using Simpson’s rule, evaluate the integral
2 2
I =
dxdy
∫ ∫ (x + y)
1 1
SOLUTION
h = 0.5 and k = 0.5.
I =
0.25
[ f (1, 1) + f (1, 2) + f (2, 1) + f (2, 2) + 4{ f (1, 1.5) + f (1.5, 1)
9
+ f (11.5, 2) + f (2, 1.5)} + 16 f (1.5, 1.5)]
or
I =
0.25 
2
2  16 
 2
0.5 + + 0.25 + 4 
+
+

 = 0.33988
9 
3
 2.5 3.5  3 
A more accurate solution is 0.339789.
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
73
Numerical solution of first-order differential equations
Euler’s method [2]
This method is very simple, but the results are not very accurate. In this
method, y1 is obtained by assuming that f(x,y) varies so little in the interval x0 ≤ x ≤ x1 that only a small error is made by replacing it by the constant value f(x0, y0). Integrating
dy
= f (x, y)
dx
from x0 to x1, we get
x1
y( x1 ) − y0 = y( x1 ) − y( x0 ) =
∫ f (x, y)dx ≈ f (x , y )(x − x )
0
0
1
0
(1.134)
x0
Because h = x1 − x0
y1 = y0 + hf(x0, y0)
The general equation is
yn+1 = yn + hf(xn, yn)
(1.135)
Improved Euler’s method
The accuracy of Euler’s method is improved by introducing the trapezoidal rule:
x1
h
∫ f (x, y)dx ≈ 2 { f (x , y ) + f (x , y(x ))}
0
0
1
1
(1.136)
x0
The value of y(x1) is not known and is estimated by Euler’s method,
which is called z1. This modification results in general equations such as
zn+1 = yn + hf(xn, yn)
(1.137a)
and
yn+1 = yn +
h
[ f ( xn , y n ) + f ( xn + 1 , zn + 1 )]
2
© 2010 Taylor & Francis Group, LLC
(1.137b)
74
Process engineering and design using visual basic®
Example 1.27
Using Euler’s method, solve the following equation:
dy
= −2xy 2,
dx
y (0 ) = 1
at the value x = 1.
SOLUTION
The value of h = 0.2 and the calculation is made by using the values of
x as 0.0, 0.2, 0.4, 0.6, 0.8, and 1.0.
z1 = y0 + hf(x0, y0) = 1
y1 = y 0 +
h
 f ( x0 , y 0 ) + f ( x1 , z1 ) = 1 + 0.1 * [0 − 0.4] = 0.96
2
z2 = y1 + hf(x1, y1) = 0.96 + 0.2(−0.3686) = 0.8863
y 2 = y1 +
h
 f ( x1 , y1 ) + f ( x2 , z2 ) = 0.96 + 0.1
2
* [−0.3686 − 0.6284] = 0.88603
and so on.
The results are presented in Table 1.8 (calculated value = 0.5034).
The exact solution is 0.5.
Runge–Kutta method [1]
The Runge–Kutta method is widely used as a numerical method to solve
differential equations. This method is more accurate than the improved
Euler’s method. This method computes the solution of the initial value
problem.
y′ = f(x,y), y(x0) = y0 at equidistance points
x1 = x0 + h,
x2 = x0 + 2h, . . .,
xN = x0 + Nh
Table 1.8 Solution of Example 1.27
xn
yn
f(xn, yn)
zn+1
f(xn+1, zn+1)
yn+1
0
0.2
0.4
0.6
0.8
1.0
1
0.96
0.8603
0.7350
0.6116
0.5034
​0
−0.3686
−0.5921
−0.6483
−0.5985
1
0.8863
0.7419
0.6053
0.4919
−0.4
−0.6284
−0.6605
−0.5862
−0.4839
0.96
0.8603
0.7350
0.6116
0.5034
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
75
The general method of the solution is
For n = 0, 1, 2, 3, . . . , N − 1
Evaluate
k1 = hf(xn, yn)
(1.138a)
k2 = hf(xn + 0.5h, yn + 0.5k1)
(1.138b)
k3 = hf(xn + 0.5h, yn + 0.5k2)
(1.138c)
k4 = hf(xn + h, yn + k3)
(1.138d)
xn+1 = xn + h
(1.138e)
1
( k1 + 2 k 2 + 2 k 3 + k 4 )
6
(1.138f)
yn+1 = yn +
This method is repeated till the solution is reached.
Example 1.28
Solve Example 1.27 using the Runge–Kutta method.
dy
= −2xy 2 , y(0) = 1
dx
at the value x = 1.
SOLUTION
For n = 0, xn = 0, yn = 1, and value of h = 0.2.
Using the preceding equations
k1 = 0.2f(0, 1) = 0
k2 = 0.2f(0.1, 1) = −0.04
k3 = 0.2f(0.1, 0.98) = −0.0384
k4 = 0.2f(0.2, 0.9616) = −0.074
x1 = 0.2
y1 = 1 +
1
(0 − 0.08 − 0.0768 − 0.074) = 0.9615
6
Similarly, other values can be calculated and tabulated as in Table
1.9 (the calculated value is 0.500007).
The exact solution is 0.5.
© 2010 Taylor & Francis Group, LLC
76
Process engineering and design using visual basic®
Table 1.9 Solution of Example 1.28
n
xn
yn
k1
k2
k3
k4
xn+1
yn+1
0
1
2
3
4
5
0
0.2
0.4
0.6
0.8
1.0
1
0.9615
0.8621
0.7353
0.6098
0.500007
​0
−0.074
−0.119
−0.13
−0.119
−0.04
−0.103
−0.129
−0.126
−0.109
−0.038
−0.099
−0.127
−0.127
−0.111
−0.074
−0.119
−0.1296
−0.119
−0.0995
0.2
0.4
0.6
0.8
1.0
0.9615
0.8621
0.7353
0.6098
0.500007
Second-order differential equations
Runge–Kutta–Nystrom method [1]
This system computes the solution of the initial value problem for an
equation y″ = f(x, y, y′), where the initial values x0, y0, and y0′, the step size
h, and the number of steps N are known.
The general method of solution is as follows:
For n = 0, 1, 2, . . . , N − 1
Evaluate
k1 = 0.5hf ( xn , y n , y n′ )
(1.139a)
K = 0.5h( y n′ + 0.5k1 )
(1.139b)
k 2 = 0.5hf ( xn + 0.5h, y n + K , y n′ + k1 )
(1.139c)
k 3 = 0.5hf ( xn + 0.5h, y n + K , y n′ + k 2 )
(1.139d)
L = h( y n′ + k 3 )
(1.139e)
k 4 = 0.5hf ( xn + h, y n + L, y n′ + 2k 3 )
(1.139f)
xn+1 = xn + h
(1.139g)
1


y n + 1 = y n + h  y n′ + (k1 + k 2 + k 3 )


3
(1.139h)
1
( k1 + 2 k 2 + 2 k 3 + k 4 )
3
(1.139i)
y n′ + 1 = y n′ +
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
77
Example 1.29
Solve the following second-order differential equation:
d2 y
= 0.5( x + y + y ′ + 2) at the value x = 1
dx 2
Initial values: x0 = 0, y0 = 0, y0′ = 0, and h = 0.2.
SOLUTION
For n = 0, xn = 0, yn = 0, yn′ = 0, and the value of h = 0.2.
Using the preceding equations
k1 = 0.1
K = 0.005
k2 = 0.11025
k3 = 0.110763
L = 0.022153
k4 = 0.122184
xn+1 = 0.2
yn+1 = 0.021401
y′n+1 = 0.221401
Similarly, other values can be calculated and tabulated as shown
in Table 1.10.
The solution up to four decimal places = 0.7183.
Partial differential equations
In chemical engineering, the solutions of partial differential equations are
often required, particularly for the problems associated with heat conduction. Different numerical methods are used for solving different types of
partial differential equations.
Table 1.10 Solution of Example 1.29
Value n
0
1
2
3
4
5
xn
0
0.2
0.4
0.6
0.8
1
yn
yn′
xn+1
yn+1
yn′+1
0
0.021401
0.091819
0.222109
0.425526
0.718258
0
0.221401
0.49182
0.82211
1.225526
1.718259
0.2
0.4
0.6
0.8
1
0.021401
0.091819
0.222109
0.425526
0.718258
0.221401
0.49182
0.82211
1.225526
1.718259
© 2010 Taylor & Francis Group, LLC
78
Process engineering and design using visual basic®
Time row j+1
k
Time row j
h
h
Figure 1.17 Mesh size.
Heat conduction problem [1]
Generally, the Crank–Nicolson method is used for the numerical solution
of heat conduction problems that are parabolic in nature. This method
solves a mesh with mesh size h in the x-direction and mesh size k in the
y-direction (time direction). It calculates the values of u at six points as
shown in Figure 1.17.
The equation used to solve the problem is
1
1
(ui , j + 1 − uij ) =
(ui + 1, j − 2uij + ui −1, j )
k
2h2
+
1
(ui + 1, j + 1 − 2ui , j + 1 + ui −1, j + 1 )
2h 2
(1.140)
Using r = k/h2, Equation 1.140 can be modified to
(2 + 2r )ui , j + 1 − r(ui + 1, j + 1 + ui −1, j + 1 ) = (2 − 2r )uij + r(ui + 1, j + ui −1, j )
(1.141)
In Equation 1.141, the three values on the left side are unknown,
whereas the three values on the right side are known. Using the initial
and boundary conditions, most of the values can be made known except
two values. Equation 1.141 is then solved for two consecutive values of m
(m = 1,2, then m = 2,3, etc.). Under this situation, we will have two equations with two unknowns that can be solved easily.
Example 1.30
Solve the problem in Example 1.17 using theoretical and numerical
methods.
The heat equation
∂u
∂ 2u
= c2 2
∂t
∂x
Initial condition: u = x when t = 0
Boundary condition: u = 0 when x = 0 and x = l
This problem is further simplified by assuming c = 1 and l = 1.
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
79
SOLUTION
Theoretical solution
The general solution assuming c = 1 and l = 1 can be obtained from
Equation 1.104c as follows:
∞
u=
2
(−1)n + 1 −( nπ )2 t
sin( nπx)
e
π n =1
n
∑
Value i = 0
Values of u = 0 for all grid points of n
Value i = 1 (x = 0.3333)
Value of j = 1 (t = 0.0278)
For n = 1
sin (nπx) = 0.866
Exp (−n2π2t) = 0.7602
Subvalue = 0.6584
For n = 2
sin (nπx) = 0.866
Exp (−n2π2t) = 0.334
Subvalue = −0.1446
For n = 3
sin (nπx) = 0
Exp (−n2π2t) = 0.0848
Subvalue = 0
For n = 4
sin (nπx) = −0.866
Exp (−n2π2t) = 0.0124
Subvalue = 0.0027
It is clear from the preceding analysis that this infinite series is
highly converging in nature, and it is adequate to calculate up to n = 4.
The sum = 0.3287.
Similarly, other values of j can be estimated as follows:
Value of j = 2 (t = 0.0556)
The sum = 0.2879
Value of j = 3 (t = 0.0833)
The sum = 0.2319
Value of j = 4 (t = 0.1111)
The sum = 0.1807 and so on.
Using the procedure given earlier, other values for i = 2, and so on
can be estimated easily.
Numerical solution
The nodal points are represented in Figure 1.18 as follows.
For the solution, the values of h and k have been assumed as h = 1/3
and k = 1/36; therefore, r = 1/4.
© 2010 Taylor & Francis Group, LLC
80
Process engineering and design using visual basic®
u0,1
u0,0
u2,2
u1,2
u0,2
u3,2
u1,1
u2,1
u3,1
u1,0
u2,0
u3,0
Figure 1.18 Representation of nodal points.
Equation 1.141 will be changed to
−0.25ui −1, j + 1 + 2.5ui , j + 1 − 0.25ui + 1, j + 1 = 0.25ui −1, j + 1.5uij + 0.25ui + 1, j (1.142)
From the initial and boundary conditions, we have
u0,0 = 0;
u1,0 = 1/3; u2,0 = 2/3
u0,1 = u0,2 = u3,0 = u3,1 = u3,2 = 0
For j = 0
For i = 1
− 0.25u0,1 + 2.5u1,1 − 0.25u2,1 = 0.25u0,0 + 1.5u1,0 + 0.25u2,0
or
2.5u1,1 − 0.25u2,1 = 0.6667
(1.143a)
For i = 2
−0.25u1,1 + 2.5u2,1 − 0.25u3,1 = 0.25u1,0 + 1.5u2,0 + 0.25u3,0
or
−0.25u1,1 + 2.5u2,1 = 1.0833
(1.143b)
Solving Equations 1.143a and 1.143b, u1,1 = 0.3131 and u2,1 = 0.4646.
For j = 1
For i = 1
−0.25u0,2 + 2.5u1,2 − 0.25u2,2 = 0.25u0,1 + 1.5u1,1 + 0.25u2,1
or
2.5u1,2 − 0.25u2,2 = 0.5858
© 2010 Taylor & Francis Group, LLC
(1.144a)
Chapter one: Basic mathematics
81
Table 1.11 Example 1.29
Node Point
u1,1
u2,1
u1,2
u2,2
Theoretical
Numerical analysis
0.3287
0.5094
0.2879
0.3494
0.3131
0.4646
0.268
0.3369
For i = 2
−0.25u1,2 + 2.5u2,2 − 0.25u3,2 = 0.25u1,1 + 1.5u2,1 + 0.25u3,1
or
−u1,2 + 2.5u2,2 = 0.7752
(1.144b)
Solving Equations 1.144a and 1.144b, u2,2 = 0.3369 and u1,2 = 0.268.
The theoretically calculated values and values obtained through
numerical analysis are presented in Table 1.11.
Alternating direction implicit method [3]
These methods are two-step methods involving the solution of the traditional systems of equations along the lines parallel to the x- and y-axes, at
the first and second steps, respectively.
In the Peaceman–Rachford alternating direction implicit (ADI)
method, in the first step, it is advanced from tm to tm+1/2, and in the second
step, it is advanced from tm+1/2 to tm+1. The ADI method is written as
uim, j+ 1/2 − uim, j
1
1
= 2 ∂ 2x uim, j+ 1/2 + 2 ∂ 2y uim, j
k/2
h
h
(1.145a)
uim, j+ 1 − uim, j+ 1/2
1
1
= 2 ∂ 2x uim, j+ 1/2 + 2 ∂ 2y uim, j+ 1
k/2
h
h
(1.145b)
and
The equations can also be written as
r 2  m + 1/2 
r 

=  1 + ∂ 2y  uim, j
 1 − ∂ x  ui , j

2
2 
(1.146a)
r 2  m+1 
r 2  m + 1/2

 1 − ∂ y  ui , j =  1 + ∂ x  ui , j
2
2
(1.146b)
© 2010 Taylor & Francis Group, LLC
82
Process engineering and design using visual basic®
The intermediate values can be obtained as
r 
r 


uim, j+ 1/2 = 0.5  1 + ∂ 2y  uim, j + 0.5  1 − ∂ 2y  uim, j+ 1



2
2 
(1.147)
The boundary conditions to obtain the solution of Equation 1.146a can
be obtained as
r 
r 


uim, j+ 1/2 = 0.5  1 + ∂ 2y  uim, j + 0.5  1 − ∂ 2y  uim, j+ 1 i = 0,, M


2 
2 
(1.148)
Eliminating uim, j+1/2 from Equations 1.146a and 1.146b
r 2 
r 2  m+1 
r 2 
r 2 m

 1 − ∂ x   1 − ∂ y  ui , j =  1 + ∂ x   1 + ∂ y  ui , j
2
2
2
2
(1.149)
Equation 1.149 can be written in the D’yakonov split form as
r 2  * m+1 
r 2 
r 2 m

 1 − ∂ x  ui , j =  1 + ∂ x   1 + ∂ y  ui , j
2
2
2
(1.150a)
r 2  m+1

* m+1
 1 − ∂ y  ui , j = ui , j
2
(1.150b)
Intermediate boundary conditions
r 

u0* m, j + 1 =  1 − ∂ 2y  ( g1 )mj + 1

2 
(1.151a)
r 2

* m+1
uM
∂ y  ( g 2 )mj + 1
,j = 1 −

2 
(1.151b)
and
Example 1.31
Solve the following two-dimensional heat conduction equation:
∂u
∂ 2u ∂ 2u
=
+
∂t
∂x 2 ∂y 2
Initial condition:
u(x, y, 0) = sin πx sin πy,
© 2010 Taylor & Francis Group, LLC
0 ≤ x, y ≤ 1
Chapter one: Basic mathematics
83
Boundary conditions:
u = 0,
on the boundary for t ≥ 0
Use the ADI method. Assume h = 1/4 and k = 1/8 and integrate
for one time step.
SOLUTION
The nodal points are
xi =
i
, 0≤i≤4
4
yj =
j
, 0≤ j≤4
4
The initial and boundary conditions become
 πj 
 πi 
ui0, j = sin   sin  
 4
 4
u0m, +j 1/2 = u4m, +j 1/2 = 0, 0 ≤ j ≤ 4
uim, 0+ 1 = 0 = uim, 4+ 1 , 0 ≤ i ≤ 4
The ADI method is given by
1 2  m + 1/2 
1 2 m

