Agro-Product Processing Technology: principles and practice 9780429487507, 9781138596689, 113859668X

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AGRO­PRODUCT PROCESSING TECHNOLOGY PRINCIPLES AND PRACTICE B. K. Bala

Agro-Product Processing Technology

Agro-Product Processing Technology Principles and Practice

B. K. Bala

Bangabandhu Sheikh Mujibur Rahman Science and Technology University Gopalganj, Bangladesh

First edition published 2020 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN © 2020 Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, LLC 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, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. ISBN: 978-1-138-59668-9 (hbk) ISBN: 978-0-429-48750-7 (ebk) Typeset in Times by Lumina Datamatics Limited

Contents Foreword........................................................................................................................................... xv Preface............................................................................................................................................xvii Author..............................................................................................................................................xix

Chapter 1 Introduction................................................................................................................... 1 1.1 Introduction.......................................................................................................... 1 1.2 Impotence of Postharvest Technology..................................................................2 1.3 Importance of Postharvest Losses........................................................................2 1.4 Postharvest Technology........................................................................................ 3 References..................................................................................................................... 5 Chapter 2 Physical, Thermal, and Chemical Properties of Food and Biological Materials..........7 2.1 Introduction.......................................................................................................... 7 2.2 Physical Properties............................................................................................... 7 2.2.1 Physical Dimensions................................................................................. 7 2.2.2 1000-Grain Weight...................................................................................7 2.2.3 Bulk Density............................................................................................. 8 2.2.4 Shrinkage.................................................................................................. 8 2.2.5 Friction.....................................................................................................9 2.2.5.1 Angle of Internal Friction and Angle of Repose.......................9 2.2.5.2 Coefficient of Friction.............................................................. 10 2.3 Thermal Properties............................................................................................. 11 2.3.1 Specific Heat........................................................................................... 11 2.3.2 Thermal Conductivity............................................................................ 13 2.3.2.1 Theory...................................................................................... 13 2.3.2.2 Apparatus and Measurement................................................... 14 2.3.3 Latent Heat of Vaporization................................................................... 15 2.3.3.1 Determination of Latent Heat of Vaporization........................ 16 2.3.4 Heat Transfer Coefficient of Product Bed.............................................. 18 2.3.4.1 Dimensional Analysis.............................................................. 19 2.3.4.2 Comparison of Theory and Experiment.................................. 19 2.3.4.3 Theory......................................................................................20 2.3.4.4 Determination of Volumetric Heat Transfer Coefficient......... 22 2.4 Chemical Properties...........................................................................................24 2.4.1 Starch......................................................................................................24 2.4.2 Protein....................................................................................................24 2.4.3 Fat...........................................................................................................24 2.4.4 Vitamin...................................................................................................24 Key to Symbols...........................................................................................................24 Exercises......................................................................................................................26 References................................................................................................................... 27

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Chapter 3 Cleaning, Grading, and Sorting.................................................................................. 29 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

Grade Factor..................................................................................................... 29 Washing............................................................................................................ 30 Sorting Fruits and Vegetables........................................................................... 30 Sorting Grain.................................................................................................... 30 Spiral Separator................................................................................................ 32 Indent Cylinder Separator................................................................................ 33 Color Separator................................................................................................. 33 Centrifugal Separation.....................................................................................34 3.8.1 Stokes’ Equation..................................................................................34 3.9 The Centrifuge................................................................................................. 35 3.10 The Cream Separator........................................................................................ 35 3.11 Cyclone Separator............................................................................................. 36 3.12 Machine Vision................................................................................................ 39 3.12.1 Image Acquisition.............................................................................. 39 3.12.1.1 Computer Vision System................................................... 39 3.12.1.2 Ultrasound and Infrared....................................................40 3.12.1.3 Tomographic Imaging.......................................................40 3.12.2 Preprocessing.....................................................................................40 3.12.3 Segmentation...................................................................................... 41 3.12.4 Feature Extraction.............................................................................. 42 3.12.4.1 Color Features................................................................... 42 3.12.4.2 Morphological Features.................................................... 43 3.12.4.3 Texture Features................................................................ 43 3.12.5 Classification......................................................................................44 Bibliography................................................................................................................ 45 Chapter 4 Psychrometry............................................................................................................... 47 4.1 4.2

4.3 4.4

Introduction...................................................................................................... 47 Psychrometric Terms........................................................................................ 47 4.2.1 Humidity Ratio................................................................................... 48 4.2.2 Relative Humidity.............................................................................. 48 4.2.3 Specific Volume................................................................................. 49 4.2.4 Vapor Pressure................................................................................... 49 4.2.5 Dry-Bulb Temperature....................................................................... 49 4.2.6 Dew Point Temperature..................................................................... 49 4.2.7 Wet-Bulb Temperature....................................................................... 49 4.2.8 Enthalpy............................................................................................. 50 4.2.9 Adiabatic Wet-Bulb Temperature....................................................... 51 4.2.10 Psychrometric Wet-Bulb Temperature............................................... 52 Construction of Psychrometric Chart............................................................... 54 Use of Psychrometric Chart............................................................................. 54 4.4.1 Sensible Heating and Cooling............................................................ 55 4.4.2 Heating with Humidification.............................................................. 55 4.4.3 Cooling with Humidification............................................................. 56 4.4.4 Cooling with Dehumidification......................................................... 56 4.4.5 Drying................................................................................................ 57 4.4.6 Mixing of Airstreams........................................................................ 59

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4.4.7 Heat Addition with Air Mixing........................................................... 61 4.4.8 Drying with Recirculation................................................................... 62 Key to Symbols...........................................................................................................66 Exercises...................................................................................................................... 68 Bibliography................................................................................................................ 68 Chapter 5 Drying of Agro Products............................................................................................. 69 5.1 Principles of Drying........................................................................................... 69 5.2 Importance of Drying......................................................................................... 69 5.3 Moisture Content................................................................................................ 70 5.3.1 Moisture Content Representation........................................................... 70 5.3.2 Determination of Moisture Content....................................................... 72 5.3.2.1 Direct Methods........................................................................ 73 5.3.2.2 Indirect Methods...................................................................... 75 5.4 Equilibrium Moisture Content........................................................................... 76 5.4.1 Determination of Static Equilibrium Moisture Content......................... 77 5.4.2 Static Equilibrium Moisture Content Models........................................ 77 5.5 Mechanism of Drying........................................................................................ 82 5.6 Thin-Layer Drying.............................................................................................84 5.6.1 Thin-Layer Drying Equations................................................................84 5.6.1.1 Empirical Drying Equations....................................................84 5.6.1.2 Theoretical Drying Equations.................................................84 5.6.1.3 Semi-Theoretical Drying Equations........................................ 85 5.6.2 Drying Rate............................................................................................ 87 5.6.3 Drying Parameters.................................................................................. 87 5.6.4 Drying Rate Constant and Diffusion Coefficient................................... 88 5.6.4.1 Drying Rate Constant.............................................................. 88 5.6.5 Half Response Time............................................................................... 91 5.7 Deep-Bed Drying...............................................................................................92 5.7.1 Logarithmic Model.................................................................................92 5.7.2 Partial Differential Equation Model....................................................... 98 5.7.2.1 Method of Solution................................................................ 102 5.7.2.2 Comparisons of Simulated and Observed Results................. 103 5.8 Fluidized Bed Drying Model........................................................................... 103 5.8.1 Heat Balance Equation......................................................................... 104 5.8.2 Drying Rate Equation........................................................................... 104 5.8.3 Mass Balance Equation........................................................................ 105 5.9 Agro-Product Drying Systems......................................................................... 106 5.9.1 Solar Drying Systems........................................................................... 106 5.9.1.1 Solar Dryers........................................................................... 106 5.9.2 Batch Drying Systems.......................................................................... 110 5.9.2.1 Flatbed Dryer......................................................................... 111 5.9.3 Continuous Flow Drying Systems........................................................ 113 5.9.3.1 Cross-Flow Dryer.................................................................. 113 5.9.3.2 Cross-Flow Batch Dryer........................................................ 113 5.9.3.3 Concurrent Flow Dryer.......................................................... 114 5.9.3.4 Counterflow Dryer................................................................. 114 5.9.3.5 Mixed-Flow Dryer................................................................. 116

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5.10 Safe Temperature for Drying Grain............................................................... 117 5.11 Selection of Dryers......................................................................................... 117 Key to Symbols......................................................................................................... 118 Exercises.................................................................................................................... 119 Bibliography.............................................................................................................. 120 Chapter 6 Parboiling of Rice..................................................................................................... 123 6.1 6.2 6.3