= 1 +
∂ y ui , j
 1 − 16 ∂ x  ui , j

16 
1 2  m+1 
1 2  m + 1/2

 1 − 16 ∂ y  ui , j =  1 + 16 ∂ x  ui , j
For m = 0, the solution for ui1,/j 2
−
1 1/2
9
1 1/2
1 0
7
1 0
ui − 1, j + ui1,/j 2 −
ui + 1, j =
ui , j − 1 + ui0, j +
ui , j + 1
16
8
16
16
8
16
For j = 1
For i = 1
−
1 1/2
1 0
7
1 0
1 1/2 9 1/2
u0 ,1 + u1,1 −
u2 , 1 =
u1, 0 + u10,1 +
u1, 2
16
8
16
16
8
16
For i = 2
−
1 1/2
1 0
7
1 0
1 1/2 9 1/2
u1,1 + u2 ,1 −
u3 , 1 =
u2 , 0 + u20,1 +
u2 , 2
16
8
16
16
8
16
© 2010 Taylor & Francis Group, LLC
84
Process engineering and design using visual basic®
For i = 3
1 1/2
1 0
7
1 0
1 1/2 9 1/2
u2 ,1 + u3 ,1 −
u4 , 1 =
u3 , 0 + u30,1 +
u3 , 2
16
8
16
16
8
16
−
Using boundary conditions
 9/8

 −1/16
 0
0  u11,/12  (7 2 + 1)/(16 2) 

 

−1/16  u12/,12  = (14 + 2 )/(16 2 )


9/8  u13/,112  (7 2 + 1)/(16 2 )


−1/16
9/8
−1/16
The solution is
u11,/12 = u13/,12 = 0.46468
u12/,12 = 0.65716
For j = 2
For i = 1
−
1 1/2
1 0
7
1 0
1 1/2 9 1/2
u0 , 2 + u1, 2 −
u2 , 2 =
u1,1 + u10, 2 +
u1, 3
16
8
16
16
8
16
For i = 2
−
1 1/2
1 0
7
1 0
1 1/2 9 1/2
u1, 2 + u2 , 2 −
u3 , 2 =
u2 ,1 + u20, 2 +
u2 , 3
16
8
16
16
8
16
For i = 3
−
1 1/2
1 0
7
1 0
1 1/2 9 1/2
u2 , 2 + u3 , 2 −
u4 , 2 =
u3 ,1 + u30, 2 +
u3 , 3
16
8
16
16
8
16
Using boundary conditions
 9/8

 −1/16
 0
−1/16
9/8
−1/16
0  u11,/22  (7 2 + 1)/16 

 

−1/16  u12/, 22  = (14 + 2 )/16 


9/8  u13/,222  (7 2 + 1)/16 


The solution is
u11,/22 = u13/, 22 = 0.65716
u12/, 22 = 0.92936
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
85
For j = 3
For i = 1
−
1 1/2
1 0
7
1 0
1 1/2 9 1/2
u0 , 3 + u1, 3 −
u2 , 3 =
u1, 2 + u10, 3 +
u1, 4
16
8
16
16
8
16
For i = 2
−
1 1/2
1 0
7
1 0
1 1/2 9 1/2
u1, 3 + u2 , 3 −
u3 , 3 =
u2 , 2 + u20, 3 +
u2 , 4
16
8
16
16
8
16
For i = 3
−
1 1/2
1 0
7
1 0
1 1/2 9 1/2
u2 , 3 + u3 , 3 −
u4 , 3 =
u3 , 2 + u30, 3 +
u3 , 4
16
8
16
16
8
16
Using boundary conditions
 9/8

 −1/16
 0
0  u11,/32  (14 + 2 )/32 

 

−1/16  u12/, 32  = (7 2 + 1)/32 


9/8  u13/,332  (14 + 2 )/32 


−1/16
9/8
−1/16
The solution is
u11,/32 = u13/, 32 = 0.46468
u12/, 32 = 0.65716
For the solution of ui1, j
−
1 1
9
1 1
1 1/2
7
1 1/2
ui , j − 1 + ui1, j −
ui , j + 1 =
ui − 1, j + ui1,/j 2 +
ui +1, j
16
8
16
16
8
16
For i = 1
For j = 1, 2, and 3
 9/8

 −1/16
 0
−1/16
9/8
−1/16
0   u11,1  0.44767 
 


−1/16   u11, 2  =  0.63310 
9/8  u11, 3  0.44767 
© 2010 Taylor & Francis Group, LLC
86
Process engineering and design using visual basic®
The solution is
u11,1 = u11, 3 = 0.43186
u11, 2 = 0.61074
This method is repeated to find the solution at each mesh point.
Unit conversions
The basic unit conversions are presented in Table 1.12. An electronic conversion table is also included in this chapter.
Programming
General notes for all programs
• All programs have separate input and output frames.
• Most input data entry cells are white—it means that user inputs are
required.
• Some input cells are yellow—it means that the values cannot
only be calculated using this program, but can also be modified
externally.
• All output cells are red—it means that the values are calculated
results and cannot be modified externally.
• Both the SI and English units can be used for all programs, with the
SI unit being the default. These units are fixed units and cannot be
modified externally.
Vessel
The vessel.exe program has been developed to estimate the total and
filled-liquid volumes and the dry and wet surface areas for horizontal,
vertical, and inclined (e.g., slug catcher) vessels. For horizontal and vertical vessels, the program also calculates the weight of the material and the
hydro test weight.
Once the vessel program executes, the form shown in Figure 1.19 will
appear. This form presents a general understanding of the program and
has three menus:
• Vessel
• Project details
• Help
© 2010 Taylor & Francis Group, LLC
free fall (g)
ft/s2
grad
min
rad
cm2
ft2
ha
in.2
km2
m2
byte
exabyte
Gb
Kb
Mb
degree
acre
bit
Con­verted to
cm/s2
Con­verted from
© 2010 Taylor & Francis Group, LLC
acre
degree
Computer
0.125
bit
1.0842E−19
1.16415E−10
0.00012207
1.19209E−7
Area
Angle
cm/s2
Con­verted from
Acceleration
40468564.2
43560
0.4046863
6272640
0.004046856
4046.856
1.11111
60
0.0174533
0.001097
0.032808
Multiplied by
Table 1.12 Unit Conversions
nibble
Petabyte
terabyte
yottabyte
zettabyte
mi2
mm2
perch
square
yd2
revolution
sec
m/s2
mm/s2
Con­verted to
0.25
1.11022E−16
1.13687E−13
1.03398E−25
1.05879E−22
continued
0.0015625
4046856422.4
160
435.6
4840
0.0027778
3600
0.01
10
Multiplied by
Chapter one: Basic mathematics
87
μM/cm3
μM/L
mmol/cm3
mmol/L
g/m3
kg/cm3
kg/m3
mg/cm3
mg/m3
cm
dm
Furlong
ft
in.
km
lightyear
m
g/cm3
angstrom
Con­verted to
kmol/m3
Con­verted from
Con­verted from
© 2010 Taylor & Francis Group, LLC
1E−8
1E−9
4.97097E−13
3.28084E−10
3.93701E−9
1E−13
1.057023E−26
1E−10
Distance
Density
angstrom
g/cm3
Concentration
kmol/m3
1000,000
0.001
1000
1000
1000,000,000
1000
1000000
1
1000
Multiplied by
Table 1.12 (continued) Unit Conversions
μm
Mil
Mi
mm
nm
nmi
parsec
yd
oz/gal (E)
oz/gal (U.S.)
lb/ft3
lb/in.3
mmol/m3
mol/cm3
mol/L
mol/m3
Con­verted to
0.0001
0.00000393701
6.213712E−14
0.0000001
0.1
5.39957E−14
3.24078E−27
1.09361E−10
160.35591
133.52381
62.42769
0.0361273
1000000
0.001
1
1000
Multiplied by
88
Process engineering and design using visual basic®
g-force
dyn
© 2010 Taylor & Francis Group, LLC
0.00101972
0.0416667
6624.5833
1.840162
0.2339458
0.000064984
34.973
1.4572083
0.0242868
0.00040478
42
1.75
10,550,600,000
0.00000105506
10,758,575.59
0.000393
1055.03
erg
GJ
gf-cm
hp-h
J
barrel/h
cm3/h
cm3/s
ft3/h
ft3/s
gal(E)/d
gal(E)/h
gal(E)/min
gal(E)/s
gal(U.S.)/d
gal(U.S.)/h
251.99
cal
barrel/day
Btu
Force
dyn
Flow
Energy
barrel/day
Btu
N
gal(U.S.)/min
gal(U.S.)/s
L/h
L/s
m3/d
m3/h
m3/s
mL/h
mL/s
oz/h
oz/s
kJ
kWh
N-m
lbf-ft
Wh
kcal
0.00001
continued
0.0291667
0.000486111
6.6245833
0.001840162
0.15899
0.00662458
0.00000184
6624.58333
1.840162
224
0.0622222
1.05506
0.00029307
1055.06
778.1693
0.29307
0.25199
Chapter one: Basic mathematics
89
fc
lx
gr
g
kg
μg
mg
Btu/min
cal/min
Flame
Carat
Btu/h
Con­verted from
© 2010 Taylor & Francis Group, LLC
0.016667
4.199833
3.08647
0.2
0.0002
200000
200
4
43.055642
Power
Btu/h
Mass
carat
Light
flame
Heat transfer coefficient
0.00027778
Btu/(h⋅ft2⋅F)
0.488251
4882.51
0.000135625
1.35625
1E−8
0.01
kN
mN
Btu/(s⋅ft2⋅F)
cal(h⋅cm2⋅C)
cal/(h⋅m2⋅C)
cal/(s⋅cm2⋅C)
cal/(s⋅m2⋅C)
0.00000101972
Multiplied by
kg-force
Con­verted to
Btu/(h⋅ft2⋅F)
Con­verted from
Table 1.12 (continued) Unit Conversions
kcal/min
kW
oz
lb
stone
t (metric)
m-cd
kcal/(h⋅m2⋅C)
kJ/(h⋅m2⋅C)
kW/(m2⋅K)
W/(m2⋅K)
lb-force
oz-force
Con­verted to
0.00419993
0.00029281
0.0070548
0.000440925
0.000031494
0.0000002
43.055642
4.88251
20.44175
0.005678263
5.678263
0.000002248
0.0000359694
Multiplied by
90
Process engineering and design using visual basic®
bar
in. Hg
kg/cm2
kg/m2
kPa
MPa
cm/s
ft/h
ft/s
km/h
km/s
mN/m
N/m
°F
K
atm
cm/h
dyn/cm
°C
hp
kcal/h
© 2010 Taylor & Francis Group, LLC
°C * 1.8 + 32
°C + 273.15
1
0.001
Speed
cm/h
Pressure
atm
Temperature
°C
Surface Tension
dyn/cm
0.000277778
0.0328084
0.000009113
0.00001
2.777778E−9
1.01325
29.9213
1.0332
10332
101.325
0.101325
0.000392665
0.251996
°R
pdl/ft
kn
m/h
m/s
mi/h
mi/s
mbar
mm Hg
lb/ft2
lb/in.2
torr
lb-ft/min
W
continued
°C * 1.8 + 491.67
0.002205
5.399E−6
0.01
2.777778E−6
6.21371E−6
1.72603E−9
1013.25
760
2116.224
14.696
760
12.958
0.2928104
Chapter one: Basic mathematics
91
Con­verted to
Btu/(s⋅ft⋅F)
cal/(h⋅cm⋅C)
cal/(h⋅m⋅C)
cal/(s⋅cm⋅C)
cal/(s⋅m⋅C)
d
decade
fortnight
h
leap year
millennium
gf-cm
kgf-m
kN-m
MN-m
mN-m
Con­verted from
Btu/(h⋅ft⋅F)
century
dyn-cm
Con­verted from
© 2010 Taylor & Francis Group, LLC
0.001019716
1.01972E−8
1.0E−10
1.0E−13
0.0001
36525
10
2608.9286
876600
25
0.1
Torque
dyn/cm
Time
century
Thermal Conductivity
0.00027778
Btu/(h⋅ft⋅F)
14.8813
1488.13
0.00413369
0.4133694
Multiplied by
Table 1.12 (continued) Unit Conversions
N-m
ouncef-ft
poundf-ft
poundf-in.
ms
min
month
s
week
year
kcal/(h⋅m⋅C)
kJ/(h⋅m⋅C)
kW/(m⋅K)
W/(m⋅K)
Con­verted to
1.0E−7
1.180097E−6
7.37561E−8
8.850728E−7
3.15576E12
52596000
1200
3155760000
5217.857
100
1.48813
6.229312
0.00173036
1.73036
Multiplied by
92
Process engineering and design using visual basic®
g/(cm⋅s)
kg/(m⋅h)
lb/(ft⋅h)
lb/(ft⋅s)
mN⋅s/m2
cm3
ft3
in.3
m3
yd3
cup
drop
gal (E)
gal (U.S.)
cP
barrel
158,990
5.6147
9702.202
0.15899
0.207952
672.0064
2,451,625.8
34.974
42
0.01
3.6
2.42
0.00067222
1
Volume
barrel
Viscosity
cP
L
μL
mL
oz (E)
oz (U.S.)
pt (E)
pt (U.S.)
tablespoon
teaspoon
milli Pas⋅s
N⋅s/m2
Pas⋅s
P
158.99
158,990,000
158,990
5595.678
5376.006
279.7834
336.0053
10,752.012
32,256.036
1
0.001
0.001
0.01
Chapter one: Basic mathematics
© 2010 Taylor & Francis Group, LLC
93
94
Process engineering and design using visual basic®
Figure 1.19 Vessel.
The vessel menu gives the following calculation options:
• Horizontal
• Inclined
• Vertical
Program limitations
The following program limitations may be noted:
• The inclined vessel calculation is typically designed for a pipe without dished ends.
• If the filled length of the inclined vessel (Figure 1.22) is less than the
total length, the value of the height must be entered as 0 (default value).
• For the inclined vessel, any value of the height will ignore the value
of the length.
The Project details menu will open a project details form as shown in
Figure 1.20. This form will enable one to enter the general details of the
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
95
Figure 1.20 Project details.
project and the calculations. This also includes the calculation done by
and the calculation checked by options along with dates. Once the details
are entered, pressing the OK button will transfer the information to the
main vessel form.
Horizontal
Once the Horizontal menu executes, the form shown in Figure 1.21 will
appear. This form is designed for horizontal vessel calculations.
This form has two menus:
• File
• Unit
The File menu is further divided into four submenus: (1) Open, (2)
Save, (3) Print, and (4) Exit.
The Unit menu is used to select a proper unit for calculations. Both the
SI and Eng units can be used for calculation, the default unit being SI. It is
not possible to change the unit of individual items.
Data entry
During data entry, care must be taken while entering the units. Because
the basic units are fixed, the data must be entered for the unit shown in the
form. The tab key is preferably used for data entry. The tab key will allow
© 2010 Taylor & Francis Group, LLC
96
Process engineering and design using visual basic®
Figure 1.21 Horizontal vessel form.
data entry in a sequential manner, and once the data entry is complete, the
tab key will select the Calculate button for the calculation.
The data entry is simple; however, the program allows the following
options:
Type of dish end:
1. Elliptical
2. Hemispherical
The selection of dish ends has a significant impact on the calculation
of the volume, the surface area, and the weight.
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
97
Material of construction:
1. Aluminum
2. Copper
3. Gold
4. Iron
5. Mild steel
6. Monel
7. Nichrome
8. Platinum
9. Stainless steel
The material of construction is only used to estimate the vessel weight
and requires plate thickness for estimation. It must be noted that only the
shell thickness is entered, and it is assumed that the dish end thickness is
the same as the shell thickness. This may not be correct for all cases, and
the designer should keep this in mind. However, for a preliminary estimate, the calculated weight will be adequate.
In the weight calculation, a 5% margin is added. This addition is
mainly to account for the weight of nozzles, man way, and so on.
For a particular material in the list, the density is automatically
selected by the program. However, it is not possible to modify the density.
In case the material is not present in the list, it is preferable to select one
with closely matching density.
Buttons
1. Calculate
2. Cancel
The Calculate button is used to calculate the parameters after the
completion of data entry. If the tab key is used for data entry, this will
automatically select the Calculate button once the data entry is complete.
Pressing the Calculate button will calculate the parameters.
The Cancel button is used to cancel the execution. Once the Cancel
button is pressed, the program will be transferred to the main menu for
further calculation options.
Inclined
The general format of the inclined vessel calculation is presented in Figure
1.22. The general data entry and calculation principles are as discussed for
the horizontal vessel calculation. However, the following points should
be considered:
• The height indicated on the left side of the form is the height of the
liquid at the furthest left side of the inclined vessel. The default value
© 2010 Taylor & Francis Group, LLC
98
Process engineering and design using visual basic®
Figure 1.22 Inclined vessel form.
of the height is zero. In most slug catcher type of calculations, the
volume is calculated for partially filled vessels, and for such calculations, the height must be zero. The calculation is based on the total
length and the length as indicated in the form. In general, the value
of the length is smaller than the value of the total length, and the
value of the height must be zero. If any positive value is entered for
the height, the program will ignore the value entered for the length
and will calculate the volume and surface area based on the total
length and height.
• Dished ends have not been used in this calculation. In general, for
the slug catcher type of calculations, the impact of dished ends is
negligible.
The inclined calculation does not include the calculation of the weight,
and material selection is not required for this. The calculation is only for
the total and filled volumes and the total and wet surface areas.
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
99
Vertical
The general format of the vertical vessel calculation is presented in Figure
1.23. The general data entry and calculation principles are as discussed for
the horizontal vessel calculation. However, the following points should
be considered:
Figure 1.23 Vertical vessel form.
© 2010 Taylor & Francis Group, LLC
100
Process engineering and design using visual basic®
Figure 1.24 Conversion.
• The height indicated in the form is the height above the tan line of
the vessel as almost all calculations are like this. The dished end
volume added is based on the type of dished ends. In case a negative
height is entered, the program will calculate the dished end volume
and subtract the negative shell volume. The result will not represent
the actual volume of the liquid in the dished end.
• For an elliptical dished end, the dished end volume is calculated
for a dished end height 0.25 times the vessel diameter, and for a
hemispherical dished end, the volume is calculated for a dished end
height 0.5 times the vessel diameter.
Conversion (Figure 1.24)
Unit conversion has been divided into 23 subgroups, namely
• Acceleration
• Angle
© 2010 Taylor & Francis Group, LLC
Chapter one: Basic mathematics
101
• Area
• Computer
• Concentration
• Density
• Distance
• Energy
• Flow
• Force
• Heat transfer coefficient
• Light
• Mass
• Power
• Pressure
• Speed
• Surface tension
• Temperature
• Thermal conductivity
• Time
• Torque
• Viscosity
• Volume
Program limitations
The following program limitations may be noted:
• The program calculates the output values automatically when the
input values are changed.
• The program does not calculate the output values when the input/
output sections are changed.
• If the section of the input/output parameters is changed, the value in
the input value cell must be re-entered to calculate the output value.
• Large numbers of commonly used units have been selected for each
subgroup for conversion.
Procedure
The procedure for conversion is as follows:
• Select the required subgroup (from 23 radio buttons).
• Available units will appear in both the input and output boxes.
• Select the input unit; this is the unit that is to be converted. The input
value will indicate the selected unit.
• Select the output unit; this is the unit to be converted to. The output
value will indicate the selected unit.
© 2010 Taylor & Francis Group, LLC
102
Process engineering and design using visual basic®
• Enter the value in the input value text box. The output values will
be calculated automatically. The output value text box is read-only
and cannot be modified externally. The input value is to be modified
every time a calculation is required. Changing unit selection will
not automatically recalculate the output value.
References
1. Kreyszig, E., Advanced Engineering Mathematics, 7th ed., John Wiley and Sons,
New York, 1993.
2. Grossman, S.I., Multivariable Calculus, Linear Algebra and Differential Equations,
3rd ed., Sunders College Publishing, 1994.
3. Jain, M.K., Iyengar, S.R.K., and Jain, R.K., Numerical Methods for Scientific and
Engineering Computation, Wiley Eastern Limited, New Delhi, 1985.
© 2010 Taylor & Francis Group, LLC
chapter two
Thermodynamics
Introduction
The thermodynamic approach was developed in the nineteenth century
for a better understanding of the change in the energy in steam engines.
Certain principles were developed, known as the first and second laws
of thermodynamics, to mathematically correlate the changes in heat
and energy during a particular process. An understanding of the above
is extremely important for all branches of science and engineering. The
mathematical correlations developed help us to understand how the
physical properties and energies are changed during operations such as
compression, expansion, and so on.
Heat, work, and energy
From our experience, we know that if two bodies of different temperatures are brought into contact with each other, the heat is transferred from
the hot body to the cold body. In fact, the transfer of heat is directly proportional to the difference in temperatures between the two bodies. A
popular unit of heat is the calorie, which is defined as the heat required
to increase the temperature of 1 g of water by 1°C. The calorie is now recognized as a unit of energy. In the SI unit, the unit of energy is the joule,
which is equal to 1 Nm or the mechanical work done when a force of 1 N
acts through a distance of 1 m. Thus, mathematically, the work can be
defined as a product of force (N) and displacement (m).
Force
Force is derived from Newton’s second law as a product of mass and acceleration. The SI unit of force is N. Mathematically
F = ma
(2.1)
While the mass of a body is a scalar quantity that is independent of
location, force is a vector quantity that depends on the acceleration due
to gravity at any particular location. For example, if an astronaut weighs
© 2010 Taylor & Francis Group, LLC
1
2
Process engineering and design using visual basic®
750 N on Earth (a = 9.81 m/s2), his mass on Earth will be 76.45 kg and his
weight on the moon will be 127.67 N (assuming that the acceleration on
the moon is 1.67 m/s2).
Kinetic and potential energy
The concept of work, kinetic energy, and potential energy was developed
through Newton’s laws of motion. The work done is defined as
dW = madl
(2.2)
Considering the definition of acceleration as velocity over time and
velocity as length over time, Equation 2.2 can be modified as
dW = mvdv
(2.3)
 v 2 − v12 
 mv 2 
W = m 2
= ∆

 2 
 2 
(2.4)
Integrating
The term ½ mv2 is called the kinetic energy.
EK =
1
mv 2
2
(2.5)
If a body of mass m is raised by a distance z, the work done can be
defined as
W = Δ(mzg)
(2.6)
The term mzg is called the potential energy:
EP = mzg
(2.7)
First law of thermodynamics
The first law of thermodynamics defines the relationship between heat
and work. According to the first law, if heat is produced from work, there
will be a relationship between the work done and heat produced or vice
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
3
versa. Alternatively, energy can exist in different forms and the total
quantity of energy is always constant. If energy disappears in one form, it
will appear in some other form simultaneously.
According to the first law of thermodynamics, the relationship among
internal energy, heat, and work in a closed system can be expressed mathematically as
dU = dQ + dW
(2.8)
Phase rule
In a heterogeneous system, phase rule defines how the degree of freedom
(F) relates to the number of components (C) and the number of phases (P),
such as
F=C−P+2
(2.9)
As an example, water exists in two phases (liquid and vapor) at
101.3 kPa and 100°C. In this case, the degree of freedom is 1, meaning that
when water exists in the vapor and liquid phases and at 101.3 kPa pressure, the temperature cannot be changed from 100°C. If we add another
component, say, glycol, the degree of freedom will be 2. This means that
even at 101.3 kPa pressure and under a two-phase condition, the temperature can be a variable depending on the composition.
Reversible process
A system can change from one state to another through either a reversible or an irreversible process. In a reversible process, the change occurs
slowly in minute quantities till the total specified change happens.
If an expansion is carried out from an initial pressure P1 to a final
pressure P2 (volume from V1 to V2), the reversible work (Wr) and irreversible work (Wir) done at constant temperature are
Wr = RT ln
V2
P
= RT ln 1
V1
P2
P

Wir = P2 (V2 − V1 ) = RT  1 − 2 

P1 
© 2010 Taylor & Francis Group, LLC
(2.10)
(2.11)
4
Process engineering and design using visual basic®
The difference between reversible work and irreversible work can be
defined as
Wr − Wir =
RT
(P1 − P2 )2
P1P2
(2.12)
The right-hand side of Equation 2.12 is a positive quantity, meaning
that the work done in a reversible process is always greater than the work
done in an irreversible process.
Example 2.1
1 mol air at 300 K is expanded from 500 to 100 kPa. Estimate the work
done in both reversible and irreversible processes.
SOLUTION
Reversible work
Wr = 8.314 * 300 * ln(5) = 4014.3 J
Irreversible work
Wir = 8.314 * 300 * (1 − 0.2) = 1995.4 J
Heat content or enthalpy
The enthalpy of a system is defined as
H = U + PV
(2.13)
dH = dU + PdV + VdP
(2.14)
In differential form
Heat capacity at constant volume and constant
pressure
The heat capacity at constant volume is defined as the change in internal
energy with the temperature at constant volume, and the heat capacity at
contact pressure is defined as the change in enthalpy with the temperature at constant pressure. Mathematically [1]
 ∂U 
CV = 
 ∂T  V
© 2010 Taylor & Francis Group, LLC
(2.15a)
Chapter two:
Thermodynamics
5
 ∂H 
CP = 
 ∂T  P
(2.15b)
The values of two specific heats are not the same, and the relationship
between the two values can be expressed as
 ∂H 
 ∂U 
CP − CV = 
−
 ∂T  P  ∂T  V
 ∂U 
 ∂U 
 ∂V 
=
+ P
−



 ∂T  P
 ∂T  P  ∂T
T V
 ∂U 
 ∂U 
 ∂U   ∂V 
 ∂V 
=
+
+ P
−
 ∂T  V  ∂V  T  ∂T  P
 ∂T  P  ∂T  V
(Since U = f(T,V))
  ∂U 
  ∂V 
= 
+ P 


  ∂V  T
  ∂T  P
For ideal gas, (∂U/∂V)T = 0
Therefore
R
 ∂V 
CP − CV = P 
= P*
=R
 ∂T  P
P
(2.16)
Isothermal process
An isothermal process is a process when the temperature remains constant. The work done to increase an ideal gas pressure from P1 to P2 is
defined as
PV = constant
W = RT ln
P2
P1
(2.17)
Adiabatic process
An adiabatic process is the process when there is no heat transfer between
the system and the surrounding area (dQ = 0). For this case, the work done
© 2010 Taylor & Francis Group, LLC
6
Process engineering and design using visual basic®
to increase the system pressure from P1 to P2 is defined as [1]
PV γ = constant
( γ − 1)/ γ

RT1  P2 
− 1
 
γ − 1  P1 

W =
(2.18)
Example 2.2
1 kmol air is compressed from 100 kPa and 25°C to 500 kPa. Calculate
the work done if the compression is (a) isothermal and (b) adiabatic.
The ratio of specific heats for air is 1.4.
SOLUTION
a. The work done for an isothermal process
W = RT ln
P2
= 8.314 * 298 * 1.609 = 3987.5 kJ
P1
b. The work done for an adiabatic process
W =
8.314 * 298
(1.584 − 1) = 3616.1 kJ
0.4
Equation of state
Boyle’s law and Charles’s law
We can assume a cubical box with each side of length l (as shown in Figure
2.1) and whose three axes are x, y, and z. The velocity v can be defined as
v 2 = x 2 + y 2 + z2
y
B
z
Figure 2.1 Pressure in a cube.
© 2010 Taylor & Francis Group, LLC
(2.19)
c
x
A
Chapter two:
Thermodynamics
7
The momentum of a gas particle, with mass m, in the x-direction will
be mx, and if we assume that the gas molecule rebounds in an elastic manner, the change in momentum will be 2mx.
Now, with the distance between the two opposite walls equal to l, the
number of collisions per unit time will be x/l. The change in momentum
per unit time will be 2mx2/l. Considering all the sides, the total momentum imparted per unit time will be
2m ( x 2 + y 2 + z 2 ) 2mv 2
=
l
l
Now, considering N number of molecules, the total force exerted on
all the walls will be
F=
2mNv 2
l
(2.20)
Again, pressure P can be defined as
P=
1 mNv 2
F
=
6l 2
3 V
(2.21a)
1
mNv 2
3
(2.21b)
or
PV =
where
V = volume of the cube
Now
PV =
2
1
N * mv 2
3
2
For a given mass of gas, 2/3 N and ½ mv2 (kinetic energy) are constant
if the temperature is constant. Otherwise, PV is constant when the temperature is constant. This is Boyle’s law.
Again, at constant pressure, V/T is constant, which is Charles’s law.
Combining Boyle’s and Charles’s laws
PV = RT
where
R = universal gas constant
© 2010 Taylor & Francis Group, LLC
(2.22)
8
Process engineering and design using visual basic®
Equation of state for real gas
Equation 2.22 is valid for a perfect gas. In practice, the equation is required
to be modified to represent the actual world. There are large numbers of
semi-empirical correlations available in the literature, and they are generally developed following identical logics. A gas molecule experiences uniform attraction in all directions when located at the center of a container.
However, the gas molecule close to the wall will experience an unbalanced attraction (pulling inward). This will reduce the momentum of the
gas molecule. If this decrease in pressure is denoted as p/, then the ideal
pressure will be P + p/, where P is the actual observed pressure.
Similarly, the gas molecule will occupy some space within the container and the actual volume will be reduced to some extent. If the
reduction in volume is b, the actual volume will be V – b. With these modifications, the general equation is modified as
(P + a)(V − b) = RT
(2.23)
The first equation of state (EOS) was developed by J.D. van der Waals
in 1873 and presented as [1]
a 

 P + 2  (V − b) = RT
V
(2.24)
where the values of a,b typically depend on the pressure and temperature
conditions. Changing the values to zero, the equation will modify into the
equation of an ideal gas.
Various EOSs were developed to establish calculation procedures for
a and b. Out of a large number of EOSs, the following two are extensively
used in process engineering calculations:
• Peng–Robinson (PR) [2]
• Soave–Redlich–Kwong (SRK) [3]
Comparison between PR and SRK EOSs
The comparison between PR and SRK EOSs is presented in Table 2.1.
Acentric factor
One important parameter presented in the EOS is the acentric factor ω.
The acentric factor is used to correlate physical and thermodynamic properties. Mathematically, the acentric factor is defined as [4]
ω = − log pr* − 1.000
© 2010 Taylor & Francis Group, LLC
(2.26)
Chapter two:
Thermodynamics
9
Table 2.1 Comparison between PR and SRK EOSs
PR
P=
SRK
RT
a
−
V − b V (V + b) + b(V − b)
Z 3 − (1 − B)Z 2 + ( A − 2B − 3B2 )Z
Equation
number
RT
a
−
V − b V (V + b)
(2.25a)
Z3 − Z2 + (A − B − B2)Z −
AB = 0
(2.25b)
P=
− ( AB − B2 − B3 ) = 0
where
b=
∑x b
∑x b
i i
i i
i =1
i =1
bi = 0.077796
(2.25c)
N
N
RTci
Pci
 N