Introduction.................................................................................................... 123 Principles of Parboiling.................................................................................. 124 Soaking........................................................................................................... 124 6.3.1 Kinetics of Soaking............................................................................ 126 6.3.2 Finite Element Modeling of Soaking of Water by Paddy................... 132 6.3.3 Half Response Time........................................................................... 137 6.3.4 Kinetics of Water Diffusion and Starch Gelatinization..................... 137 6.4 Steaming......................................................................................................... 140 6.5 Drying............................................................................................................. 140 6.6 Effect of Parboiling on Milling, Nutritional, and Cooking Qualities of Rice............................................................................................................ 141 6.7 Parboiling Methods........................................................................................ 142 6.7.1 Traditional Methods........................................................................... 142 6.7.1.1 Single Stage Parboiling....................................................... 142 6.7.1.2 Double Stage Parboiling...................................................... 142 6.7.2 Modern Methods................................................................................ 142 6.7.2.1 CFTRI (Central Food Technological Research Industries) Method.............................................................. 143 6.7.2.2 Jadavpur University Method............................................... 143 6.7.2.3 Malek Process..................................................................... 143 6.7.2.4 Schule Process..................................................................... 143 6.7.2.5 Crystal Rice Process........................................................... 144 6.7.2.6 Rice Conversion Process..................................................... 144 6.7.2.7 Avorio Process..................................................................... 144 6.7.3 Estimation of Heat Required for Parboiling....................................... 144 6.7.3.1 Soaking Operation.............................................................. 144 6.7.3.2 Steaming Operation............................................................ 144 6.7.3.3 Drying Operation................................................................ 145 Exercises.................................................................................................................... 159 Bibliography.............................................................................................................. 160 Chapter 7 Milling of Rice and Wheat........................................................................................ 163 7.1 7.2 7.3

7.4

Introduction.................................................................................................... 163 Rice Milling................................................................................................... 163 Traditional Methods....................................................................................... 163 7.3.1 Home Pounding.................................................................................. 163 7.3.2 Huller Mills........................................................................................ 164 7.3.3 Sheller Mills....................................................................................... 164 7.3.4 Rubber Roll Sheller Mills................................................................... 164 The Modern Rice Milling Process................................................................. 165

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7.5 Modern Rice Milling Machinery..................................................................... 167 7.5.1 Paddy Cleaner....................................................................................... 167 7.5.2 Stoner.................................................................................................... 167 7.5.3 Rubber Roll Sheller.............................................................................. 167 7.5.4 Paddy Separator.................................................................................... 169 7.5.5 Whitening or Polishing........................................................................ 171 7.5.5.1 Cone-Type Polisher................................................................ 171 7.5.5.2 Horizontal Abrasive-Type Polisher........................................ 172 7.5.5.3 Friction-Type Polisher............................................................ 172 7.5.6 Bran and Polished Rice Separator........................................................ 173 7.5.7 Rice Grader.......................................................................................... 173 7.5.8 Rice Mixing.......................................................................................... 174 7.6 Wheat Milling.................................................................................................. 174 7.6.1 Conditioning/Hydrothermal Treatment................................................ 174 7.6.2 Milling.................................................................................................. 174 7.6.3 Storage of Finished Products................................................................ 175 7.7 Size Characteristics.......................................................................................... 175 7.7.1 Sieve...................................................................................................... 175 7.7.2 Fineness Modulus................................................................................. 176 7.7.3 Energy Requirements........................................................................... 177 Bibliography.............................................................................................................. 179 Chapter 8 By-Product Utilization.............................................................................................. 181 8.1 Introduction...................................................................................................... 181 8.2 Fuels and Combustion...................................................................................... 181 8.2.1 Furnaces............................................................................................... 182 8.3 Pyrolysis and Gasification................................................................................ 183 8.3.1 Pyrolysis (Destructive Distillation) and Gasification........................... 183 8.3.2 Types of Gasifiers................................................................................. 184 8.3.2.1 Countercurrent Moving Bed Gasifiers................................... 185 8.3.2.2 Concurrent Moving Bed Gasifiers......................................... 185 8.3.2.3 Crosscurrent Moving Bed Gasifiers...................................... 185 8.3.2.4 Fluidized Bed Gasifiers......................................................... 185 8.3.3 Gasification Process............................................................................. 185 8.3.3.1 Oxidation............................................................................... 186 8.3.3.2 Reduction............................................................................... 186 8.3.3.3 Drying.................................................................................... 186 8.3.4 Gasifier Units........................................................................................ 186 8.4 Liquefaction...................................................................................................... 187 8.5 Hydrolysis Followed by Fermentation.............................................................. 189 8.6 Biochar Production and Utilization.................................................................. 190 8.6.1 Biochar Carbonizer.............................................................................. 190 8.6.2 Types of Carbonizers............................................................................ 191 8.6.2.1 Application............................................................................ 192 8.7 Rice Husk Pelletizing and Briquetting............................................................. 192 8.7.1 Need for Briquetting............................................................................. 193 8.7.2 Principle and Technology..................................................................... 193

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8.7.3

Types of Briquetting Machines........................................................ 193 8.7.3.1 High- and Medium-Pressure Compaction........................ 194 8.7.3.2 Screw Press....................................................................... 194 8.7.3.3 Piston Press....................................................................... 195 8.7.3.4 Low-Pressure Compaction................................................ 195 8.7.3.5 Hand-Molded Briquettes................................................... 195 8.7.4 Applications...................................................................................... 195 8.7.5 Limitations....................................................................................... 196 8.7.6 Future Prospective............................................................................ 196 8.8 Biogas Digesters............................................................................................. 196 8.8.1 Anaerobic Digestion Process............................................................ 197 8.8.2 Indian-Type Biogas Digester............................................................ 197 8.8.3 The Chinese Biogas Digester........................................................... 198 8.8.4 Digester Sizing................................................................................. 199 8.9 Composting.....................................................................................................202 8.9.1 Process of Composting..................................................................... 203 8.9.2 Mixing of Materials in the Compost................................................ 203 8.9.3 Starting a Composter........................................................................204 8.9.4 Operating a Compost........................................................................204 8.9.5 Simple Thermophile Composting Procedure...................................204 8.9.6 Types of Composters........................................................................205 8.10 Utilization of Rice Bran – Stabilizer Design and Oil Extraction...................207 8.11 Bran – Stabilizer Design................................................................................207 8.12 Oil Extraction.................................................................................................207 8.12.1 Batch Extraction Method..................................................................208 Bibliography..............................................................................................................208 Chapter 9 Storage of Agro Products..........................................................................................209 9.1 9.2 9.3 9.4

9.5

Principles of Storage.......................................................................................209 Interactions of Physical, Chemical, and Biological Variables in the Deterioration of Stored Grains....................................................................... 211 Computer Simulation Modeling for Stored Grain Pest Management............ 212 Grain Storage Systems................................................................................... 213 9.4.1 Traditional Storage Systems............................................................. 213 9.4.2 Modern Storage Systems.................................................................. 214 9.4.2.1 Bagged Storage System..................................................... 214 9.4.2.2 Silo Storage System.......................................................... 215 9.4.2.3 Airtight Grain Storage...................................................... 216 9.4.2.4 Aerated Storage System.................................................... 218 9.4.2.5 Low-Temperature Storage System (Grain Chilling by Refrigeration)............................................................... 221 9.4.2.6 Controlled Atmosphere Storage Systems.........................224 9.4.2.7 Damp Grain Storage System with Chemicals.................. 226 Design of Grain Storages............................................................................... 228 9.5.1 Structural Requirements................................................................... 228 9.5.2 Janssen’s Equation............................................................................ 229 9.5.3 Rankine’s Equation.......................................................................... 232 9.5.4 Airy’s Equation................................................................................. 233 9.5.5 Construction Materials..................................................................... 236

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Key to Symbols.........................................................................................................240 Exercises.................................................................................................................... 241 Bibliography.............................................................................................................. 241 Chapter 10 Heating and Cooling of Agro Products..................................................................... 245 10.1 Introduction..................................................................................................... 245 10.2 Heat Conduction.............................................................................................246 10.2.1 The Differential Equation of Heat Conduction in Cartesian and Cylindrical Coordinate Systems................................................246 10.2.1.1 The Differential Equation of Heat Conduction in Cartesian Coordinate System...........246 10.2.1.2 The Differential Equation of Heat Conduction in Cylindrical Coordinate System........248 10.2.2 The Composite Wall......................................................................... 249 10.2.3 Cylinder and Sphere.......................................................................... 251 10.3 Convection...................................................................................................... 252 10.3.1 Forced Convection............................................................................ 252 10.3.2 Natural or Free Convection............................................................... 255 10.3.3 Heat Exchangers............................................................................... 256 10.4 Radiation......................................................................................................... 261 10.4.1 Radiation Intensity and Shape Factor............................................... 263 10.4.2 Radiation Exchange between Black Surfaces...................................264 10.4.3 Heat Exchange by Radiation between Gray Surfaces...................... 265 10.5 Cooling............................................................................................................ 267 10.5.1 Cooling Rate..................................................................................... 267 10.6 Freezing.......................................................................................................... 269 10.6.1 Freezing Point Depression................................................................ 269 10.7 Heating............................................................................................................ 271 10.7.1 Boiling-Point Elevation..................................................................... 272 Key to Symbols........................................................................................................ 275 Exercises.................................................................................................................. 276 Bibliography............................................................................................................. 277 Chapter 11 Refrigeration and Cold Storage................................................................................. 279 11.1 Introduction..................................................................................................... 279 11.2 Vapor Compression Refrigeration Cycle........................................................280 11.3 Pressure–Enthalpy (p-h) Chart.......................................................................280 11.3.1 Unit of Refrigeration......................................................................... 281 11.4 Refrigerants..................................................................................................... 281 11.4.1 Desirable Characteristics of Refrigerants......................................... 281 11.5 Construction of Psychrometric Chart............................................................. 287 11.6 Moisture Control and Storage of Vegetables Crops........................................ 291 11.6.1 Potatoes............................................................................................. 291 11.6.1.1 Adequate Volume.............................................................. 291 11.6.1.2 Adequate Strength............................................................ 291 11.6.1.3 Storage Environment........................................................ 292 11.6.1.4 Suberization Period........................................................... 292