xi ai 0.5 
a=
 i =1

∑
2
0.08664
RTci
Pci
 N

xi ai 0.5 

 i =1

∑
2
aciαi
ai = aciαi
(2.25d)
(2.25e)
(2.25f)
(RTci )2
Pci
(2.25g)
α 0i .5 = 1 + mi (1 − T 0ri.5 )
1 + mi (1 − T ri0.5 )
(2.25h)
mi = 0.37464 + 1.54226 ω i − 0.26992ω 2i
0.48 + 1.574ω i − 0.176ω 2i
(2.25i)
aci = 0.457235
(RTci )2
Pci
0.42747
A=
aP
(RT )2
aP
(RT )2
(2.25j)
B=
bP
RT
bP
RT
(2.25k)
Note: T = absolute temperature, P = absolute pressure, V = specific volume, Z = compressibility, Tc = critical temperature, Pc = critical pressure, Tr = reduced temperature,
ω = acentric factor, a,b = parameters used in the equations, a(T) = parameter used in
the equations, ac = value of a(T) at T = Tc, α = correction factor, a/ac, m = slope of α0.5
against Tr0.5, i = component i in the mixture, N = number of components.
© 2010 Taylor & Francis Group, LLC
10
Process engineering and design using visual basic®
where
pr* = reduced vapor pressure, p*/pc
p* = vapor pressure at T = 0.7Tc, kPaA
pc = critical pressure, kPaA
T = temperature, K
Tc = critical temperature, K
For hydrocarbon mixtures, the acentric factor is calculated as
N
ω=
∑x ω
i
i
(2.27)
i =1
where
N = number of components
ωi = acentric factor of component i
xi = mole fraction of component i
Other important factors used to establish the physical and thermodynamic properties of hydrocarbon mixtures are the Watson characterization factor K and the critical compressibility factor zc. Table 2.2 presents
the values of the acentric factor, the Watson characterization factor, and
the critical compressibility factor. A complete list is available in the API
Technical Data Book [4].
Vapor pressure of pure components
The vapor pressure of pure components can be calculated using the
Antoine equation. The Antoine equation defines the vapor pressure of
pure components as [5]
ln(P) = A −
B
T+C
(2.28)
where
P = vapor pressure, mmHg
T = temperature, K
A,B,C = constants in the Antoine equation
For the commonly used hydrocarbon vapor, the values of constants A,
B, and C along with critical pressures and temperatures are presented in
Table 2.3. A more complete list is available in the literature [1,5].
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
11
Table 2.2 Acentric Factors, Watson Characterization Factors, and Critical
Compressibility Factors of Pure Substances
Watson
characterization
factor (K)
Critical
compressibility
factor (zc)
Components
Acentric
factor (ω)
Hydrogen
Nitrogen
Oxygen
Carbon monoxide
Carbon dioxide
Sulfur dioxide
Hydrogen sulfide
Water
Ammonia
Nonhydrocarbons
−0.2153
0.04
0.0218
0.0663
0.23894
0.2451
0.081
0.344
0.252
0.305
0.292
0.288
0.295
0.274
0.269
0.284
0.229
0.242
Methane
Ethane
Propane
n-Butane
i-Butane
n-Pentane
i-Pentane
2,2-Dimethylpropane
n-Hexane
i-Hexane
3-Methylpentane
2,2-Dimethylbutane
2,3-Dimethylbutane
n-Heptane
i-Heptane
3-Methylhexane
3-Ethylpentane
2,2-Dimethylpentane
2,3-Dimethylpentane
2,4-Dimethylpentane
3,3-Dimethylpentane
2,2,3-Trimethylbutane
n-Octane
Paraffins
0.011498
19.54
0.0986
19.49
0.1524
14.69
0.201
13.50
0.18479
13.78
0.25389
13.03
0.2224
13.01
0.1964
13.36
0.3007
12.78
0.2781
12.82
0.2773
12.65
0.2339
12.77
0.2476
12.62
0.3494
12.68
0.3282
12.72
0.3216
12.55
0.3094
12.36
0.2879
12.60
0.2923
12.35
0.3018
12.72
0.2672
12.42
0.2503
12.38
0.3962
12.68
0.288
0.284
0.280
0.274
0.282
0.269
0.270
0.269
0.264
0.267
0.273
0.272
0.269
0.263
0.261
0.255
0.268
0.267
0.256
0.265
0.273
0.266
0.259
continued
© 2010 Taylor & Francis Group, LLC
12
Process engineering and design using visual basic®
Table 2.2 (continued) Acentric Factors, Watson Characterization Factors, and
Critical Compressibility Factors of Pure Substances
Watson
characterization
factor (K)
Critical
compressibility
factor (zc)
Components
Acentric
factor (ω)
i-Octane
n-Nonane
n-Decane
0.3768
0.4368
0.4842
12.64
12.64
12.64
0.261
0.255
0.249
Cyclopropane
Cyclobutane
Cyclopentane
Methylcyclopentane
Ethylcyclopentane
Cyclohexane
Methylcyclohexane
Ethylcyclohexane
Cyclooctane
Napthenes
0.1348
11.93
0.1866
11.45
0.1943
10.94
0.2302
11.32
0.2715
11.39
0.2149
11.00
0.2350
11.31
0.2455
11.36
0.2536
10.89
0.274
0.274
0.273
0.272
0.269
0.273
0.269
0.270
0.270
Ethylene
Propylene
1-Butene
cis-2-Butene
trans-2-Butene
i-Butylene
1-Pentene
cis-2-Pentene
trans-2-Pentene
1-Hexene
cis-2-Hexene
trans-2-Hexene
1-Heptene
trans-2-Heptene
1-Octene
trans-2-Octene
1-Nonene
1-Decene
0.0852
0.1424
0.1867
0.2030
0.2182
0.1893
0.2330
0.2406
0.2373
0.2800
0.2722
0.2613
0.3310
0.3389
0.3747
0.3384
0.4171
0.4645
Olefins
© 2010 Taylor & Francis Group, LLC
14.21
13.04
12.61
12.93
13.01
12.65
12.48
12.61
12.50
12.30
12.45
12.41
12.39
12.40
12.36
12.42
12.46
0.277
0.275
0.276
0.272
0.274
0.275
0.270
0.279
0.279
0.265
0.266
0.267
0.262
0.256
0.256
0.260
0.249
0.247
Chapter two:
Thermodynamics
13
Table 2.2 (continued) Acentric Factors, Watson Characterization Factors, and
Critical Compressibility Factors of Pure Substances
Watson
characterization
factor (K)
Critical
compressibility
factor (zc)
Components
Acentric
factor (ω)
Propadiene
1,2-Butadiene
1,3-Butadiene
1,2-Pentadiene
1,3-Pentadiene
2,3-Pentadiene
0.1594
0.2509
0.1932
0.2235
0.0837
0.2194
12.58
12.16
12.51
11.90
12.21
11.91
0.271
0.267
0.270
0.256
0.285
0.253
Acetylene
Methylacetylene
Ethylacetylene
Acetylenes
0.1873
16.72
0.2161
12.34
0.0500
12.14
0.271
0.276
0.270
Benzene
Toluene
Ethylbenzene
o-Xylene
m-Xylene
p-Xylene
n-Propylbenzene
i-Propylbenzene
n-Butylbenzene
i-Butylbenzene
Napthalene
Aromatics
0.2108
9.74
0.2641
10.11
0.3036
10.33
0.3127
10.27
0.3260
10.41
0.3259
10.44
0.3462
10.59
0.3377
10.54
0.3917
10.82
0.3811
10.84
0.3019
9.32
0.271
0.264
0.263
0.263
0.259
0.260
0.265
0.262
0.261
0.256
0.269
Diolefins
Source: Adapted from Technical Data Book—Petroleum Refining, 4th ed., American Petroleum
Institute, Washington, D.C., 1982.
Vapor pressure of water
A much better correlation can be developed for the estimation of water
vapor pressure by using the following equation:
P = 10(7.97−1668.2/(T−45.2))
where
P = vapor pressure of water, mmHg
T = temperature, K
© 2010 Taylor & Francis Group, LLC
(2.29)
14
Process engineering and design using visual basic®
Table 2.3 ​Values of Antoine Constants and Critical Pressures and Temperatures
Constant in Antoine
equation
Temperature
limits (°C)
Compo­
nents
A
C
Min
Max
Pressure Temperature
(kPa)
(K)
Methane
15.2243
597.84
−7.16
−180
−153
4640.7
190.7
Ethane
Propane
n-Butane
i-Butane
n-Pentane
n-Hexane
n-Heptane
n-Octane
n-Nonane
n-Decane
Carbon
monoxide
Carbon
dioxide
Hydrogen
sulfide
Nitrogen
Oxygen
Hydrogen
Water
15.6637
15.726
15.6782
15.5381
15.8333
15.8366
15.8737
15.9426
15.9671
16.0114
14.3686
1511.42
1872.46
2154.9
2032.73
2477.07
2697.55
2911.32
3120.29
3291.45
2456.8
530.22
−17.16
−25.16
−34.42
−33.15
−39.94
−48.78
−56.51
−63.63
−71.33
−78.67
−13.15
−143
−109
−78
−86
−53
−28
−3
19
39
57
−210
−74
−24
17
7
57
97
127
152
179
203
−165
4883.9
4256.7
3796.6
3647.6
3375.1
3031.6
2736.8
2490
2290
2110
3499
305.4
369.9
425.2
408.1
469.6
507.9
540.2
568.7
594.6
617.7
132.9
22.5898
3103.39
−0.16
−119
−69
7370
304.1
16.1040
1768.69
−26.06
−83
−43
9007.8
373.6
14.9542
15.4075
13.6333
18.3036
588.72
734.55
164.9
3816.44
−6.6
−6.45
3.19
−46.13
−219
−210
−259
11
−183
−173
−248
168
3394.4
5043
1313
22,120
126.2
154.6
33.19
647.3
B
Critical conditions
Vapor pressure calculation using EOSs
The vapor pressure of pure components can be calculated by estimating
the fugacity of the component. The general equation of fugacity for pure
components
SRK EOS
f
A  Z + B
= Z − 1 − ln(Z − B) − ln 

P
B  Z 
(2.30)
f
Z + 2.414B
A
= Z − 1 − ln(Z − B) −
ln
P
2 2B Z − 0.414B
(2.31)
ln
PR EOS
ln
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
15
Vapor pressure is the equilibrium pressure, where
fL = fV
where
f L = fugacity of the liquid phase
f V = fugacity of the vapor phase
P = pressure
Z = compressibility
B = parameter defined in Table 2.1
Example 2.3
Estimate the vapor pressure of n-hexane at 31.6°C (304.75 K).
SOLUTION
Use the Antoine equation
A = 15.8366
B = 2697.55
C = −48.78
ln(P) = 15.8366 −
2697.55
= 5.29806
304.75 − 48.78
or
P = 199.95 mmHg = 26.66 kPa
Use PR EOS
Critical pressure = 3025 kPa
Critical temperature = 507.6 K
Acentric factor = 0.3047
Temperature = 304.75 K
Parameters are calculated as
ac = 0.457235
(8.314 * 507.6)2
= 2692.04
3025
mi = 0.37464 + 1.54226 * 0.33047 − 0.26992 * 0.3047 2 = 0.8195
Tri = 304.75/507.6 = 0.6
α = (1 + 0.8195 * (1 − 0.6 0.5 ))2 = 1.4031
a = ac * α = 3777.15
b = 0.077796
8.314 * 507.6
= 0.1085
3025
© 2010 Taylor & Francis Group, LLC
(2.32)
16
Process engineering and design using visual basic®
The vapor pressure will be the pressure at which the fugacity of
the liquid and vapor phases will be the same. This is calculated by
assuming an initial value and then modifying the value suitably. This
calculation converges very rapidly.
Assumption 1: Vapor pressure is 1 kPa
A=
3777.15 * 1
= 0.0005884
(8.314 * 304.75) 2
B=
0.1085 * 1
= 0.000043
304.75 * 8.314
Using the above values of A and B, the cubic equation will be
Z3 − 0.99996Z2 + 0.000503Z − 2.337 * 10−8 = 0
The above equation has three real roots; the highest value is the
compressibility of the vapor phase and the lowest one is the compressibility of the liquid phase:
Compressibility of vapor phase = 0.99945
Compressibility of liquid phase = 0.000052
The fugacity can be calculated as
Liquid-phase fugacity = 26 kPa
Vapor-phase fugacity = 1 kPa
Since the values are not the same, the initial pressure is to be
modified and recalculated.
This calculation is repeated till the liquid and vapor phase fugacity difference is <0.00001.
The estimated vapor pressure is 26.7 kPa (200.3 mmHg).
The calculated vapor pressure of hexane at 31.6°C:
Using the Antoine equation = 26.66 kPa (199.95 mmHg)
Using PR EOS = 26.7 kPa (200.3 mmHg)
Literature value [6] = 200 mmHg
Second law of thermodynamics
The first law of thermodynamics states that the energy is interchangeable,
but it did not provide any indication whether change will occur at all or to
what extent it will occur. The second law provides the essential conditions
to make a change possible.
• The change from heat to work is possible through a thermodynamic
engine, which works in a reversible cyclic process.
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
17
• The thermodynamic engine must work between two temperatures,
taking up heat from the higher-temperature source, converting a
portion into work, and giving up the rest of the heat to the lowertemperature sink.
Carnot’s cycle
Carnot’s cycle (Figure 2.2) can be used to explain how and to what extent
work is obtained from heat. In this cycle, the engine starts at a point A and
comes back to the same point. Alternatively, the engine must operate in
a complete cycle. Also, maximum work can be obtained when every step
operates in a reversible fashion.
In the above cycle, the following operations can be analyzed.
Operation 1: Gas is allowed to expand, isothermally and reversibly,
from point A to point B (volume changes from V1 to V2). For an ideal gas,
the heat absorbed is equal to the work done.
Q1 = RT1 ln
V2
V1
(2.33)
Operation 2: The gas is then allowed to expand from V2 to V3 adiabatically and reversibly till the temperature drops to the temperature of the
sink. Owing to the adiabatic process, the heat exchange is 0. The work
done (W1) by the gas can be calculated as
W1 = CV (T1 − T2)
(2.34)
Operation 3: The gas is compressed isothermally and reversibly from V3
to V4. For an ideal gas, the heat given out is equal to the work done by the gas:
Q2 = RT2 ln
A V1
V4
V3
T1
(2.35)
B
V2
P
D
V4
T2
V
Figure 2.2 Carnot’s cycle.
© 2010 Taylor & Francis Group, LLC
C V3
18
Process engineering and design using visual basic®
Operation 4: The gas is compressed adiabatically and reversibly from
V4 to V1 and comes back to the initial state.
W2 = −CV (T1 − T2)
(2.36)
The net work (W) done by the gas in the complete cycle
W = RT1 ln
V2
V
+ CV (T1 − T2 ) + RT2 ln 4 − CV (T1 − T2 )
V1
V3
(2.37)
V2
V
+ RT2 ln 4
V1
V3
(2.38)
or
W = RT1 ln
Now, from the first law
V2
V
= 3
V1
V4
Therefore, the net work done can be defined as
W = R (T1 − T2 )ln
V2
V1
(2.39)
The efficiency (η) of the process is given as
W
T
= 1− 2
Q1
T1
(2.40)
T 

W = Q1  1 − 2 

T1 
(2.41)
η=
And the work is given as
This relationship is the mathematical form of the second law.
Entropy
With some extension of the above analysis, it can be concluded that a thermodynamic change defined as dQ/T is independent of the path of the
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
19
transformation of the system. This function is called entropy and is normally denoted by S.
Mathematically, the change in entropy is measured by the ratio of
the heat change and temperature at which the heat change occurs and is
defined as
dS =
dQrev
T
(2.42)
The term dQrev indicates that the heat change occurs through a reversible process.
Sensible heat
Enthalpy is a function of temperature and pressure
H = f(T,P)
or
 ∂H 
 ∂H 
dH = 
dT + 
dP
 ∂ T  P
 ∂ P  T
or
 ∂H 
dH = CP dT + 
dP
 ∂ P  T
(2.43a)
In the above equation, the second term is zero for the following
conditions:
a. When the pressure is constant
b. When enthalpy is independent of the pressure, for example, lowpressure gas
Considering the above conditions, the change in enthalpy can be
defined as
dH = CP dT
where CP is the isobaric specific heat.
© 2010 Taylor & Francis Group, LLC
(2.43b)
20
Process engineering and design using visual basic®
Thermodynamic properties
The following thermodynamic properties are extremely important in all
process engineering calculations. This section will discuss the method to
estimate these thermodynamic properties.
• Isobaric specific heat of hydrocarbon ideal gases
• Isobaric specific heat of hydrocarbon real gases
• Isobaric specific heat of hydrocarbon gas mixtures
• Isobaric specific heat of ideal liquids
• Isobaric specific heat of real liquids
• Enthalpy of gases
• Enthalpy of gas mixtures
• Entropy of ideal gases
• Entropy of real gases
• Entropy of hydrocarbon gas mixtures
• Viscosity of ideal liquids
• Viscosity of ideal hydrocarbon vapors
• Liquid viscosity of defined mixtures at low pressure
• Vapor viscosity of defined mixtures at low pressure
• Thermal conductivity of pure hydrocarbon liquids at low pressure
• Thermal conductivity of pure hydrocarbon vapors at low pressure
Isobaric specific heat of hydrocarbon ideal gases
Specific heat at a constant pressure is an important parameter for engineering calculations. This value is largely influenced by the temperature
and the actual pressure of the fluid. The specific heat of pure components
for an ideal gas is calculated as [1]
CP/R = A + BT + CT2 + DT−2
(2.44)
where
CP = specific heat of ideal gas, kJ/(kmol⋅K)
R = universal gas constant, kJ/(kmol⋅K)
T = temperature, K
A,B,C,D = constants
Constants for the calculation of the specific heat of commonly used
ideal gases are presented in Table 2.4.
Example 2.4
Estimate the heat required to raise the temperature of 1 kmol methane from 100°C to 400°C.
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
21
Table 2.4 Constants for the Calculation of Specific Heat—Ideal Gases
Ideal Gases
MW
Tmax (K)
A
Methane
16.043
1500
Ethane
30.070
1500
Propane
44.097
n-Butane
i-Butane
D * 10−5
B * 103
C * 106
1.702
9.081
−2.164
1.131
19.225
−5.561
1500
1.213
28.785
−8.824
58.123
1500
1.935
36.915
−11.402
58.123
1500
1.677
37.853
−11.945
n-Pentane
72.150
1500
2.464
45.351
−14.111
n-Hexane
86.177
1500
3.025
53.722
−16.791
n-Heptane
100.204
1500
3.570
62.127
−19.486
n-Octane
114.231
1500
4.108
70.567
−22.208
Air
28.851
2000
3.355
0.575
Nitrogen
28.014
2000
3.280
0.593
0.040
Oxygen
31.999
2000
3.639
0.506
−0.227
Hydrogen
2.016
3000
3.249
0.422
0.083
Hydrogen sulfide
34.082
2300
3.931
1.49
−0.232
Carbon monoxide
28.010
2500
3.376
0.557
−0.031
Carbon dioxide
44.01
2000
5.457
1.045
−1.157
Water
18.015
2000
3.47
1.45
0.121
−0.016
Source: Adapted from Smith, J.M., Van Ness, H.C., and Abbott, M.M., Introduction to Chemical
Engineering Thermodynamics, 6th ed., McGraw-Hill, New York, 2001.
Note: Values for other components are available in the literature [1,5,7]. Tmax is the maximum temperature, K.
SOLUTION
The heat required can be calculated as
T2
Q = dH =
∫ C dT
P
T1
k
CP
∫ R dT
=R
1


T2
T3
= R  AT1 (k − 1) + B 1 (k 2 − 1) + C 1 (k 3 − 1)
2
3


where
T1 = initial temperature = 373.15 K
k = temperature ratio = 673.15/373.15 = 1.804
A,B,C = constants as per Table 2.4
© 2010 Taylor & Francis Group, LLC
22
Process engineering and design using visual basic®
Solving the above equation for methane

373.152
Q = 8.314 1.702 * 373.15 * 0.804 + 9.801 * 10 −3
*
2

2.254 − 2.164 * 10 −6

373.153
* 4.871
3

= 15515 kJ
Isobaric specific heat of hydrocarbon real gases
Isobaric specific heat is an important thermodynamic parameter and is
used in several calculations, including sensible heat effect, Joule–Thomson
cooling, and so on [4].
Isobaric specific heat is calculated as
 C 0 − CP 
CP = C P0 − R  P
R 

(2.45)
where
CP = isobaric specific heat of the real gas, kJ/(kmol·K)
C P0 = isobaric specific heat of the ideal gas, kJ/(kmol·K)
R = gas constant = 8.314 kJ/(kmol·K)
(C P0 − CP )/R = dimensionless correction factor
The dimensionless correction factor is calculated as
0
0
h
 C P0 − CP  
CP0 − CP  CP0 − CP 
ω  C P0 − CP 

=
+ h 
−
R
R 
R 
R  
ω 




(2.46)
where
((C P0 − CP )/R)0 = effect of the pressure for simple fluid (use Equation
2.47 to calculate)
((C P0 − CP )/R)h = effect of the pressure for the heavy reference fluid
(n-octane) (use Equation 2.47 to calculate)
The pressure effect is calculated as
i
 CP0 − CP 
T (∂Pr /∂Tr )V2 r  ∆CV 
= 1+ r
+


 R 
R 
( ∂P /∂V )
r
© 2010 Taylor & Francis Group, LLC
r Tr
i
(2.47)
Chapter two:
Thermodynamics
23
where
1 
 ∂ Pr 
2c
b1 + b3 /Tr2 + 2b4 /Tr3 c1 − 2c3 /Tr3
d
+
+ 15 − 3 4 2
 ∂ T  = V 1 +
2
Tr Vr
Vr
Vr
Vr
r Vr
r 

γ 
 γ   
×  β + 2  exp  − 2   
 Vr   
Vr 

(2.48)

Tr 
c4 
γ  γ 
2B 3C 6D
 ∂ Pr 

 ∂ V  = − V 2 1 + V + V 2 + V 5 + T 3V 2  3β + 5 − 2  β + V 2   V 2 
r Tr
r 
r
r
r
r r 
r

 r 

 γ 
× exp  − 2  
 Vr  
(2.49)
i
2 (b3 + 3b4 /Tr )
3c
 ∆CV 
+ 3 3 2 + 6E

 = −
2
R
Tr Vr
Tr Vr
(2.50)
where
Pr = reduced pressure
Tr = reduced temperature
Vr = reduced volume = ZTr/Pr
Z = compressibility
B = b1 −
b2
b
b
− 32 − 43
Tr Tr
Tr
(2.51a)
c2
c
+ 33
Tr Tr
(2.51b)
d2
Tr
(2.51c)
C = c1 −
D = d1 +
E=
γ 
c4 

 γ 
β + 1 −  β + 1 + 2  exp  − 2  
3 

 Vr  
2Tr γ 
Vr 
(2.51d)
Other constants used in the above equations for both simple and
heavy reference fluids are tabulated in Table 2.5.
© 2010 Taylor & Francis Group, LLC
24
Process engineering and design using visual basic®
Table 2.5 ​Values of Constants
Constants
Simple fluid
Heavy reference
fluid
b1
b2
b3
b4
c1
c2
c3
c4
d1 * 104
d2 * 104
β
γ
0.1181193
0.265728
0.154790
0.030323
0.0236744
0.0186984
0
0.042724
0.155488
0.623689
0.65392
0.060167
0.2026579
0.331511
0.027655
0.203488
0.0313385
0.0503618
0.016901
0.041577
0. 48736
0.0740336
1.226
0.03754
Source: Adapted from Technical Data Book—Petroleum
Refining, 4th ed., American Petroleum Institute,
Washington, D.C., 1982.
Isobaric specific heat of hydrocarbon gas mixtures
The above method can also be used to calculate the isobaric specific heat of
real hydrocarbon gas mixtures [4]. However, the pseudocritical properties
are to be used for the calculation. Pseudocritical properties are calculated as
n
Ppc =
∑x P
i ci
(2.52)
∑xT
(2.53)
i =1
n
Tpc =
i ci
i =1
n
ω=
∑x ω
i
i
(2.54)
i =1
CP =
∑w C
i
pi
where
Ppc = mixture’s pseudocritical pressure, kPa
Pci = critical pressure of component i, kPa
Tpc = mixture’s pseudocritical temperature, K
Tci = critical temperature of component i, K
© 2010 Taylor & Francis Group, LLC
(2.55)
Chapter two:
Thermodynamics
25
n = number of components
xi = mole fraction of component i
ω = mixture’s acentric factor
ωi = acentric factor of component i
CP = mixture’s isobaric specific heat
CPi = isobaric specific heat of component i
wi = weight fraction of component i
Example 2.5
Estimate the isobaric specific heat for
i. Methane at 10,000 kPag and 20°C
ii. Ethane at 3000 kPag and 50°C
iii. Nitrogen at 8000 kPag and 30°C
iv. Mixed gas (nitrogen 10 mol%, methane 45 mol%, and ethane 45 mol%, at 3000 kPag and 30°C) (ignore the interaction
parameters)
SOLUTION
i. Isobaric specific heat of ideal gas = 34.74 kJ/(kmol·K)
Compressibility = 0.81698
 ∂ Pr 
 ∂ T  = 2.3698
r Vr
 ∂ Pr 
 ∂ V  = −3.4043
r Tr
i
 ∆CV 
 R  = −0.2418
0
 C P0 − CP 
= −1.77765

R 
 C P0 − CP 
= −1.7572

R 
CP = 34.74 + 8.314 * 1.7572 = 49.35 kJ/(kmol·K)
(HYSYS® calculated value is 49.32 kJ/(kmol·K))
ii. Isobaric specific heat of ideal gas = 56.226 kJ/(kmol·K)
Compressibility = 0.77283
 C P0 − CP 
= −1.6363

R 
CP = 69.83 kJ/(kmol·K)
(HYSYS calculated value is 69.98 kJ/(kmol·K))
© 2010 Taylor & Francis Group, LLC
26
Process engineering and design using visual basic®
iii. Isobaric specific heat of ideal gas = 29.13 kJ/(kmol·K)
Compressibility = 0.98869
 C P0 − CP 
= − 0.3971

R 
CP = 32.43 kJ/(kmol·K)
(HYSYS calculated value is 32.99 kJ/(kmol·K))
iv. Pseudocritical pressure = 4625.48 kPa
Pseudocritical temperature = 235.88 K
Acentric factor = 0.05354
Compressibility factor = 0.87746 (Note-1)
Isobaric specific heat of methane at 30°C = 35.38 kJ/(kmol·K)
Isobaric specific heat of ethane at 30°C = 53.61 kJ/(kmol·K)
Isobaric specific heat of nitrogen at 30°C = 29.13 kJ/(kmol·K)
Weight fraction of methane = 0.3064
Weight fraction of ethane = 0.5743
Weight fraction of nitrogen = 0.1193
Isobaric specific heat of ideal mixtures = 45.11 kJ/(kmol·K)
 CP0 − CP 
= − 0.7559

R 
CP = 51.4 kJ/(kmol·K)
(HYSYS calculated value is 49.64 kJ/(kmol·K), Note-1)
Note-1: The compressibility factor is calculated without considering the interaction parameters. As a result, both the compressibility factor and the isobaric specific heat of real gas
mixtures will be different from the calculated value using the
interaction parameters.
Joule–Thomson coefficient
If we consider a case where volume V1 of a gas at pressure P1 is allowed to
pass to a lower pressure P2, through a semipermeable membrane slowly
and reversibly, the volume at the lower pressure side will increase to V2
(V2 > V1) as shown in Figure 2.3. If we assume that the piston is frictionless
and the system is thermally insulated and there is no heat change (dQ = 0),
then the work done can be defined as
Work done by the high-pressure side = P1V1
Work done by the low-pressure side = P2V2
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
27
V1
P1
V2
P2
Figure 2.3 Joule–Thomson effect.
Now, from the first law of thermodynamics
dU = −dW = P1V1 − P2V2
or
U1 + P1V1 = U2 + P2V2
or
H1 = H2
or
ΔH = 0
This indicates that an adiabatic expansion is also isenthalpic.
However, during this expansion, there will be a change in the temperature, and this variation of temperature is called the Joule–Thomson effect.
The rate of change in temperature with pressure is called the Joule–
Thomson coefficient.
 dT 
µ=
 dP  H
(2.56)
Generally, the value of μ is positive, meaning that with a drop in pressure, the temperature will also drop. However, in some cases (hydrogen,
helium at room temperature), the value of μ is negative, meaning that
there will be an increase in temperature with a decrease in pressure.
Now, H = f(T,P)
1  ∂H 
∂T 
 ∂T   ∂H 

 = − 
 
 = −


∂P H
∂H P ∂P T
CP  ∂ P  T
or
µ=−
© 2010 Taylor & Francis Group, LLC
1  ∂H 


CP  ∂ P  T
(2.57)
28
Process engineering and design using visual basic®
Again
RT 2  ∂ Z 
 ∂H 
 ∂V 
=
−
=
−
V
T






∂P  T
∂T  P
P  ∂T  P
where
Z = compressibility factor
The general equation of the Joule–Thomson coefficient can be
defined as [1]
µ=
RT 2  ∂ Z 


CP P  ∂ T  P
Example 2.6
Estimate the Joule–Thomson cooling temperature if methane at
10,000 kPag and 20°C is expanded to the atmospheric pressure.
SOLUTION
This can be calculated by reducing the pressure in steps and estimating
the temperature at the end of each step. The isobaric specific heat of a
real gas is calculated using the procedure described in Example 2.5.
The PR EOS is used for this calculation.
Let the number of steps be 50; therefore, the reduction in pressure
at the end of each step will be 200 kPa.
Constant parameters
Critical pressure = 4640.68 kPa
Critical temperature = 190.7 K
Acentric factor = 0.0115
Step 1 calculation
Calculated the value of mi (Equation 2.25i) = 0.3923
Calculated the value of aci (Equation 2.25g) = 247.67
Calculated the value of bi (Equation 2.25d) = 0.0266
Compressibility (Z) = 0.81792
Value of (∂Z/∂T)P = 0.00239
Isobaric heat capacity of a real gas = 49.35 kJ/(kmol·K)
Joule–Thomson coefficient (Equation 2.58) = 0.00342
Drop in temperature (Equation 2.56) = 0.68°C
Temperature after first step = 19.32°C
The above step is repeated 50 times to calculate the final
temperature.
© 2010 Taylor & Francis Group, LLC
(2.58)
Chapter two:
Thermodynamics
29
Table 2.6 ​Joule–Thomson Temperature Calculation for Single Component
Pressure (kPag)
Final temperature
(oC)
Component
Initial
Final
Initial
temperature (oC)
HYSYS
Manual
Methane
Methane
Methane
Ethane
Propane
i-Butane
Carbon dioxide
Nitrogen
Nitrogen
Nitrogen
10,000
8000
8000
3000
2000
1000
5000
8000
7000
6000
0
0
3000
0
0
0
0
0
0
2000
20
−7
−10
60
100
150
60
−7
10
0
−32.3
−60.5
−39.9
29.1
78.6
141.6
8.1
−30.3
−8.2
−10.6
−32.0
−61.3
−40.4
30.1
78.8
141.6
6.1
−30.6
−8.4
−10.7
Calculated final temperature = −32.0°C (Note-1)
(HYSYS calculated temperature = −32.3°C)
Note-1: This calculation is not valid if the liquid phase is detected
at any step.
The results of a few typical calculations along with the HYSYS
calculated values are presented in Table 2.6.
Isobaric specific heat of ideal liquids
Similarly, specific heat at a constant pressure for liquids can be calculated as
CP/R = A + BT + CT2
(2.59)
Constants for the calculation of specific heat of commonly used liquids
are presented in Table 2.7.
Isobaric specific heat of real liquids
The calculation of the isobaric specific heat for a real liquid is identical
to the procedure used for a real gas [4]. The isobaric specific heat of real
hydrocarbon liquid mixtures is calculated as [4]
CP =
∑w C
i
pi
where
CP = isobaric specific heat of mixtures
wi = weight fraction of component i
CPi = isobaric specific heat of component i
© 2010 Taylor & Francis Group, LLC
(2.60)
30
Process engineering and design using visual basic®
Table 2.7 Constants for the Calculation of Specific Heat—Liquids
Liquids
A
B * 103
C * 106
Ammonia
22.626
−100.75
192.71
Aniline
15.819
29.03
−15.80
Benzene
−0.747
67.96
−37.78
1,3-Butadiene
22.711
−87.96
205.79
Carbon tetrachloride
21.155
−48.28
101.14
−31.90
Chlorobenzene
11.278
32.86
Chloroform
19.215
−42.89
83.01
Cyclohexane
−9.048
141.38
−161.62
Ethanol
33.866
−172.60
349.17
Ethylene oxide
21.039
−86.41
172.28
Methanol
13.431
−51.28
131.13
n-Propanol
41.653
−210.32
427.20
Sulfur trioxide
−2.930
137.08
−84.73
Toluene
15.133
6.79
16.35
Water
8.712
1.25
−0.18
Source: Adapted from Smith, J.M., Van Ness, H.C., and Abbott, M.M., Introduction to
Chemical Engineering Thermodynamics, 6th ed., McGraw-Hill, New York, 2001.
Enthalpy of gases
The enthalpy of a gas is estimated by estimating the enthalpy of an ideal
gas. The enthalpy of an ideal gas is calculated using the following equation:
0
T
H =H
0
25
R
+
M
T
CP
∫ R dT
(2.61)
298.15
where
H T0 = ideal gas enthalpy of formation at temperature T, kJ/kg
0
H 25
= ideal gas enthalpy of formation at 25°C, kJ/kg
R = gas constant, kJ/(kmol·K)
M = molecular weight
CP = isobaric specific heat, kJ/(kg·°C)
T = temperature, K
The ideal gas enthalpy of formation at 25°C is presented in Table 2.8 [4].
The enthalpy of a real gas is estimated by using the following equation [1]:
HT = H T0 + H R
© 2010 Taylor & Francis Group, LLC
(2.62)
Chapter two:
Thermodynamics
31
Table 2.8 Ideal Gas Enthalpy of Formation at 25°C
MW
Heat of formation
at 25°C (kJ/kg)
Methane
16.043
−4645.1
Ethane
30.070
−2787.5
Propane
44.097
−2373.9
n-Butane
58.123
−2164.2
i-Butane
58.123
−2308.5
n-Pentane
72.150
−2034.1
n-Hexane
86.177
−1937.2
n-Heptane
100.204
−1872.7
n-Octane
114.321
−1827.4
Component
Source: Adapted from Technical Data Book—Petroleum Refining,
4th ed., American Petroleum Institute, Washington,
D.C., 1982.
Note: Values for other components are available in the literature [4].
where
HT = enthalpy of real gas, kJ/kg
HR = residual enthalpy, kJ/kg
Residual enthalpy is calculated as [1]
 d ln α(Tr )
 