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11.6.1.5 Short-Term Storage........................................................... 292 11.6.1.6 Long-Term Storage........................................................... 292 11.7 Cooling Requirement...................................................................................... 293 Exercises.................................................................................................................... 297 Bibliography.............................................................................................................. 298 Chapter 12 Separation.................................................................................................................. 299 12.1 Introduction..................................................................................................... 299 12.2 Contact Equilibrium Process.......................................................................... 299 12.2.1 Absorption........................................................................................ 299 12.2.2 Extraction..........................................................................................300 12.2.2.1 Rate of Extraction.............................................................300 12.2.2.2 Leaching........................................................................... 301 12.2.3 Distillation........................................................................................ 303 12.2.3.1 Vapor-Liquid Equilibrium................................................ 303 12.2.3.2 Flash Vaporization............................................................ 305 12.2.3.3 Batch Distillation..............................................................306 12.2.3.4 Fractionation.....................................................................307 12.2.3.5 Steam Distillation............................................................. 311 12.3 Mechanical Separation Process...................................................................... 312 12.3.1 Filtration............................................................................................ 312 12.3.2 Sedimentation................................................................................... 314 12.3.2.1 Sedimentation for Low Concentration Suspensions........... 314 12.3.2.2 Sedimentation for High Concentration Suspensions.......... 315 12.3.3 Centrifugation................................................................................... 316 12.3.3.1 Rate Separation................................................................. 316 12.3.3.2 Liquid-Liquid Separation................................................. 317 12.3.3.3 Particle Gas Separation.................................................... 318 Exercises.................................................................................................................... 319 Bibliography.............................................................................................................. 319 Chapter 13 Materials Handling and Conveying.......................................................................... 321 13.1 13.2 13.3 13.4

Introduction..................................................................................................... 321 Principles of Materials Handling.................................................................... 322 Classification of Materials Handling Equipment............................................ 322 Belt Conveyors................................................................................................ 323 13.4.1 The Capacity of the Conveyor Belt Needed..................................... 325 13.4.2 Belt Width......................................................................................... 325 13.4.3 Belt Speed......................................................................................... 326 13.4.4 Belt Tension....................................................................................... 326 13.4.5 Selection of Belt Carcass.................................................................. 327 13.4.6 Selection of Driving and Other Pulleys............................................ 327 13.4.7 Motor Power...................................................................................... 327 13.4.8 Selection of Idlers............................................................................. 328 13.5 Chain Conveyors............................................................................................. 329 13.5.1 Trolley Chain Conveyor.................................................................... 329 13.5.2 Scraper Chain Conveyor................................................................... 330 13.5.3 Apron Chain Conveyor..................................................................... 330

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13.6

Screw Conveyors.......................................................................................... 331 13.6.1 Power Requirement for Screw Conveyors...................................... 332 13.7 Bucket Conveyors......................................................................................... 333 13.8 Pneumatic Conveyors................................................................................... 334 13.9 Hydraulic Conveyors.................................................................................... 337 13.10 Gravity Conveyors........................................................................................ 338 13.11 Cranes.......................................................................................................... 338 13.12 Lift and Carrying Trucks and Cart.............................................................. 338 13.13 Robotic Handling System............................................................................ 338 Exercises....................................................................................................................348 Bibliography.............................................................................................................. 349 Chapter 14 Process Dynamics and Control................................................................................. 351 14.1 14.2 14.3 14.4

Introduction.................................................................................................. 351 Feedback Control Systems........................................................................... 351 Block Diagrams............................................................................................ 351 Dynamic Behavior of First- and Second-Order Systems............................. 352 14.4.1 First-Order System......................................................................... 352 14.4.2 Step Input........................................................................................ 353 14.4.3 Ramp Input..................................................................................... 354 14.4.4 Sinusoidal Input.............................................................................. 354 14.4.5 Second-Order System..................................................................... 355 14.5 The Laplace Transform................................................................................ 356 14.6 Transfer Function......................................................................................... 358 14.6.1 Routh–Hurwitz Stability Criterion................................................. 359 14.7 Transient Response...................................................................................... 361 14.7.1 Performance of Second-Order System........................................... 361 14.7.2 Performance Indices....................................................................... 362 14.8 Frequency Response..................................................................................... 362 14.8.1 Bode Plot........................................................................................ 363 K 14.8.1.1 First-Order Process (τs+1 ) . .............................................. 363 14.8.1.2 Integrating Process G(s) = Ks .........................................364 14.8.1.3 Second-Order Process G(s) = τ2s2 +K2ζτs +1 .........................364 14.8.1.4 Process Zero τs +1......................................................... 365 14.8.1.5 Time Delay e − jωθ............................................................ 366 14.8.1.6 Bode Stability Criterion................................................. 367 14.8.1.7 Gain Margin and Phase Margin.................................... 367 14.8.2 Nyquist Criterion............................................................................ 368 14.8.2.1 Gain Margin and Phase Margin.................................... 369 14.9 Frequency Response of Controllers............................................................. 369 14.9.1 Proportional Controller.................................................................. 369 14.9.2 Proportional–Integral Controller.................................................... 369 14.9.3 Ideal Proportional–Derivative Controller...................................... 370 14.9.4 Proportional–Derivative Controller with Filter.............................. 370 14.9.5 Parallel PID Controller................................................................... 370 14.9.6 Series PID Controller..................................................................... 371 14.9.7 Series PID Controller with a Derivative Filter............................... 371 Exercises.................................................................................................................... 373 Bibliography.............................................................................................................. 374

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Chapter 15 Emerging Technologies............................................................................................. 375 15.1 Introduction..................................................................................................... 375 15.2 Neural Network Modeling.............................................................................. 375 15.2.1 Structure of ANN Model of a Dryer................................................. 376 15.2.2 Training of ANN Model................................................................... 377 15.3 Energy and Exergy Analysis of Drying Process............................................ 378 15.3.1 Drying Efficiency.............................................................................. 378 15.3.2 Exergy Analysis through the Analysis of Second Law of Thermodynamics.......................................................................... 378 15.3.3 Exergy Calculation for Dryer............................................................ 379 15.4 Finite Element Modeling of Single Kernel..................................................... 380 15.5 Computational Fluid Dynamics (CFD) Modeling.......................................... 385 15.5.1 CFD Model Formulation.................................................................. 385 15.5.1.1 Continuity Equation......................................................... 386 15.5.1.2 Momentum Conservation Equations................................ 386 15.5.1.3 Energy Conservation Equation......................................... 386 15.5.1.4 User-Defined Function...................................................... 386 15.5.2 CFD Analysis.................................................................................... 387 15.5.2.1 Preprocessing.................................................................... 387 15.5.2.2 Processing......................................................................... 388 15.5.2.3 Post-processing................................................................. 388 15.6 Optimal Design Using Genetic Algorithm..................................................... 390 15.6.1 Optimal Design of Agro-Product Processing Systems..................... 390 15.6.2 Genetic Algorithm............................................................................ 390 15.6.2.1 Basic Principles of Genetic Algorithm............................. 391 15.6.2.2 Fitness Function................................................................ 391 15.6.2.3 Genetic Algorithm Procedure for Optimal Drying Parameters............................................................ 391 Key to Symbols......................................................................................................... 393 Bibliography.............................................................................................................. 394 Index............................................................................................................................................... 397

Foreword How can we ensure access to safe, nutritious, and affordable food in a sustainable manner? This question of food security must be addressed in the context of a world population that will reach 9 billion by 2040, rapid climate changes and its catastrophic effects on ecosystems and agriculture, loss of biodiversity, and increasing demand for fuel and energy. A multipronged approach is needed. Efforts are being made to use climate change-resilient cultivars and to implement best practices to increase production without the need to bring new land into agricultural use. Another approach is to address the significant postharvest losses due to damage, poor storage conditions, rot, pests, and vermin, which are estimated to be 27%–40% of total food production. Minimizing such losses is thus important for ensuring food security. To address the challenges noted above, one must have a good understanding of the unit operations applicable to the agro-products arena. The 15 chapters covered in this book by Professor Bala give good grounding on the basic principles and the current practice of agro-processing technology. The book is suitable for both researchers and instructors who are expected to make contributions to the advancement of knowledge and contribute to the design of new systems in both farming and industrial sectors. It not only covers heating and drying aspects but also refrigeration and cooling. Another key feature is the inclusion of technical aspects along with the biological attributes of commodities. I am particularly pleased to note the chapters on materials handling and conveying, process control and dynamics, and emerging technologies. The concept of a circular bio-economy is absolutely vital in today’s environmental connectivity. The by-product utilization chapter contributes to this area to some extent. It is an excellent initiative. I appreciate the request by the author to write this foreword. The  book is an excellent addition for the classroom use, and I encourage this aspect to the publishers. Sincerely Vijaya Raghavan, PhD, FRSC, FNAAS, FIE, FISAE, FCSBE, ASABE, FASME James McGill Professor and Graduate Program Director Bioresource Engineering, Faculty of Agricultural and Environmental Sciences McGill University, Ste-Anne-de-Bellevue, Québec, Canada

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Preface This book has been written primarily for undergraduate and graduate students in agricultural engineering, food engineering, agro-product processing technology, and food technology. It is the outcome of several years of teaching and research work carried out by the author. This book covers the wide spectrum of the principles and practices of agricultural process engineering and agro-product processing technology, which is probably not  available in a single book. Chapters  1 through 10 deal with an introduction of the importance of agro-product processing technology; physical, thermal and chemical properties; cleaning and sorting of agro products; psychrometry; drying of agro products; parboiling and milling; by-product utilization; heating and cooling; and comprehensive treatment of modern storage systems. Chapter 11 has been primarily devoted to refrigerated cooling and cold storage. Chapters 12 through 15 deal with separation, materials handling, process dynamics and control, and emerging technologies. A good number of problems have been solved to help understand the relevant theory. At the end of each chapter, unsolved problems have been provided for further practices. An extensive bibliography at the end of each chapter will help the reader to find detailed information on various topics of his or her interest. I have great pleasure in acknowledging what I owe to many persons in writing this book. I am deeply indebted to my teacher, Professor J. R. O’Callaghan of Newcastle University of Newcastle upon Tyne, UK, who inspired me to write books on agricultural processing and wrote the foreword of the first edition of my book titled Drying and Storage of Cereal Grains. Also, I sincerely express my acknowledgments to Professor Vijaya Raghavan, McGill Universality, Canada, for writing the foreword of this book. At the Bangabandhu Sheikh Mujibur Rahman Science and Technology University, I acknowledge the encouragement and assistance received from Professor Dr. Khondkar Md. Nasiruddin, the Vice Chancellor, Bangabandhu Sheikh Mujibur Rahman Science and Technology University, Gopalganj, Bangladesh. I am grateful to Professor Serm Janjai, Department of Physics, Silpakorn University, Nakhon Pathom, Thailand, for his cooperation in the preparation of the manuscript. And I owe thanks to MR. Sattra Sirikaew, PhD student in the Department of Physics, Silpakorn University, Nakhon Pathom, Thailand, for the assistance in graphics and in drawing the beautiful figures. B. K. Bala Bangabandhu Sheikh Mujibur Rahman Science and Technology University

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Author Professor B. K. Bala is the dean of the Faculty of Engineering, Bangabandhu Sheikh Mujibur Rahman Science and Technology University, Gopalganj, Bangladesh, and was a professor in the Department of Farm Power and Machinery, Bangladesh Agricultural University, Mymensingh, Bangladesh, where he was engaged in teaching and research for over 42 years starting from 1970. He has supervised more than a dozen PhD students in the areas of drying and storage of agro products, modeling of food security, modeling of integrated energy systems and energy policy, renewable energy, modeling of climate change impacts, and emerging technologies such as neural network, exergy, genetic algorithm, and computational fluid dynamics. Professor Bala received a B.Sc. (Eng.) degree from Bangladesh University of Engineering and Technology, Dhaka, Bangladesh, in 1969 and an M.Eng. degree from the Asian Institute of Technology, Bangkok, Thailand, in 1975. Professor Bala also received a Ph.D. degree from the University of Newcastle upon Tyne, UK, in 1983 and worked on modeling and simulation of heat and mass transfer during industrial deep-bed drying of malt and provided the foundation and basis for energy conservation and online control of industrial deep bed drying of malt. He also worked on solar drying as an EC postdoctoral fellow at the University of Newcastle upon Tyne, UK, from 1992 to 1993. He was on a DAAD study visit at the Institute of Agricultural Engineering in the Tropics and Subtropics, University of Hohenheim, Germany, for research on solar drying in 1995. Professor Bala also served as a research fellow at the Institute of Agricultural and Food Policy Research, Universiti Putra Malaysia from 2012 to 2014. Professor Bala has published 207 scientific papers mostly in high-impact factor journals, and he is also the author of nine textbooks. He is the author of Drying and Storage of Cereal Grains published by Wiley-Blackwell in 2017, and he is the co-author of System Dynamics Modeling and Simulation published by Springer in 2017.

xix

1

Introduction

This chapter introduces the concepts of agricultural engineering and agro-product processing technology and their importance, postharvest losses, and postharvest technology. Postharvest losses include weight loss, nutritional loss, viability loss, and indirect loss. Selection of unit operations and management of postharvest technology using a systems approach for production of high-quality, value-added products are also presented.

1.1 INTRODUCTION Global food security is a challenging issue to meet the food and nutritional requirements and has become an issue of national policy and public concern (Majumder et al., 2016). Cereal grains, fruits, and vegetables are the major food crops to meet energy and nutritional requirements of the growing population of the world. We need to enhance agricultural production as well as to reduce the postharvest loss, improve the quality of the processed products, and add value to the products to make more quality food available. We need nutritional quality food available for consumption and sale, not the production statistics. This demands application of engineering principles to agriculture at large to ensure higher production and fewer losses and better quality thus making more quality food available. Agro-product processing technology plays the major role to reduce the postharvest losses, improve the quality of the processed products, and hence add value to the products. Essentially, it starts at harvesting and involves the primary postharvest operations such as cleaning, grading, sorting, drying, parboiling, milling, packaging, and storage and continues until consumption. This also would generate employment and hence ultimately would contribute to food security. Southeast Asia and South Asia witnessed the green revolution in the late sixties on account of adoption of improved seeds and packages of other inputs in conjunction with irrigation that resulted in increased productivity. This was followed by a quantum jump in the production of fruits, vegetables, milk, egg, fish, and meat—thus changing the production scenario of this region. Although there have been some problems of equitable distribution, still many countries in this region, including Bangladesh, have achieved self-sufficiency in food with some export. It is to be noted, however, that 15%–30% of the product is lost every year due to lack of scientific postharvest management systems, i.e., proper utilization of agro-product processing technology (Bala, 2017). To overcome this problem and for overall development, we need to develop postharvest technologies to promote processing and value addition for reduction in losses and better utilization of produces and by-products. The role of postharvest technology in increasing productivity, income, and employment can be illustrated as shown in Figure 1.1. Farm income can be maximized by increasing farm production as well as loss reduction, value addition, and employment. Farm productivity can be increased by adequate inputs supported by assets and appropriate technology transfer. Loss reduction, value addition, and employment generation can be implemented through proper utilization of postharvest technology, which largely depends on capital investment, technology transfer, and research and development. Thus, postharvest technology transfer plays a major role in enhancing the farm income and hence increases the farm assets—i.e., it contributes to economic growth. Proper use of farm inputs and adoption of proper postharvest technology contribute to increased farm production and postharvest loss reduction, respectively, and the process is regenerative as indicated in Figure 1.1. This emphasizes importance of postharvest technology to reduce the losses of farm production to make more food available rather than the statistics of more farm production.

1

2

Agro-Product Processing Technology Technology Transfer influences

Postharvest research and extension

+ Farm assets +

+ Use of farm inputs

+

R

Farm productivity

+ Investment capacity R + + + Use of postharvest technology

+ Farm production

-

Farm income Loss reduction + + + ++ Value addition +

Price of farm products

Employment

FIGURE 1.1  Postharvest technology in increasing farm productivity, income, and employment.

1.2  IMPOTENCE OF POSTHARVEST TECHNOLOGY Postharvest technology has the capability to reduce losses, provide value addition, and generate employment. Agricultural development has two major activities: production and postharvest processing, as indicated in Figure 1.1. Postharvest processing refers to the processing and treatments that are carried out after harvesting, and it starts at harvesting and ends at marketing. All the processes such as harvesting, threshing, drying, storage, parboiling, milling, and preserving in cold storage are major postharvest operations for cereal grains while harvesting, handling, transportation, cooling, and preserving in cold storage are major postharvest operations for fruits and vegetables. The purposes of postharvest processing are to reduce the losses, maintain or enhance quality of the products, and make it marketable.