HR
1 
=
− 1 qI 
Z − 1 + 
RT
M 
 d ln(Tr )
 
(2.63)
Different equations are used for different EOSs to calculate the value
of α(Tr) are indicated below [1].
For the PR EOS
α (Tr ) = [1 + (0.37464 + 1.54226ω − 0.26992ω 2 )(1 − Tr0.5 )]2
(2.64a)
For the SRK EOS
α (Tr ) = [1 + (0.480 + 1.574 ω − 0.176ω 2 )(1 − Tr0.5 )]2
where
ω = acentric factor
© 2010 Taylor & Francis Group, LLC
(2.64b)
32
Process engineering and design using visual basic®
The value of q is calculated using the following equation [1]:
For the PR EOS
q=
0.457235α(Tr )
0.077796Tr
(2.65a)
q=
0.42747α (Tr )
0.08664(Tr )
(2.65b)
For the SRK EOS
The value of I is calculated using the following equations [1]:
 Z + β
I = ln 
 Z 
(2.66)
Pr
Tr
(2.67a)
Pr
Tr
(2.67b)
where
Z = compressibility
and for the PR EOS
β = 0.077796
For the SRK EOS
β = 0.08664
where
Pr = reduced pressure
Tr = reduced temperature
Enthalpy of gas mixtures
For hydrocarbon gas mixtures, a similar procedure is used to estimate
the enthalpy of gas mixtures with some modifications to the pressure and
temperature parameters [4].
The critical temperature is replaced by a pseudocritical temperature
using the following equation:
n
Tpc =
∑xT
i ci
i =1
© 2010 Taylor & Francis Group, LLC
(2.68)
Chapter two:
Thermodynamics
33
The critical pressure is replaced by a pseudocritical pressure using
the following equation:
n
∑x P
Ppc =
(2.69)
i ci
i =1
The mixture acentric factor is calculated as
n
ω=
∑xω
i
(2.70)
i
i =1
The ideal gas enthalpy of the mixture is calculated as
H0 =
∑x H
wi
0
i
where
Tpc = pseudocritical temperature of the mixture, K
Tci = critical temperature of component i, K
xi = mole fraction of component i
Ppc = pseudocritical pressure of the mixture, kPa
Pci = critical pressure of component i, kPa
ω = mixture acentric factor
ωi = acentric factor of component i
H0 = ideal gas enthalpy of the mixture, kJ/kg
xwi = weight fraction of component i
H i0 = ideal gas enthalpy of component i
n = number of components
Example 2.7
Estimate the enthalpy of methane at 293.15 K and 10,101.3 kPa and
n-butane at 500 K and 5000 kPa. Use the PR EOS for the calculation.
SOLUTION
Enthalpy of methane
The enthalpy of methane at 25°C is –4645.1 kJ/kg.
The change in enthalpy for a temperature change from 298.15 to
293.15 K is calculated using the following equation:
T2
dH =
∫ C dT
P
T1
=

R 
T12 2
T13 3
(k − 1)
 AT1 (k − 1) + B (k − 1) + C
2
3
M

© 2010 Taylor & Francis Group, LLC
(2.71)
34
Process engineering and design using visual basic®
where
R = gas constant, kJ/(kmol·K) = 8.314
T1 = initial temperature = 298.15
k = temperature ratio = 0.983
A = constant = 1.702
B = constant = 9.081 * 10−3
C = constant = −2.164 * 10−6
M = molecular weight = 16.043
Differential enthalpy = −10.9 kJ/kg
The enthalpy of ideal methane at 20°C is −4656 kJ/kg
Calculation of residual enthalpy
The compressibility of methane at 10,101.3 kPa and 293.5 K is 0.8179.
(The calculation of compressibility has been discussed before.)
Acentric factor = 0.011498
Critical pressure = 4640.7 kPa
Critical temperature = 190.7 K
Reduced pressure = 2.177
Reduced temperature = 1.573
Value of α(Tr) = 0.82065
Value of q = 3.1376
Value of I = 0.1263
Residual enthalpy = −120.2 kJ/kg
Enthalpy of real gas = −4776.2 kJ/kg
(HYSYS calculated enthalpy = −4795 kJ/kg)
Enthalpy of n-butane
The enthalpy of methane at 25°C is –2164.2 kJ/kg.
The change in enthalpy for a temperature change from 298.15 to
500 K is calculated using the following equation:
T2
dH =
∫ C dT
P
T1
=

R 
T12 2
T13 3
(k − 1)
 AT1 (k − 1) + B (k − 1) + C
2
3
M

where
R = gas constant, kJ/(kmol·K) = 8.314
T1 = initial temperature = 298.15
k = temperature ratio = 1.6773
A = constant = 1.935
B = constant = 36.915*10−3
C = constant = −11.402*10−6
M = molecular weight = 58.123
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
35
Differential enthalpy = 427.7 kJ/kg
The enthalpy of ideal methane at 500 K = −1736.5 kJ/kg
Calculation of residual enthalpy
The compressibility of methane at 5000 kPa and 500 K is 0.6908.
(The calculation of compressibility has been discussed before.)
Acentric factor = 0.201
Critical pressure = 3796.6 kPa
Critical temperature = 425.2 K
Reduced pressure = 1.317
Reduced temperature = 1.176
Value of α(Tr) = 0.8895
Value of q = 4.446
Value of I = 0.1188
Residual enthalpy = −89.1 kJ/kg
Enthalpy of real gas = −1825.6 kJ/kg
(HYSYS calculated enthalpy = −1827 kJ/kg)
Entropy of ideal gases
The entropy of an ideal gas at 25°C can be calculated using the following
equation [4]:
0
S25
= 4.1867 * (6.2854B + 1.0734C + 0.43207 D + 0.0206E + 0.001037 F + G)
(2.72)
where
0
S25
= entropy of ideal gas at 25°C, kJ/(kg·K)
B,C,D,E,F,G = constants
The constants used in the above equation are presented in Table 2.9.
The ideal gas entropy at temperature T can be calculated using the
following equation [4]:
T
0
ST0 = S25
+
∫
298.15
C P0
dT
T
(2.73)
or
T
R 
CT 2 DT −2 
S =S +
A ln T + BT +
−

M
2
2  298.15
0
T
0
25
© 2010 Taylor & Francis Group, LLC
(2.74)
36
Process engineering and design using visual basic®
Table 2.9 Constants for the Calculation of Entropy—Ideal Gases
Ideal Gases
B
C
D
E
F
G
Methane
0.53829 −0.21141
0.33928 −1.16432
1.38961 −0.50287
Ethane
0.26461 −0.02457
0.29140 −1.28103
1.81348
0.08335
Propane
0.16030
0.12608
0.18143 −0.91891
1.35485
0.26090
n-Butane
0.09969
0.26655
0.05407 −0.42927
0.66958
0.34597
i-Butane
0.09907
0.23874
0.09159 −0.59405
0.90965
0.30764
n-Pentane
0.11183
0.22852
0.08633 −0.54465
0.81845
0.18319
n-Hexane
0.08971
0.26535
0.05778 −0.45221
0.70260
0.21241
n-Heptane
0.08978
0.26092
0.06345 −0.48471
0.75546
0.15776
n-Octane
0.07780
0.27936
0.05203 −0.46312
0.75074
0.17417
Nitrogen
0.25410 −0.01662
0.01530 −0.03100
0.01517
0.04868
Oxygen
0.22172 −0.02052
0.03064 −0.10861
0.13061
0.14841
Hydrogen
3.19962
Hydrogen sulfide
0.23745 −0.02323
0.03881 −0.11329
0.11484 −0.04064
Carbon monoxide
0.25284 −0.01540
0.01608 −0.03434
0.01757
0.10562
Carbon dioxide
0.15884 −0.03371
0.14811 −0.96620
2.07383
0.15115
0.39279 −0.29345
1.09007 −1.38787 −3.93825
Source: Adapted from Smith, J.M., Van Ness, H.C., and Abbott, M.M., Introduction to Chemical
Engineering Thermodynamics, 6th ed., McGraw-Hill, New York, 2001.
Note: Values for other components are available in the literature [4].
where
ST0 = entropy of ideal gas at temperature T, kJ/(kg·K)
0
S25
= entropy of ideal gas at 25°C, kJ/(kg·K)
C P0 = specific heat of ideal gas, kJ/(kg·K)
T = temperature, K
A,B,C,D = constants as per Table 2.4
Entropy of real gases
The entropy of a pure real gas can be calculated using the following equation [4]:
S = ST0 −
R  S0 − S 
M  R 
where
S = entropy of real gas, kJ/(kg·K)
ST0 = entropy of ideal gas at temperature T, kJ/(kg·K)
© 2010 Taylor & Francis Group, LLC
(2.75)
Chapter two:
Thermodynamics
37
R = gas constant, kJ/(kmol·K)
M = gas molecular weight
((S0 − S)/R) = dimensionless effect of pressure on entropy
The dimensionless effect of pressure on entropy can be calculated
as [4]
 f
 S0 − S 
 p 
R M
 R  = − H RT + ln  p  + ln  6.8947 
(2.76)
c
where
HR = residual enthalpy, kJ/kg; calculated using the previous method
f/p = fugacity correction
p = pressure, kPa
T = temperature, K
Tc = critical temperature, K
M = molecular weight
Fugacity correction
Fugacity correction can be calculated using the following equation [4]:
0
0
h
 f
 f 
f
ω  f 
 ln
ln =  ln  +
−  ln  
ω h  p 
p  p
 p 


(2.77)
where (ln(f/p))i can be calculated using the following equation:
i
 f
B
C
D
 ln p  = Z − 1 − ln Z + V + 2V 2 + 5V 5 + E
r
r
(2.78)
r
where
Z = compressibility
Other parameters are calculated using the method described under
specific heat calculation and constants in Table 2.5.
© 2010 Taylor & Francis Group, LLC
38
Process engineering and design using visual basic®
Entropy of hydrocarbon gas mixtures
The entropy of a defined hydrocarbon gas mixture can be calculated using
the following equation [4]:
n
0
S =
∑ x S − M (x ln x )
0
wi i
R
i
i
i =1
where
S0 = entropy of the gas mixture, kJ/(kg·K)
xwi = weight fraction of component i
R = gas constant, kJ/(kmol·K)
M = molecular of the gas mixture = ∑ i xi Mi
xi = mole fraction of component i
Mi = molecular weight of component i
Example 2.8
Estimate the entropy of the following pure components using the
PR EOS.
a. Methane at 10,000 kPag and 20°C
b. Ethane at 4800 kPag and 30°C
SOLUTION
a. Methane at 10,000 kPag and 20°C
Phase = vapor
Entropy at 25°C = 11.629 kJ/(kg·K)
Entropy at 20°C = 11.593 kJ/(kg·K)
Residual enthalpy calculated using the previous method =
−120.2 kJ/kg
Entropy pressure correction = 6.27
8.314
Entropy at 10,000 kPag, 20°C = 11.593 −
* 6.27
16.04
= 8.343kJ/(kg·K)
(HYSYS calculated value is 8.733 kJ/(kg·K))
b. Ethane at 4800 kPag and 30°C
Phase = liquid
Entropy at 25°C = 7.626 kJ/(kg·K)
Entropy at 30°C = 7.656 kJ/(kg·K)
Residual enthalpy calculated using the previous method =
−300.8 kJ/kg
Entropy pressure correction = 9.803
© 2010 Taylor & Francis Group, LLC
(2.79)
Chapter two:
Thermodynamics
39
Entropy at 4800 kPag, 30°C = 7.6558 −
8.314
* 9.803
30.07
= 4.945 kJ/(kg⋅K )
(HYSYS calculated value is 4.664 kJ/(kg·K))
Viscosities of ideal liquids
The viscosity of pure components depends primarily on the temperature
and can be calculated using the following general equation [5]:

 1 1
µ = exp  A *  −  
 T B 

(2.80)
Table 2.10 Constants for the Calculation of Liquid Viscosity—Pure
Components at Low Pressure
Ideal Components
A
B
Methane
Ethane
Propane
n-Butane
i-Butane
n-Pentane
n-Hexane
n-Heptane
n-Octane
Nitrogen
Oxygen
Hydrogen
Hydrogen sulfide
Carbon monoxide
Carbon dioxide
Methanol
Ethanol
Propanol
Benzene
Toluene
Water
114.14
156.60
222.67
265.84
302.51
313.66
362.79
436.73
473.70
90.3
85.68
13.82
342.79
94.06
578.08
555.30
686.64
951.04
545.64
467.33
658.25
57.60
95.57
133.41
160.20
170.20
182.48
207.09
232.53
251.71
46.14
51.50
5.39
165.54
48.90
185.24
260.64
300.88
327.83
265.34
255.24
283.16
Source: Adapted from Coulson, J.M. and Richardson, J.F., Chemical
Engineering, Volume 6 (SI Unit), Pergamon Press, USA, 1986.
Note: Values for other components are available in the literature [5].
© 2010 Taylor & Francis Group, LLC
40
Process engineering and design using visual basic®
where
μ = viscosity, cP
A,B = constants as per Table 2.10
T = temperature, K
Viscosity of water
The viscosity of water at low pressure can also be calculated using the
­following equation. The following equation is valid for a temperature
range of 0–150°C.
609.246 

µ = exp −3.827 +

138.89 + t 

(2.81)
where
μ = viscosity, cP
t = temperature, °C
Viscosity of ideal hydrocarbon vapors
The viscosity of ideal vapors can be calculated using the following equation [4]:
µ=
1000 ATRB
(1 + (C/TR ) + (D/TR2 ))
(2.82)
where
μ = viscosity, cP
TR = temperature, R
A,B,C,D = constants as per Table 2.11
Liquid viscosity of defined mixtures at low pressure
The liquid viscosities of defined mixtures at low temperature can be calculated using the following equation [4]:
 n