1.3  IMPORTANCE OF POSTHARVEST LOSSES Postharvest loss of an agro product is the total loss of the different operations starting from harvesting to consumption, and it is the summation of losses of different postharvest operations such as harvesting, drying, transportation, and storage with correction for moisture content (Bala, 1978; Majumder et al., 2016). This loss may be quantity loss, quality loss, or even goodwill loss. The following brief descriptions are intended to demonstrate different types of losses: Weight loss: Weight or quantity loss over a period of investigation. Nutritional loss: Any loss of edible dry matter. Viability loss: It is apparent through reduced germination. Indirect loss: Goodwill loss. Postharvest loss of an agro product, for example, rice for Asian countries, is shown in Table 1.1 and ranges from 10% to 37%. It is estimated that postharvest losses of rice at a farm level in Bangladesh are 9.16%, 10.10%, and 10.17% for Aman, Boro, and Aus, respectively. Also, it is estimated that 15%–50% of loss of horticultural crops such as vegetables and fruits results from improper processing and storage. Proper methods of processing, storage, transport, and marketing are required for export crops such as jute, tea, mango, jackfruit, and spices. Table 1.2 shows the range of postharvest loss of fruits and vegetables.

3

Introduction

TABLE 1.1 Postharvest Loss of Rice in Southeast Asian Countries Item No.

Operation

Range of Losses, %

1 2 3 4 5 6

Harvesting Threshing Drying Handling Milling Storage

1–3 2–6 1–5 2–7 2–10 2–6 Range 10–37

TABLE 1.2 Postharvest Loss of Fruits and Vegetables Fruit Mango Banana Litchi Papaya Guava

%, Loss

Vegetable

%, Loss

17.1–36.7 20–80 14–26 37–80 14–22

Tomato Cauliflower Cabbage Potato Onion

5–80 49 37 10–13 16.35

1.4  POSTHARVEST TECHNOLOGY Postharvest technology is of recent origin. It is capable of making more food, feed, fiber, and fuel available. Scientific conservation reduces losses and makes available more nutritive food by processing, packaging, transportation, and marketing. It enables agro-based industries to provide valueadded products and generate employment. Postharvest technology uses solutions that are unique, and these are commodity and location specific. There are numerous unit operations and these are threshing, drying, cleaning, grading, packaging, transportation, storage, and marketing. The following aspects may be given priority for cereal grains (Pandey, 2007):

1. Popularization of methods and techniques reduces losses. 2. More attention should be given to the primary postharvest operations such as cleaning, grading, sorting, drying, parboiling, milling, packaging, and storage. 3. Rubber rolls are used in place of haulers for rice milling. 4. Simple and effective storage systems are implemented. 5. Extension service is established.

The following aspects need attention for fruits and vegetables (Ramaswamy, 2015):

1. Harvesting at the optimum stage of maturity is followed by quick cooling. 2. Packaging and transfer to a controlled atmosphere storage are done where the temperature, relative humidity, air velocity, and atmospheric composition are the most appropriate for the produce. 3. Development of postharvest technology for mango, litchi, and jackfruit. 4. Extension service. 5. Design and development of innovative agro products from fruits and vegetables which add value and generate employment, resulting in agro-processing centers/industries in the rural sector.

4

Agro-Product Processing Technology

Finally, postharvest technology suitable for one crop and one location may not  be suitable for another location and another crop. For example, a dryer suitable for rice in Thailand may not be suitable for drying of rice in Bangladesh. Furthermore, a reduction of postharvest loss at one stage of the postharvest loss may have a far-reaching effect on the overall reduction of postharvest loss. For example, overdrying of paddy will increase the storage life, but this will increase the breakage percentage of the rice during milling. This suggests that a systems approach is essential for increasing the efficiency of the production system. Hence research and development of postharvest technology should be conducted to design and develop efficient technology/systems based on a systems approach for production of high-quality, value-added products. Example 1.1 Postharvest loss of rice in Bangladesh is 12%, and the total production of rice is 34.56  million metric tons. If 5% of the loss could be saved, how many days could we feed the population of Bangladesh? The population of Bangladesh is 163.08 million; assume per capita consumption of rice to be 180 kg.

Solution The annual food requirement of Bangladesh is given by Annual food requirement =



163.08 × 106 × 180 = 29.3544 million tons 103

The amount of rice lost for 5% postharvest loss of rice in Bangladesh is given by 5% loss of rice =



34.56 × 106 × 5 = 1.72 million tons 100

The number of days this rice can feed the population of Bangladesh is given by Number of days =



1.72 × 365 = 21.39 days 29.3544

Example 1.2 Postharvest loss of maize in Bangladesh is 5%, and the total production of maize is 2.3 million metric tons. If the price of maize is TK 15,000 per ton, calculate the annual monetary loss due to postharvest loss of maize in Bangladesh.

Solution The annual production of maize in Bangladesh is



Annual maize production = 2.30 million tons

5

Introduction The amount of maize lost for 5% postharvest loss of maize in Bangladesh is given by

5% loss of maize =



2.30 × 106 × 5 = 0.115 million tons 100

The monetary loss of maize due to postharvest loss of maize in Bangladesh is given by Monetary loss for maize = 0.115 × 15, 000 = 1725 million taka

Example 1.3

The  total production of rice is 34.56  million metric tons. The  population of Bangladesh is 163.08 million; assume per capita consumption of rice to be 180 kg. Compute the food security of rice in Bangladesh.

Solution The annual food requirement of Bangladesh is given by

Annual food requirement =



163.08 × 106 × 180 = 29.3544 million tons 103

The annual food production is Annual production of rice = 34.56 million tons



The food security in rice of Bangladesh is given by



Food security = in rice

34.56 = 1.18 29.3544

REFERENCES Bala, B. K. 1978. Post harvest losses of paddy in Bangladesh. AMA, 9(4): 54–56. Bala, B. K. 2017. Drying and Storage of Cereal Grains, 2nd edition. Wiley-Blackwell, Chichester, UK. Majumder, S., Bala, B. K., Arshad, F. M., Haque, M. A. and Hossain, M. A. 2016. Food security through ­increasing technical efficiency and reducing post harvest losses of rice production systems in Bangladesh. Food Security, 8: 36–372. Pandey P. H. 2007. Principles and Practices of Postharvest Technology. Kalyani Publishers, New Delhi, India. Ramaswamy, H. S. 2015. Postharvest Technologies of Fruits  & Vegetables. Destech Publications Inc, Lancaster, PA.

2

Physical, Thermal, and Chemical Properties of Food and Biological Materials

This chapter presents physical, thermal, and chemical properties of food and biological materials. The physical properties presented are physical dimension, 1000-grain weight, bulk density, shrinkage, angle of internal friction and angle of repose, and coefficient of friction. The thermal properties included are specific heat, thermal conductivity, volumetric heat transfer, and determination of volumetric heat transfer coefficient. Chemical properties discussed are starch, protein, fat, and vitamin.

2.1 INTRODUCTION This chapter describes the basic properties of food and biological materials that are required for simulating the heat and mass transfer phenomena during processing and storage. The  physical dimensions and 1000-grain weight are used to describe the physical characteristics of grains and their influence on drying. When the drying is simulated by a model using the diffusion equations to describe internal moisture movement, the above parameters are essential in order to select geometry and specific size. Bulk density, specific heat, and latent heat of vaporization of agro products as well as the shrinkage and heat transfer coefficient of a product bed are essential in any simulation of the heat and mass transfer during drying. In addition to these properties, a knowledge of thermal conductivity is essential for simulating heat and moisture movement during storage. The coefficient of friction of cereal grains in themselves and that on various surfaces are essential for rational design of storage structure. To address the quality during processing and storage, a knowledge of chemical properties is essential.

2.2  PHYSICAL PROPERTIES 2.2.1  Physical Dimensions Physical dimensions of a cereal grain are of vital importance in the design of cleaning and grading equipment. Furthermore, the dimensions of a grain have an important influence on its drying characteristics. Grain breeders need a guide in their work for developing new varieties of desirable size and shape. The length and width of a cereal grain are usually obtained by direct measurement through an accurate microscope, and the thickness is measured by a micrometer.

2.2.2  1000-Grain Weight The 1000-grain weight is also of importance in the design of cleaning and grading equipment. One thousand-grain weight also has application in the determination of effective diameter of the grain. The 1000-grain weight is usually determined by multiplying the weight of 100 randomly selected grains by 10, and the weight of 100 grains is measured by an electronic precision balance.

7

8

Agro-Product Processing Technology

Example 2.1 The 1000-kernel weight of fababeans with a moisture content of 8.5% (d.b.) is 405 g. Develop an expression to compute 1000-kernel weight at any moisture content on a dry basis.

Solution We can write W1000 = Wd (1+ Md )



The above equation can be rewritten as Wd =



W1000 (1+ Md )

Substituting W1000 = 405 g and Md = 0.085, we have Wd =



405 = 373.27 g 1+ 0.085

Hence required expression is



W1000 = 373.27(1+ Md )

2.2.3  Bulk Density Bulk density is one measure of quality. The bulk density of an agro product is usually determined by measuring the weight of an agro-product sample of known volume. The sample is placed in a cylindrical container of known volume, and the uniform density in the cylinder is obtained by gently tapping the cylinder vertically down onto a table several times in the same manner. The excess on the top of the cylinder is removed by sliding a string along the top edge of the cylinder. After the excess has been completely removed, the weight of the sample is measured by an analytical balance. Dry weight of the agro product is determined from the weight of the agro product and the moisture content.

2.2.4  Shrinkage Shrinkage of an agricultural product during drying is an observable phenomenon and has a ­significant effect on drying rate and temperature distribution. Although several researchers have reported that when agricultural products such as malt are dried from very high moisture content to very low moisture content, the shrinkage is not a linear function of moisture reduction, but the rate of shrinkage decreases with the increase in moisture reduction. The rate of change of shrinkage of grain bed with respect to the reduction in moisture content from initial moisture content is proportional to the difference between the maximum possible shrinkage and the actual shrinkage (Bala, 1983). Mathematically,

dy ∞ (y 0 − y) dx

(2.1)

9

Physical, Thermal, and Chemical Properties of Food and Biological Materials

The Equation (2.1) can be written as dy = ks (y 0 − y) dx



(2.2)

This equation requires the determination of the shrinkage coefficient and the maximum possible shrinkage from experimental data. The shrinkage of an agro-product bed is usually determined by simultaneously monitoring the changes in weight and depth during drying in a bin at constant temperature and mass flow rate of air. Equation 2.3 illustrates a typical example of shrinkage of malt during drying (Bala, 1983). S = 1.591(1 − exp( − 0.0966(M w0 − M w ))

S.E = 0.6871



(2.3)

Physically, the nonlinearity of Equation 2.3 can be interpreted as follows: The shrinkage of a malt bed at any instant during drying is the cumulative effect of the free shrinkage of the cells due to the loss of moisture and elastic shrinkage, if any, due to constraints on the free shrinkage exerted by the adjacent cells of the grains in the bed. The rate of shrinkage of the cells in the grains decreases as the moisture content approaches a low value. This explains why the rate of shrinkage in the grain bed gradually decreases to almost zero at very low moisture content. Example 2.2 Malt with moisture content of 45% (w.b.) was loaded to a depth of 1.0 m in a static bed dryer and it was dried to a moisture content of 3% (w.b). Determine the final depth of the grain bed.

Solution Here, Mw0 = 45% and Mw = 3% Using Equation 2.3, we have S = 15.91(1− exp( −0.0966 ( 45 − 3))) = 15.63



The final depth of the grain bed is



(1− 1× (15.63 / 100)) = 0.84 m

2.2.5  Friction The coefficients of friction of cereal grains on themselves and that on various surfaces are essential for rational design of grain bins, silos, and other storage structures. These properties are also important in the design of handling and processing equipment. 2.2.5.1  Angle of Internal Friction and Angle of Repose The coefficient of friction between granular materials is the tangent of the angle of internal friction for that material. The angle of repose is the angle that the side of the piled materials makes with the horizontal (Figure 2.1). For any material, the angle of repose varies with moisture content and amount of foreign materials present and increases with the increase in either. Engineers generally assume that the angle of internal friction is approximately the same as the angle of repose. But some researchers have reported that there is a difference between the two.

10

Agro-Product Processing Technology

FIGURE 2.1  Angle of repose of grains.

There are two types of angle of repose: (i) static angle of repose and (ii) dynamic angle of repose. The static angle of repose is the angle of friction taken up by a granular solid about to slide by itself, and the dynamic angle of repose is the angle of repose when the bulk of material is in motion—such as movement of solids from bins. Static angle of repose of cereal grains can be determined by using a wooden frame full of grains mounted on a tilting table. The top of the table is tilted until the grain begins to move along an inclined surface of the grains. Dynamic angle of repose can be determined by using a specially constructed box that contains grains. Then the front panel is quickly removed, and this allows the grains to flow to their natural slope. Angle of internal friction is measured by means of a special shearing box in which bulk material particles slide on each other along a plane. 2.2.5.2  Coefficient of Friction When one body is sliding on another, the tangential force resisting the motion is directly proportional to the normal force between the surfaces in contact. Thus, if F is the friction force and N is the normal force as shown in Figure 2.2, then

F = µ′ N

(2.4)

The coefficient of friction is approximately independent of the area in contact, the sliding velocity, and the intensity of pressure. There are two types of friction: (i) static friction and (ii) dynamic friction. A static friction is encountered at the start of motion whereas dynamic friction is present during motion.

FIGURE 2.2  Forces and angle of friction.

Physical, Thermal, and Chemical Properties of Food and Biological Materials

11

FIGURE 2.3  Device for measuring the angle of static friction.

Let W be weight of the frame containing grains and ϴr the tilting angle. The resulting force F is given by

F = W sin θr − µ′ W cos θr

(2.5)

At the point of sliding

0 = W sin θr − µ′ W cos θr µ′ = tan θ



(2.6)

Equation 24.6 shows that static friction is the tangent of an angle of static friction. To determine static friction of coefficient, the material to be tested is fastened to a tilting table and a small wooden frame filled with grains is placed on the table (Figure  2.3). The  frame is raised slightly so that it does not  touch the material. The  table is slowly tilted until the friction force between the grains and the material is overcome by gravity and downward movement begins. The coefficient of friction is calculated from the slope angle. The coefficient of static and dynamic friction can be determined by placing the grains in a small box in contact with a positively driven surface and employing force transducers.

2.3  THERMAL PROPERTIES 2.3.1  Specific Heat The specific heat at constant pressure is normally used for studying the heat transfer problems during drying and storage of agricultural products. The pressure dependence of specific heat is very little for both solids and liquid until extremely high pressure is encountered. Specific heat is also a function of temperature. However, it has been established that at ordinary temperatures and over temperature intervals that are not too great, specific heat may be considered as a constant physical property. The need for data on specific heat of food materials has been recognized as early as 1892. Siebel (1892) proposed that specific heat of food materials can be expressed as equal to the sum of the specific heat of the solid matter and that of water associated with the solid dry matter. Siebel proposed the following equations for food materials such as eggs, meat, fruits, and vegetables: For values above freezing

C pg = 0.837 + +0.03349 M w

(2.7)

12

Agro-Product Processing Technology

For values below freezing C pg = 0.837 + +0.01256 M w



(2.8)

Thermoflask calorimeters and the method of mixtures are now widely used for the measurement of specific heats of agro products. Whatever the type of calorimeter used and whether the calorimeter fluid is cooled or heated, the use of agro products at room temperature avoids the heat gain or loss during the transfer of the agro products into the calorimeter, and the only correction needed is for the final temperature in the calorimeter. There exists a general agreement among the researchers on the observations of specific heats of common agricultural crops that the specific heat of wet grain increases linearly with moisture content. The specific heat of a grain is usually determined by the method of mixtures using distilled water as a calorimeter fluid. The technique consists of determining the temperature change of water contained in the calorimeter when a known quantity of the product is added to it at a known different temperature. The specific heat is calculated by solving the following heat balance equation: Heat loss by product = heat gained by water and calorimeter or C pg Wg (Tg − Tf ) = WcC pc (Tf − Ti ) + WwC pl (Tf − Ti )



C pg Wg (Tg − Tf ) = C pl (WE + Ww )(Tf − Ti )



(2.9)

An ordinary thermoflask can be used as a calorimeter, and fiberglass insulation should be added between the vacuum bottle and the outer metal walls of the container. The  specific heat can be determined by dropping the product directly into the calorimeter. The water equivalent of the thermoflask calorimeter can be determined by using materials of known specific heat such as lead shot and distilled water instead of the product, and then for the material of known specific heat, WE becomes the only unknown in Equation 2.9. Example 2.3 Note that 25 g of rough rice (moisture content of 13.5% w.b.) at a temperature of 22°C was dropped into a calorimeter containing 46.1 g of ice-cold water at a temperature of 4.66°C. The final temperature of the mixture is 7.84°C. The water equivalent of the calorimeter is 17.76 g and determines the specific heat of rough rice at a moisture content of 13.5% on a wet basis.