µm = 
xiµ 1i /3

 i =1
∑
where
μm = viscosity of the mixture, cP
μi = viscosity of component i, cP
n = number of components
xi = mole fraction of component i
© 2010 Taylor & Francis Group, LLC
3
(2.83)
Chapter two:
Thermodynamics
41
Table 2.11 Constants for the Calculation of Vapor Viscosity—Pure
Components at Low Pressure
Ideal Components
A
B
C
D
Methane
Ethane
Propane
n-Butane
i-Butane
n-Pentane
n-Hexane
n-Heptane
n-Octane
Nitrogen
Oxygen
Hydrogen
Hydrogen sulfide
Carbon monoxide
Carbon dioxide
Water
3.715E−7
1.737E−7
1.670E−7
1.528E−7
5.090E−7
3.853E−8
1.155E−7
4.100E–8
1.806E–8
4.588E–7
7.906E–7
1.201E–7
3.223E–8
8.133E–7
1.639E–6
4.152E–7
5.901E–1
6.799E–1
6.861E–1
6.944E–1
5.214E–1
8.476E–1
7.074E–1
8.284E–1
9.292E–1
6.081E–1
5.634E–1
6.850E–1
1.017
5.338E–1
4.600E–1
6.778E–1
190.3
178.0
322.7
409.9
412.2
75.10
282.8
154.4
99.16
98.48
173.3
−1.06
670.3
170.5
522.0
1525.0
0
0
−2.67E4
−4.73E4
0
0
0
0
0
0
0
4.536E2
−2.08E5
0
0
−2.4E5
Source: Adapted from Technical Data Book—Petroleum Refining, 4th ed., American Petroleum
Institute, Washington, D.C., 1982.
Note: Values for other components are available in the literature [4].
Vapor viscosity of defined mixtures at low pressure
The viscosity of defined hydrocarbon mixtures at low pressure can be
­calculated using the following equation [4,8]:
n
µm =
∑
i =1
µi
n
1+
∑φ x
j =1
j≠i
φ ij =
xj
ij
i
[1 + (µ i/µ j )0.5 ( M j /Mi )0.25 ]2
8 [1 + ( Mi /M j )]0.5
where
μm = viscosity of the mixture, cP
μi = viscosity of component i, cP
ϕij = interaction parameter of component i with respect to j
xi = mole fraction of component i
xj = mole fraction of component j
Mi = molecular weight of component i
Mj = molecular weight of component j
n = number of components in the mixture
© 2010 Taylor & Francis Group, LLC
(2.84)
(2.85)
42
Process engineering and design using visual basic®
Example 2.9
Estimate the viscosity of a gas mixture at 0 kPag and 30°C.
Component
Mole fraction
Methane
Ethane
Carbon monoxide
Carbon dioxide
Nitrogen
0.9
0.02
0.02
0.03
0.03
SOLUTION
The viscosity of individual components can be calculated as
A
Constants
B
C
D
Viscosity (cP)
3.715E−7
1.737E−7
8.133E−7
1.639E−6
4.588E−7
0.5901
0.6799
0.5338
0.4600
0.6081
190.3
178.0
170.5
522.0
98.48
0
0
0
0
0
0.0114
0.0095
0.0179
0.0152
0.0179
Σϕij (xj/xi)
Component
Methane
Ethane
Carbon monoxide
Carbon dioxide
Nitrogen
The interaction parameters can be calculated as
Interaction
1-2
1-3
1-4
1-5
2-1
2-3
2-4
2-5
3-1
3-2
3-4
3-5
4-1
4-2
4-3
4-5
5-1
5-2
5-3
5-4
Mi
Mj
ϕij
ϕij (xj/xi)
16.043
16.043
16.043
16.043
30.070
30.070
30.070
30.070
28.010
28.010
28.010
28.010
44.010
44.010
44.010
44.010
28.014
28.014
28.014
28.014
30.070
28.010
44.010
28.014
16.043
28.010
44.010
28.014
16.043
30.070
44.010
28.014
16.043
30.070
28.010
28.014
16.043
30.070
28.010
44.010
1.4824
1.0340
1.3500
1.0332
0.6623
0.7228
0.9526
0.7223
0.9344
1.4619
1.3561
0.9992
0.6592
1.0411
0.7328
0.7322
0.9352
1.4633
1.0008
1.3573
0.03294
0.02298
0.04500
0.03444
29.80284
0.72275
1.42885
1.08341
42.04872
1.46190
2.03408
1.49880
19.77659
0.69409
0.48850
0.73224
28.05564
0.97552
0.66720
1.35726
© 2010 Taylor & Francis Group, LLC
0.13536
33.03785
47.04350
21.69142
31.05562
Chapter two:
Thermodynamics
43
Calculated viscosity = 0.0119 cP
(HYSYS calculated value is 0.0117 cP)
Thermal conductivity of pure hydrocarbon liquids at low pressure
The thermal conductivity of a pure hydrocarbon liquid varies linearly
with temperature. With known thermal conductivities at two different
temperatures, the value can be calculated at any temperature using linear
interpolation.
The thermal conductivity of common hydrocarbon liquids can be presented in Table 2.12.
The thermal conductivity of pure hydrocarbon liquid mixtures can
be calculated as
km =
∑ ∑ φφ k
(2.86)
i j ij
i
j
1
1
k ij = 2  + 
 ki k j 
−1
(2.87)
Table 2.12 Thermal Conductivity of Pure Hydrocarbon Liquids
at Low Pressure
Temperature (oC)
Component
Freezing
Methane
Propane
n-Butane
n-Pentane
n-Hexane
n-Heptane
n-Octane
Benzene
Toluene
o-Xylene
m-Xylene
p-Xylene
−182.5
−187.7
−138.3
−129.7
−95.3
−90.6
−56.8
5.5
−95.0
−25.2
−47.9
13.3
Normal boiling
−161.5
−42.0
−0.5
36.1
68.7
98.4
125.7
80.1
110.6
144.4
139.1
138.4
Thermal conductivity (W/(m.K))
At freezing
temperature
At normal boiling
temperature
0.2246
0.2130
0.1869
0.1782
0.1622
0.1598
0.1519
0.1493
0.1615
0.1430
0.1475
0.1326
0.1883
0.1289
0.1176
0.1086
0.1042
0.1025
0.0981
0.1265
0.1116
0.1040
0.1035
0.1031
Source: Adapted from Technical Data Book—Petroleum Refining, 4th ed., American Petroleum
Institute, Washington, D.C., 1982.
Note: Values for other components are available in the literature [4].
© 2010 Taylor & Francis Group, LLC
44
Process engineering and design using visual basic®
φi =
xiVi
∑x V
(2.88)
∑φ = 1
(2.89)
j
j
j
i
i
where
km = thermal conductivity of the mixture, W/(m.K)
ϕi, ϕj = volume fraction of pure components i and j
ki, kj = thermal conductivity of pure components i and j, W/(m.K)
Vi, Vj = molar volume of pure components i and j, m3 per kg·mol
xi, xj = mole fraction of pure components i and j
Example 2.10
Calculate the thermal conductivity of the following liquid mixture at
0 kPag and 30°C.
n-Pentane = 40 vol%
n-Hexane = 30 vol%
n-Heptane = 30 vol%
The following data are available for the calculation:
Component
Density (kg/m3)
MW
Molar volume (m3/kg·mol)
n-Pentane
n-Hexane
n-Heptane
615.6
652.4
673.8
72.15
86.18
100.2
0.1172
0.1321
0.1487
SOLUTION
Thermal conductivity of n-pentane at 30°C = 0.1112 W/(m.K)
Thermal conductivity of n-hexane at 30°C = 0.1179 W/(m.K)
Thermal conductivity of n-heptane at 30°C = 0.1232 W/(m.K)
k11 = k1 = 0.1112
k 22 = k 2 = 0.1179
k 33 = k 3 = 0.1232
k12 = k 21 = 0.1145
k13 = k 31 = 0.1169
k 23 = k 32 = 0.1205
φ 1 = 0.4, φ 2 = 0.3, φ 3 = 0.3
Thermal conductivity of the liquid mixture = 0.1167 W/(m.K)
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
45
Table 2.13 Constants for the Calculation of Hydrocarbon Vapor
Thermal Conductivity
Ideal components
A × 103
B × 105
C × 108
Methane
Ethane
Propane
n-Butane
n-Pentane
n-Hexane
n-Heptane
n-Octane
16.77
8.74
7.54
6.91
6.76
6.42
4.95
3.59
4.358
4.343
3.362
2.809
2.337
2.214
2.444
2.710
1.335
1.364
1.971
2.841
2.778
2.180
1.685
1.175
Source: Adapted from Technical Data Book—Petroleum Refining, 4th ed., American
Petroleum Institute, Washington, D.C., 1982.
Note: Values for other components are available in the literature [4].
Thermal conductivity of pure hydrocarbon vapors at low pressure
The thermal conductivity of pure hydrocarbon gas at low pressure can be
calculated using the following equation [4]:
k = A + BT + CT2
(2.90)
where
k = thermal conductivity, Btu/(h.ft.°F)
A,B,C = constants (refer to Table 2.13)
T = temperature, °F
The thermal conductivity of pure hydrocarbon vapor mixtures can be
calculated as [4]
n
km =
∑1
i =1
ki
n
∑A y
yi j =1
ij
(2.91)
j
where
km = thermal conductivity of the mixture, Btu/(h.ft.°F)
ki = thermal conductivity of component i, Btu/(h.ft.°F)
n = number of components in the mixture
yi, yj = mole fraction of components i and j
2
0.5

 µ i  M j  0.75 (1 + (Si /T ))   (1 + (Sij /T ))
1
Aij = 1 +  
 

4
 µ j  Mi  (1 + (Sj /T ))   1 + (Si /T )


© 2010 Taylor & Francis Group, LLC
(2.92)
46
Process engineering and design using visual basic®
Si, Sj = 1.5 Tb
Sij = SiSj
μi, μj = viscosities of components i and j, cP
Mi, Mj = molecular weight of component i and j
Tb = normal boiling point, R
T = temperature, R
Flash calculation
Flash calculation is probably the most important unit operation to solve
process engineering problems. Any mixture at a defined pressure and
temperature can exist either in a liquid phase or in a vapor phase, or in
two phases. Flash calculation is required to establish the state of the fluid
under a specific condition.
Vapor–liquid equilibrium
When a fluid exists in two phases, the mole fraction of any component in
the vapor and liquid phases differs depending on the vapor–liquid equilibrium constant. The higher the value of the vapor–liquid equilibrium,
the higher will be the ratio of mole fraction between the vapor and liquid
phases. Mathematically
Ki =
yi
xi
(2.93)
where
Ki = vapor–liquid equilibrium constant of component i
yi = mole fraction of component i in vapor phase
xi = mole fraction of component i in liquid phase
Once the value of the equilibrium constant is known, it is easy to do
a flash calculation and the bubble and dew points can be calculated using
the following equations:
Bubble point
∑ y = ∑ K x = 1.0
(2.94)
∑ x = ∑ K = 1.0
(2.95)
i
i i
Dew point
i
yi
i
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
47
It is clear from the above analysis that a flash calculation can proceed
only when an accurate equilibrium constant is available. Any flash calculation is initiated with the preliminary estimated values for equilibrium
constants. A large number of correlations and graphs are available in the
literature and a preliminary estimate can be made using the following
equation [9]:
Ki =

Pci
T 

exp 5.37(1 + ω i )  1 − ci  

P
T 

(2.96)
where
Ki = vapor–liquid equilibrium constant of component i
Pci = critical pressure of component i
Tci = critical temperature of component i
ωi = acentric factor of component i
P = operating pressure
T = operating temperature
The mole fraction of component i in the liquid and vapor phases can
be estimated as [9]
xi =
zi
1 + V (K i − 1)
(2.97)
yi = x i K i
(2.98)
An overall material balance can result:
N
zi (K i − 1)
∑ 1 + V(K − 1) = 0
i =1
(2.99)
i
where
zi = overall mole fraction of component i
V = vapor phase mole fraction
Other terms are as defined before.
Using the above equations, the compositions in the vapor and liquid phases can be calculated using a set of equilibrium constants. The
adequacy of the calculation needs to be checked through the calculation
of the fugacity of each component. For the correct equilibrium condition,
the fugacity of each component in both the vapor and liquid phases will
be the same. If the fugacity is different, the equilibrium constants are
© 2010 Taylor & Francis Group, LLC
48
Process engineering and design using visual basic®
to be modified and recalculated. For the PR EOS, the fugacity of each
component in both the liquid and vapor phases is calculated using the
following equation [2]:
∑
2

xi aik
fk
bk
A  i
bk 
ln
=
(Z − 1) − ln (Z − B) −
− 

xk P
b
a
b
2 2B 


 Z + 2.414B 
× ln 
 Z − 0.414B 
(2.100)
where
∑∑x x a
i
j
(2.101)
b=
∑x b
(2.102)
aij = (1 − δ ij ) ai0.5 a 0j .5
(2.103)
a=
i
j ij
i i
δij = interaction parameter, generally determined experimentally
f k = fugacity of component k
Other parameters are as defined in Table 2.1.
Equation 2.100 is solved for both the liquid and vapor phase parameters to calculate the fugacity in both phases.
Programming
Calculation of JT effect due to drop in pressure
This program has been developed to estimate the impact of the Joule–
Thomson effect when a gas is expanded through a pressure reduction
unit. The following basis has been used in this estimation:
• The calculation is based on the PR EOS.
• It is assumed that the expansion is adiabatic and not isentropic.
Generally, the temperature drop is larger in isentropic expansion in
comparison to adiabatic expansion.
• This calculation assumes only the vapor phase expansion and does
not perform any enthalpy balance due to a change in phase.
© 2010 Taylor & Francis Group, LLC
Chapter two:
Thermodynamics
49
Figure 2.4 Solution of Example 2.6.
• If the liquid phase is detected at any step in the calculation, the
­program generates a note “Note-1: Liquid phase detected. Results may
not be accurate.”
• This calculation is performed using 50 steps and a downstream
­pressure–temperature relationship can be developed from the results.
Checking Example 2.6
The Visual Basic® solution of Example 2.6 is presented in Figure 2.4.
Nomenclature
a
CP
CV
EK
EP
F
gc
H
l
K
m
MW
acceleration, m/s2
heat capacity at constant pressure, kJ/(kg·°C)
heat capacity at constant volume, kJ/(kg·°C)
kinetic energy
potential energy
force, N
acceleration due to gravity, m/s2
enthalpy, kJ
displacement, m
vapor–liquid equilibrium constant
mass, kg
molecular weight
© 2010 Taylor & Francis Group, LLC
50
Process engineering and design using visual basic®
P
Q
R
S
T
U
v
V
W
z
Z
pressure, kPa
heat, kJ
universal gas constant, kJ/(kmol·K)
entropy, kJ/(kmol·K)
temperature, K
internal energy, kJ
velocity, m/s
volume, m3
work, kNm
elevation, m
compressibility
Greek characters
γ
η
ω
μ
ratio of specific heats
efficiency
acentric factor
Joule–Thomson coefficient
References
1. Smith, J.M., Van Ness, H.C., and Abbott, M.M., Introduction to Chemical
Engineering Thermodynamics, 6th ed., McGraw-Hill, New York, 2001.
2. Peng, D.Y. and Robinson, D.B., A two constant equation of state, I.E.C.
Fundamentals, 15, 59–64, 1976.
3. Soave, G., Equilibrium constants from a modified Redlich-Kwong equation
of state, Chemical Engineering Science, 27(6), 1197, 1972.
4. Technical Data Book—Petroleum Refining, 4th ed., American Petroleum
Institute, Washington, D.C., 1982.
5. Coulson, J.M. and Richardson, J.F., Chemical Engineering, Volume 6 (SI Unit),
Pergamon Press, USA, 1986.
6. Perry, R.H. and Green, D., Perry’s Chemical Engineering Handbook, 6th ed.,
McGraw-Hill, Malaysia, 1984.
7. Moran, M.J. and Shapiro, H.N., Fundamentals of Engineering Thermodynamics,
5th ed., John Wiley, USA, 2004.
8. Bromley, A. and Wilke, C.R., Viscosity behavior of gases, Industrial and
Engineering Chemistry Chemical Engineering Science, 43(7), 1641, 1951.
9. Naji, H.S., Conventional and rapid flash calculations for the Soave-RedlichKwong and Peng-Robinson equations of state, Emirates Journal for Engineering
Research, 13(3), 81–91, 2008.
© 2010 Taylor & Francis Group, LLC
References
1 Chapter 1 - Basic mathematics
1. Kreyszig, E., Advanced Engineering Mathematics, 7th ed.,
John Wiley and Sons, New York, 1993.
2. Grossman, S.I., Multivariable Calculus, Linear Algebra
and Differential Equations, 3rd ed., Sunders College
Publishing, 1994.
3. Jain, M.K., Iyengar, S.R.K., and Jain, R.K., Numerical
Methods for Scientific and Engineering Computation, Wiley
Eastern Limited, New Delhi, 1985.
2 Chapter 2 - Thermodynamics
1. Smith, J.M., Van Ness, H.C., and Abbott, M.M.,
Introduction to Chemical Engineering Thermodynamics, 6th
ed., McGraw-Hill, New York, 2001.
2. Peng, D.Y. and Robinson, D.B., A two constant equation
of state, I.E.C. Fundamentals, 15, 59–64, 1976.
3. Soave, G., Equilibrium constants from a modified
Redlich-Kwong equation of state, Chemical Engineering
Science, 27(6), 1197, 1972.
4. Technical Data Book—Petroleum Refining, 4th ed.,
American Petroleum Institute, Washington, D.C., 1982.
5. Coulson, J.M. and Richardson, J.F., Chemical
Engineering, Volume 6 (SI Unit), Pergamon Press, USA,
1986.
6. Perry, R.H. and Green, D., Perry’s Chemical Engineering
Handbook, 6th ed., McGraw-Hill, Malaysia, 1984.
7. Moran, M.J. and Shapiro, H.N., Fundamentals of
Engineering Thermodynamics, 5th ed., John Wiley, USA,
2004.
8. Bromley, A. and Wilke, C.R., Viscosity behavior of
gases, Industrial and Engineering Chemistry Chemical
Engineering Science, 43(7), 1641, 1951.
9. Naji, H.S., Conventional and rapid flash calculations
for the Soave-RedlichKwong and Peng-Robinson equations of
state, Emirates Journal for Engineering Research, 13(3),
81–91, 2008.
3 Chapter 3 - Fluid mechanics
1. American Society of Mechanical Engineers Standard,
Measurement of Fluid Flow in Pipes Using Orifice, Nozzle,
and Venturi.
2. Fluid Meters Their Theory and Application, 6th ed.,
Report of the American Society of Mechanical Engineers
Research Committee on Fluid Meters, 1971.
3. Crane Technical Paper No. 410, Flow of Fluids through
Valves, Fittings, and Pipe, Metric edition—SI Unit, 1995.
4. Miller, R.W., Flow Measurement Engineering Handbook, 2nd
ed., McGraw-Hill, USA, 1989.
5. Perry, R.H. and Green, D., Perry’s Chemical Engineers’
Handbook, 6th ed., McGraw-Hill, Malaysia, 1984.
6. Tuve, G.L. and Sprenkle, R.E., Instruments, 6, 201, 1933.
7. Coulson, J.M. and Richardson, J.F., Chemical
Engineering, 4th ed., Vol. 1, Pergamon Press, Great
Britain, 1990.
8. ISO 5167-2, Measurement of Fluid Flow by Means of
Pressure Differential Devices Inserted in Circular Cross
Section Conduits Running Full, Part 2, Orifice Plate,
International Standard, Switzerland, 2003.
9. ASME B31.3, Process Piping, ASME Code for Pressure
Piping, American Society of Mechanical Engineers, USA,
2002.
10. ASME B16.21, Nonmetallic Flat Gaskets for Pipe Flanges,
American Society of Mechanical Engineers, USA, 2005.
11. ASME B16.5, Pipe Flanges and Flanged Fittings, American
Society of Mechanical Engineers, USA, 2003.
12. McCabe, W.L. and Smith, J.C., Unit Operations of
Chemical Engineering, 3rd ed., McGraw-Hill, Tokyo, 1976.
13. Hooper, W.B., Calculated head loss caused by change in
pipe size, Chemical Engineering, November 7, 89–92, 1988.
14. Hooper, W.B., The two-K method predict head loss in
pipe fittings, Chemical Engineering, August 24, 96–100,
1981.
15. Zenz, F.A., Minimize manifold pressure drop,
Hydrocarbon Processing and Petroleum Refiner, 41(12),
125–130, 1962.
16. Fisher ED full port equal percentage case, Fisher
Catalog-10, USA, pp. 1–73.
17. API RP 14E, Recommended Practice for Design and
Installation of Offshore Production Platform Piping
Systems, 5th ed., American Petroleum Institute, 1991.
18. Massey, B.S., Mechanics of Fluids, 6th ed., Van
Nostrand Reinhold, chap. 12, London, 1989.
19. External Extruded High-Density Polyethylene Coating
System for Pipes, Australian/New Zealand Standard AS/NZS
1518:2002.
20. External Fusion-Bonded Epoxy Coating for Still Pipes,
Australian/New Zealand Standard AS/NZS 3862:2002.
21. Kern, D.Q., Process Heat Transfer, International
edition, McGraw-Hill, Singapore, 1988.
22. Hausen, H., Darstellung des Warmeuberganges in Rohren
durch verallgemeinerte Potenzbeziehungen, Zeitsche, V.D.I.
Beihefte Verfahrenstechnik, No. 4, 1943.
23. Carslaw, H.S. and Jaeger, J.C., Conduction of Heat in
Solids, Oxford University Press, Great Britain, 1959.
24. Parker, J.D., Boggs, J.H., and Blick, E.F.,
Introduction to Fluid Mechanics and Heat Transfer,
Addison-Wesley, Reading, MA, 1969.
25. Hilpert, R., Warmeabgabe von geheizen Drahten und
Rohren, Forsch Gebiete Iingenieurw, 4, 1933.
26. Beggs, H.D. and Brill, J.P., A study of two-phase flow
in inclined pipe, Journal of Petroleum Technology, May,
607–617, 1973.
27. Mukherjee, H. and Brill, J.P., Liquid holdup
correlations for inclined twophase flow, Journal of
Petroleum Technology, May, 1003–1008, 1983.
28. Mukherjee, H. and Brill, J.P., Pressure drop
correlations for inclined twophase flow, Journal of Energy
Resources Technology, 107, December, 549–554, 1985.
29. Gregory, G.A., Mandhane, J., and Aziz, K., Some design
consideration for two-phase flow in pipes, Journal of
Canadian Petroleum Technology, January– March, 65–71, 1975.
30. Bendiksen, K.H., Malnes, D., Moe, R., and Nuland, S.,
The dynamic twofluid model OLGA: Theory and application,
SPE Production Engineering, May, 171–180, 1991.
31. Baker, A., Nielsen, K., and Gabb, A., Pressure loss,
liquid-holdup calculations developed, Oil and Gas Journal,
March 14, 55–59, 1988.
32. Baker, A., Nielson, K., and Gabb, A., Holdup,
pressure-loss calculation confirmed, Oil and Gas Journal,
March 28, 44–49, 1988.
33. Mukherjee, H. and Brill, J.P., Empirical equations to
predict flow patterns in two-phase inclined flow,
International Journal of Multiphase Flow, 11(3), 299–315,
1985.
34. Waard, C. and Lotz, U., Corrosion 93, The NACE Annual
Conference and Corrosion Show, Paper No. 69.
35. CO 2 Corrosion Rate Calculation Model, Norsok
Standard, M-506, Rev2, June 2005.
36. Waard, C., Lotz, U., and Millams, D.E., Corrosion 91,
The NACE Annual Conference and Corrosion Show, Paper No.
577.
37. NACE Standard MR0175-2001, Item No. 21304.
38. Waard, C., Lotz, U., and Dugstad, A., Corrosion 95, The
NACE International Annual Conference and Corrosion Show,
Paper No. 128.
4 Chapter 4 - Heat transfer
1. Kern, D.Q., Process Heat Transfer, International ed.,
McGraw-Hill, Singapore, 1988.
2. Holman, J.P., Heat Transfer, 8th ed., McGraw-Hill, USA,
1997.
3. Rohsenow, W.M. and Hartnett, J.P., Handbook of Heat
Transfer, McGraw-Hill, New York, 1973.
4. Carslaw, H.S. and Jaeger, J.C., Conduction of Heat in
Solids, Oxford University Press, Great Britain, 1959.
5. Andrews, R.V., Solving conductive heat transfer problems
with electricalanalogue shape factors, Chemical Engineering
Progress, 51(2), 67, 1955.
6. Kreyszig, E., Advanced Engineering Mathematics, 7th ed.,
John Wiley and Sons, New York, 1993.
7. Gray, A. and Mathews, G.B., A Treatise on Bessel
Functions and Their Applications to Physics, Macmillan,
London, 1895.
8. Coulson, J.M. and Richardson, J.F., Chemical
Engineering, Vol. 1, 4th ed., Pergamon Press, Great
Britain, 1990.
9. Bird, R.B., Stewart, W.E., and Lightfoot, E.N.,
Transport Phenomenon, International ed., John Wiley and
Sons, USA, 1976.
10. McAdams, W.H., Heat Transmission, 3rd ed., McGraw-Hill,
New York, 1954.
11. Perry, R.H. and Green, D., Perry’s Chemical Engineers’
Handbook, 6th ed., McGraw-Hill, Malaysia, 1985.
12. Parker, J.D., Boggs, J.H., and Blick, E.F.,
Introduction to Fluid Mechanics and Heat Transfer,
Addison-Wesley, Reading, MA, 1969.
13. Holland, F.A., Moores, R.M., Watson, F.A., and
Wilkinson, J.K., Heat Transfer, Heinemann Educational
Books, London, 1970.
14. Chopey, N.P. (Editor), Handbook of Chemical Engineering
Calculations, 3rd ed., McGraw-Hill, New York, 2004.
15. Welty, J.R., Wicks, C.E., and Wilson, R.E.,
Fundamentals of Momentum, Heat and Mass Transfer, 3rd ed.,
John Wiley and Sons, USA, 1984.
16. Standards of the Tubular Exchanger Manufacturers
Association, 8th ed., TEMA Inc, New York, 1999.
17. Perrotin, T. and Clodic, D., Fin Efficiency Calculation
in Enhanced Fin-and-Tube Heat Exchangers in Dry
Conditions, International Congress of Refrigeration,
Washington, DC, 2003.
18. Siegel, R. and Howell, J.R., Thermal Radiation Heat
Transfer, McGraw-Hill, New York, 1981.
5 Chapter 5 - Distillation
1. Treybal, R.E., Mass Transfer Operation, 3rd ed.,
McGraw-Hill International Student Edition, Tokyo, 1980.
2. Coulson, J.M. and Richardson, J.F., Chemical
Engineering, Vol. 6, 6th ed., Pergamon Press, Great
Britain, 1986.
3. McCabe, W.L. and Smith, J.C., Unit Operations of
Chemical Engineering, 3rd ed., McGraw-Hill International
Student Edition, Tokyo, 1976.
4. Kurihara, K., Nakamichi, M., and Kojima, K., Isobaric
vapor-liquid equilibria for methanol + ethanol + water and
the three constituent binary systems, J. Chem. Eng. Data,
38, 446–449, 1993.
5. Smoker, E.H., Analytical determination of plates in
fractioning columns, Trans. Am. Inst. Chem. Eng., 34, 165,
1938.
6. Fair, J.R., How to predict sieve tray entrainment and
flooding, Petro./Chem. Eng., 33, 45, 1961.
7. Chopey, N.P., Handbook of Chemical Engineering
Calculation, 3rd ed., McGrawHill, New York, 2004.
8. Murphree, E.V., Rectifying column calculations with
particular reference to N component mixtures, Ind. Eng.
Chem., 17(7), 747, 1925.
9. Lewis, W.K., Rectification of binary mixtures, Ind. Eng.
Chem., 28, 399, 1936.
10. Couper, J.R. et al., Chemical Process
Equipment—Selection and Design, 3rd ed., Burlington, MA,
Oxford, UK, 2010.
11. Perry, R.H. and Green, D., Perry’s Chemical Engineers’
Handbook, 8th ed., McGraw-Hill, New York, 2008.
12. AIChE Research Committee, Bubble Tray Design Manual,
New York, 1958.
13. Gerster, J.A., Hill, A.B., Hochgraf, N.N., and
Robinson, D.G., Tray Efficiencies in Distillation Columns.
Final Report, University of Delaware, AIChE, 1958.
14. Smith, B.D., Design of Equilibrium Stage Processes,
McGraw-Hill, New York, 1963.
15. Fuller, E.N., Schettler, P.D., and Giddings, J.C., A
new method for the prediction of gas-phase diffusion
coefficients, Ind. Eng. Chem., 58(5), 19, 1966.
16. Wilke, C.R. and Chang, P., Correlation of diffusion
coefficients in dilute solutions, AIChE, 1955.
17. Lockett, M.J., Distillation Tray Fundamentals,
Cambridge University Press, England, 2009.
18. Kister, H.Z., Distillation Design, McGraw-Hill, USA,
1992.
19. Robbins, L.A., Chem. Eng. Prog., May, 87, 1991.
20. Lockett, M.J., Chem. Eng. Prog., 94(1), 60, 1998.
21. Zuiderweg, F.J., Sieve trays—A view on the state of
art, Chem. Eng. Sci., 37(10), 1441, 1982.
22. Lockett, M.J. and Banik, S., Weeping from Sieve Trays,
AIChE Meeting, November, San Francisco, 1984.
6 Chapter 6 - Separators
1. Gas Processors Suppliers Association, GPSA Engineering
Databook, 11th ed. GPSA, SI, Oklahoma, 1998.
2. Perry, R.H., Ed., Chemical Engineers’ Handbook, 5th ed.,
chap. 5, McGraw-Hill Book, Tokyo, 1973.
3. Fabian, P., Cusack, R., Hennessey, P., and Neuman, M.,
Demystifying the selection of mist eliminators, Part I,
Chemical Engineering, November, 148–156, 1993.
4. API Specification 12J, Specification for Oil and Gas
Separators, 7th ed., American Petroleum Institute,
Washington DC, 1989.
5. API Specification 11P, Specification for Packaged
Reciprocating Compressors for Oil and Gas Production
Services, 2nd ed., American Petroleum Institute,
Washington DC, 1989.
6. API RP 521, Guide for Pressure-Relieving and
Depressuring Systems, 4th ed., American Petroleum
Institute, Washington DC, 1997.
7. Design guide, Separations & Mass-transfer Products, ACS
Industries, LP, Houston, USA.
8. Campbell, J.M., Gas Conditioning and Processing, 8th
ed., John M. Campbell and Company, Vol. 2, Oklahoma, 2004.
9. Scott, S.L., Shoham, O., and Brill, J.P., Prediction of
slug length in horizontal, large-diameter pipes, SPE
Production Engineering, August, 335–340, 1989.
10. CO 2 Corrosion Rate Calculation Model, NORSOK M-506,
Revision 1, June, Norway, 1998.
7 Chapter 7 - Overpressure protection
1. API 520, Sizing, Selection, and Installation of
Pressure-Relieving Devices in Refineries, Part-1—Sizing
and Selection, 8th ed., American Petroleum Institute,
Washington, D.C., 2008.
2. API 520, Sizing, Selection, and Installation of
Pressure-Relieving Devices in Refineries,
Part-II—Installation, 4th ed., American Petroleum
Institute, Washington, D.C., 1994.
3. API 521, Pressure-Relieving and Depressuring Systems,
5th ed., American Petroleum Institute, Washington, D.C.,
2007.
4. ASME Section VIII, Rules for Construction of Pressure
Vessels, Division 1, American Society of Mechanical
Engineers, New York, 2003.
5. ASME B31.3, Process Piping, American Society of
Mechanical Engineers, New York, 2002.
6. AS 1271, Safety Valves, Other Valves, Liquid Level
Gauges and Other Fittings for Boiler and Unfired Pressure
Vessels, Australian Standard, Sydney, 1997.
7. API Std 598, Valve Inspection and Testing, 8th ed.,
American Petroleum Institute, Washington, D.C., 2004.
8. Yaws, C.L., Calculate thermal-expansion coefficients,
Chemical Engineering, August 1995.
9. Perry, R.H. and Green, D., Perry’s Chemical Engineers’
Handbook, 6th ed., McGraw-Hill, Malaysia, 1984.
10. Lapple, C.E., Isothermal and adiabatic flow of
compressible fluids, Transactions of AIChE, 39, 1943.
11. Mak, H.Y., New method speeds pressure relief manifold
design, Oil and Gas Journal, November 20, 1978.
12. API Std. 2000, Venting Atmospheric and Low-Pressure
Storage Tanks, 5th ed., American Petroleum Institute,
Washington, D.C., 1998.
13. AS 1940, The Storage and Handling of Flammable and
Combustible Liquids, Australian Standard, Sydney, 1993.
14. Fisher, H.G., DIERS research program on emergency
relief systems, CEP, August, 33–36, 1985.
15. Fauske, H.K., Emergency relief system (ERS) design,
CEP, August, 53–56, 1985.
16. Leung, J.C. and Grolmes, M.A., A generalized
correlation for flashing choked flow of initially
subcooled liquid, AIChE, April, 688–691, 1988.
17. Grolmes, M.A., Leung, J.C. and Fauske, H.K.,
Large-scale experiments of emergency relief systems, CEP,
57–62, 1985.
18. Fauske, H.K. and Leung, J.C., New experimental
technique for characterizing runaway chemical reactions,
CEP, August, 39–46, 1985.
19. Brzustowski, T.A. and Sommer, E.C., Jr., Predicting
radiant heating from flares, Proceeding—Division of
Refining, 53, 865–898, 1973.
20. BS 5345, Selection, Installation and Maintenance of
Electrical Apparatus for Use in Potentially Explosive
Atmospheres (Other than Mining Applications or Explosives
Processing and Manufacture) Part 1: General
Recommendations, 1989.
21. IEC 61508, Functional Safety of
Electrical/Electronics/Programmable Electronic
Safety-related Systems (Part 1 to Part 7).
22. IEC 61511, Functional Safety—Safety Instrumented
Systems for the Process Industry Sector (Part 1 to Part
3).
23. ANSI/ISA—S84.01, Application of Safety Instrumented
Systems for the Process Industries, 1996.
8 Chapter 8 - Glycol dehydration
1. API Specification 12GDU, Specification for Glycol-Type
Gas Dehydration Units, 1st ed., December 15, 1990.
2. Campbell, J.M., Gas Conditioning and Processing, John M.
Campbell and Company, 8th ed., Vol. 2, Oklahoma, 2004.
3. Gas Processors Suppliers Association, GPSA Engineering
Data Book—SI, GPSA, Oklahoma, 1998.
4. Glycol Pumps for the Next Generation, General Product
Information, Rotor-Tech.
5. Glycol Pumps and Accessories, Kimray Inc.
6. AS 3814—2005, Australian Standard, Industrial and
Commercial Gas-Fired Applications.
Figure 8.15 Message box.