Solution Equation 2.9 can be written as Cpg =



Cpl(WE + Ww )(Tf − Ti ) WE (Tg − Tf )

We have here Ti = 4.66°C, Tg = 22°C, Tf = 7.84°C, WE = 17.76 g. Ww = 46.1 g, Wg = 25 g, and Cpl = 4.186 kJ/kg·K Substituting these values to the above equation gives



Cpg =

4.186 × (17.76 + 46.1)(7.84 − 4.66) = 2.40 kJ/kg K 25 × (22 − 7.84)

Physical, Thermal, and Chemical Properties of Food and Biological Materials

13

2.3.2 Thermal Conductivity Temperature of cereal grains changes during storage as a result of day-to-day variation in weather conditions, and it is one of the most important factors controlling the rate of deterioration during storage. To design a storage system rationally, one must predict the temperature changes during storage. A knowledge of thermal conductivities is essential for the prediction of the product temperature. Thermal conductivity of product is also important in the engineering design of drying, cooling, and aeration systems. There are mainly two methods for determining thermal conductivity of agricultural products. They are:

1. Steady-state method 2. Transient heat flow method

In steady-state method, either a cylindrical apparatus or spherical apparatus (Oxley, 1944) is used. The  transient heat flow method using a line heating source apparatus developed by Hooper and Lepper (1950) has been used extensively in the determination of thermal conductivity of cereal grains. The  main objections to the steady-state method are: (i) A  long time is required to achieve steady-state condition, and (ii) there is a possibility of moisture migration along the temperature gradient during a long test period. Both these factors are minimized in the transient heat flow method because of the reduced test period. 2.3.2.1 Theory The basic equation for the heat flow from a line source is



 ∂ 2 T 1 ∂T  ∂T = α 2 +  ∂t r ∂r   ∂r

(2.10)

The solution of the change in temperature at a point close to the line heat source between time t1 and t2 can be written in the following form:

Q loge (t 2 /t1 ) 4πK

(2.11)

Q loge (t 2 /t1 ) 4 π (T2 − T1 )

(2.12)

T2 − T1 =

Thus, the equation for thermal conductivity is

K=

There are two sources of error in the use of these equations: (i) In the derivation of this equation, only the first two terms of an infinite series have been considered. But researchers (Hooper and Lepper, 1950; Kazarian and Hall, 1965) have shown that the error involved due to neglecting the higher order terms is negligible; (ii) in actual test apparatus, the line heat source used has a finite length and diameter. Again, the axial heat flow is considered negligible. To compensate the heat source that in effect replaces a small core of grain, Vander Held and Van Drumen (1949) have shown that the difference in heat absorption between the heater and displaced core can be considered as

14

Agro-Product Processing Technology

heat production before the start of measured time. That is time t0 is subtracted from each observed time. By differentiation, Equation 2.12 gives dT 4 π K = t c dt Q



(2.13)

at dt/dT = 0 tc = t0 the time correction. The corrected equation for thermal conductivity thus becomes



K=

 t −t  Q ln  2 0  4 π (T2 − T1 )  t1 − t 0 

(2.14)

Underwood and McTaggart (1960) proposed another method that is simple and less time-­consuming. By using this method, there is no need for evaluating the correction time. Temperature rise is plotted against time on a semilog paper. Once the straight-line portion is established, values obtained from the straight-line portion of the curves are used in the following equation to calculate the bulk thermal conductivity.



K=

I2R ln (θ2 /θ1 ) 4 π (T2 − T1 )

(2.15)

Time θ2 and θ1 corresponds to temperature T2 and T1 on the straight-line portion of the curve on a semilog paper. 2.3.2.2  Apparatus and Measurement The  apparatus basically consists of a cylinder made usually of aluminum with a heating wire stretched between copper leads on the axis of a cylinder. The cylinder is insulated. A schematic diagram of an apparatus is shown in Figure 2.4. Heat is supplied by a constant DC power through a variable resistance. The current is measured by an ammeter, and the temperature of the heating wire is measured by a thermocouple. To measure thermal conductivity, the cylinder is filled with grain and tapping is performed to obtain uniform density. The temperature at the center of the cylinder is checked. Once the temperature is stabilized, current is turned on and the temperature is recorded at regular intervals. The bulk thermal conductivity is computed using Equation 2.15.

FIGURE 2.4  Thermal conductivity measurement apparatus.

15

Physical, Thermal, and Chemical Properties of Food and Biological Materials

Example 2.4 The temperature time of a test for determining the bulk thermal conductivity of rough rice with a moisture content of 13.5% (w.b.) was plotted on a semilog paper. The temperature difference corresponding to time 130 and 60 sec on the linear portion of the semilog plot was 1.8°C. The resistance and the current in the heating wire were 2.458 ohm/m and 1.3 A, respectively. Determine the bulk thermal conductivity.

Solution Here we have θ2 = 130 s, θ1 = 60 s, I = 1.3 A, R = 2.458 ohm/m, and T2−Tl = 1.8°C Substituting these values in Equation 2.15 yields



K=

2.458 × (1.3)2 ln(130/60) = 0.14 W/m C 4 π × 1.8

2.3.3 Latent Heat of Vaporization The  latent heat of vaporization of agro-product moisture is defined as the energy required to vaporize moisture from the product. The energy required to evaporate moisture from the product especially at low moisture content is higher than that of free water and depends on the type of product. Othmer (1940), starting with the Clapeyron equation, developed the following equation: 

log P′ = (L′/L) logP + C

(2.16)

where P and P′ are vapor pressures, L and L′ are molal heats of the two compounds, respectively, taken at the same temperatures, and C is a constant. Equation 2.16 states that if the log of the pressure of any substance is plotted against the log of the pressure of any other substance, a straight line results that will have for its slope the ratio of the molal latent heats. Plots were made to illustrate the utility of this relation with various materials in checking and correlating vapor pressure data, and this matched the data. Gallaher (1951), using the equilibrium moisture content data for wheat (published by Gay), determined the latent heat of vaporization of wheat and found that when the ratios of the latent heat of wheat to the latent heat of free water were plotted against moisture content (d.b.), the resulting curve could be described by the following equation:

L wheat = 1 + 23exp( − 0.40 Md ) L water

(2.17)

Bala (1983), using the equilibrium moisture content data for malt from the unpublished data of Pixton and Henderson (1981), developed Othmer plots for malt and also developed an equation to describe the ratio of the latent heat of vaporization of malt to the latent heat of vaporization of free water as a function of moisture content and of the form used by Gallaher (Equation 2.17). The following regression equation was developed.

L malt = 1 + 0.5904 exp( − 0.1367 Md ) L water

(2.18)

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Agro-Product Processing Technology

Equations 2.17 and 2.18 show that the amount of heat required to vaporize moisture from the grain increases considerably as the moisture content of the grain decreases. The theoretical basis for the determination of latent heat of a grain is the Othmer plot based on the Clapeyron equation. The Clapeyron equation is given by Rogers and Mayhew (1980),

dP L = dTab (V − v) Tab

(2.19)

If v is disregarded in comparison to V, and if it is possible to assume a perfect gas, the following equation may be used for the volume term:

V=

R 0 Tab P

(2.20)

From the combination of Equations 2.19 and 2.20 the Clapeyron equation results:

dP LP = dTab R 0 Tab 2

(2.21)



1 dP dTab = L P R 0 Tab 2

(2.22)

The same equation may be written for water in grains at the same temperature:

1 dP′ dTab = L′ P′ R 0 Tab 2

(2.23)

dP′/P′ L′ = dP/P L

(2.24)

From Equations 2.22 and 2.23

It is also apparent that Equation 2.24 can be integrated to give

log P′ =

L′ log P + C L

(2.25)

The latent heat of the material can be established from the slope of the log P′ versus log P curve. 2.3.3.1  Determination of Latent Heat of Vaporization The latent heat of vaporization of an agro product is determined by equilibrium moisture content data. For each temperature, the saturated vapor pressure is found from the steam tables. The vapor pressure of an agro product at each moisture content is determined by multiplying the corresponding

Physical, Thermal, and Chemical Properties of Food and Biological Materials

17

relative humidity by the saturation vapor pressure for the given temperature. The logarithm of the vapor pressure of the product is plotted against the logarithm of the vapor pressure of free water. These should give well-defined straight lines. The slope of the lines, namely, the ratios of the latent heat of the product to the latent heat of free water, are plotted against moisture content (d.b.). An equation can be developed to describe the ratio of the latent heat of grain to the latent heat of free water as a function of moisture content and may be of the form used by Gallaher. Example 2.5 The  equilibrium moisture content-relative humidity data of malt for different temperatures are given below. Draw an Othmer plot and determine the ratio of latent heat of the moisture of malt to latent heat of free water at each moisture content. Moisture content

Relative Humidity %

% (d.b.)

5°C

25°C

45°C

47.92 38.12 19.18 10.74 7.41 5.59

95.9 95.0 76.1 37.2 17.7 11.2

97.0 95.0 79.3 42.4 21.8 14.9

95.3 94.0 82.0 53.6 30.5 18.6

Solution The saturation vapor pressures for each of the given temperatures are obtained from steam tables. By multiplying the equilibrium relative humidity by the saturation vapor pressure for the given temperature, the values of the grain vapor pressure are obtained. These vapor pressures are plotted against the vapor pressure of free water on log paper, and the resulting Othmer plot is shown in Figure 2.5. They give reasonably well-defined straight lines. The slopes of the lines in Figure 2.5 are the ratio of the latent heat of grain moisture to the latent heat of free water. The vapor pressure and the ratios of the latent heats are given in the tabular form below: Equilibrium Vapor Pressure, Milibar

Ratio of Latent Heat of Malt to

5°C

25°C

45°C

Latent Heat of Water

8.3617 8.2832 6.6353 3.2453 1.5432 0.9765 8.7192

30.7208 30.0874 25.1151 13.4285 6.9042 4.7189 31.6710

91.3498 90.1037 78.6011 51.3782 29.2357 17.8290 95.8550

1.0000 1.0000 1.0410 1.1503 1.2222 1.2500 -

Moisture % (d.b.) 47.92 38.12 19.18 10.74 7.41 5.59 Saturation Vapor pressure Milibar

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FIGURE 2.5  Othmer plots for equilibrium moisture content data of malt.

2.3.4 Heat Transfer Coefficient of Product Bed The rate of heat transfer between a solid and fluid may be computed from the following relation:

q c = h c A ∆T

(2.26)

Equation 2.26 was originally proposed by Isaac Newton in 1770. Engineers have used this equation for many years, although it is a definition of hc rather than a phenomenological law of convection. The heat transfer coefficient is actually a complicated function of the fluid flow, the thermal properties of the fluid medium, and geometry of the system. There are mainly two general methods available for determination of convection heat transfer coefficients in a packed bed of granular materials:

1. Dimensional analysis correlation existing data 2. Direct measurement of heat transfer coefficient of a grain bed: comparing the temperature curves with Schumann’s exact solution

Dimensional analysis is mathematically simple and has a wide range of application. This method is useless and incomplete without sufficient experimental data and contributes little to our understanding of the transfer process, but it facilitates the interpretation and extends the range of application of experimental data by correlating them in terms of dimensionless groups. The second method presupposes that the physical mechanisms are sufficiently well understood to be described in the form of partial differential equations. Mathematical solution of the set of partial differential equation is quite complicated. The solution in the dimensionless form is compared with the experimental data to compute convective heat transfer coefficient. Exact solutions are more important because the assumption made in this course of analysis can be specified accurately, and the validity can be checked by experiment.

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Physical, Thermal, and Chemical Properties of Food and Biological Materials

2.3.4.1  Dimensional Analysis The  correlating equation  developed by Yoshida et  al. (1962) is selected. This  equation  was ­developed by correlation of the data of Wilke and Hougen; Gamson, Thodos, and Hougen; and Wakao, Oshaima, and Yagi. Thus, the equation  has been supported by sufficient experiment data. For  50 < Re 10 L

The product of Re and Pr numbers normally occurs in laminar flow correlations. It is called the Peclet number, viz,.

Pe = Re Pr

(10.38)

For fully developed turbulent flow in smooth tubes and ducts, we have found the following correlation is widely used (Ditus and Boelter, 1930). Ditus and Boelter correlation

Nu = 0.023 (Re)0.8 Pr n

(10.39)

where: n = 0.4 for heating n = 0.3 for cooling Note that the properties are evaluated at the bulk mean temperature. This equation is applicable under the condition

L < 60 d



23 × 102 < Re < 12 × 10 4



0.7 < Pr < 120

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For heat transfer coefficient between two parallel plates, the following correlation is observed is derived from Kay’s data for fully developed turbulent flow with one side heated and the other side insulated. Nu = 0.0158 Re 0.08



(10.40)

Malik and Buelow suggest the following equation

(

)

Nu = 0.01344 Re 0.75 / 1 − 1.586 Re −0.125



(10.41)

The heat loss from flat plates exposed to outside winds is more important in the study of solar collectors. Sparrow et al. (1979) did wind tunnel studies on rectangular plates at various orientations and found the following correlation over Reynolds number range of 2 × 104 to 9 × 104. Nu = 0.86 Re1/ 2 Pr1/ 3



(10.42)

Where the characteristic length is four times the collector area divided by the plate perimeter. Example 10.2 Compute the heat transfer coefficient for water flowing at an average temperature of 100°C and at a velocity of 0.232 m/s in a 2.54 cm bore pipe using the following Prandtl-Taylor equation: Nu =



0.0386Re3/4Pr 1+ 2.44Re −1/8 (Pr − 1)

At 100°C, Pr = 1.74, K = 0.68 × 103 kW/(mK) and v = 0.02984 ×10−5 m2/s

Solution Reynolds number is given by



Re =

v d 0.232 × 0.0254 × 10 4 = = 20,000 0.0294 υ

Using Prandtl–Taylor equation, we have



Red1/8 = 3.45



Red0.75 = 1643





Nu =

h=

0.0386 × 1643 × 1.74 = 72.4 1+ (2.44/3.45) × 0.74

72.4 × 0.68 × 10 −3 = 1.937 kW/(m2K) 0.0254

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Heating and Cooling of Agro Products

10.3.2 Natural or Free Convection Over the years, it has been found that the average free convection heat transfer coefficient can be represented in the following functional form for a variety circumstances.

Nu = c (Gr Pr)m

(10.43)

where c and m are constant. The properties in the dimensionless groups are evaluated at the film temperature Tf =



Ta + Tw 2

where Ta is the air temperature and Tw is the wall temperature. The  product of the Grashof and Prandlt numbers is called the Rayleigh number: Ra = Gr Pr



The heat transfer coefficient between the air and the horizontal surface is given by the following correlation (Ozisik, 1983):

105 < Gr Pr < 2 × 107



Nu = 0.54 (Gr Pr)1/ 4



2 × 107 < Gr Pr < 3 × 1010



Nu = 0.14 (Gr Pr)1/ 3

(10.44)

(10.45)

The heat transfer coefficient between the air and the vertical surface is given correlation (Ozisik, 1972):

1 > Gr Pr < 1012 Nu = 0.825 +

0.387 (Gr Pr)1/ 6 9/6   1 + (0.492 / Pr) 

8 / 27



(10.46)

For free convection from the inclined surfaces at an angle β with the horizontal, the following correlations are given (Fujii and Imura, 1972). Inclined surface facing downward

β > 12° and105 < Gr Pr sin β < 1011



Nu = 0.56 (Gr Pr sin β)1/ 4

(10.47)

Almost horizontal plates facing downward

0° < β < 12° and 106 < Gr Pr β < 1011



Nu = 0.58(Gr Pr)1/ 5

(10.48)

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Inclined plates facing upward

−30° < β < 15° and 105 < Gr Pr sin β < 1011



Nu = 0.145 (Gr Pr)1/ 3 − (Grc Pr)1/ 3  + 0.56 (Gr Pr sin β)1/ 4

(10.49)

Where Grc is the critical Grashof number. If Gr < Grc , the terms within square brackets are dropped. The value of Grc is dependent on the tilt angle β according to the following correlation Grc = Exp(8.991 + 0.3795β − 2.733 × 10 −2 × β2 )



(10.50)

where β is in degrees and for 15° m cC pc and Equation  (10.52) applies. The temperatures diverge at the inlet end when m h C ph < m cC pc and Equation (10.53) applies. In parallel flow, it is obvious that Tc2 will approach Th2 for an infinitely long heat exchanger but can never exceed Th2. In counterflow, it is it is quite normal for Tc2 to exceed Th2 and consequently the counterflow heat exchanger is more effective. Effectiveness is the ratio of energy actually transferred to the maximum theoretically possible. Again, the definition depends on the relative thermal capacities of the streams. The maximum theoretical heat transfer will take place in a counterflow heat exchanger of infinite length and in such a case Tc2 → Th1 when m h C ph > m cC pc , and Th2 → Tc1 when m h C ph < m cC pc . Thus, the maximum transfers in the two cases are: m cC pc (Th1 − Tc1 ) when m h C ph > m cC pc m h C ph (Th1 − Tc1 ) when m h C ph < m cC pc The actual transfers in the two cases are m cC ph (Tc2 − Tc1 ) and m h C ph (Th1 − Th2 ) and hence E, the effectiveness becomes

E=

Tc2 − Tc1 when m h C ph > m cC pc Th1 − Tc1

(10.54)

E=

Th1 − Th2 when m h C ph < m cC pc Th1 − Tc1

(10.55)

These definitions may be used in either counterflow or parallel flow; the value of E will be lower in parallel flow. The primary purpose of a heat exchanger is to achieve the required transfer rate using the smallest possible transfer area and fluid pressure drop. A large exchanger can mean unnecessary capital outlay, and a high-pressure drop means a reduced efficiency of the plant. Generally, a smaller exchanger can be produced by finning surfaces to increase the overall heat transfer coefficient. However, this leads to a higher fluid pressure and the best design is often a compromise between requirements. The fluids may flow either in parallel or counter parallel and the temperature profiles for such flows are shown in Figure  10.8. We can calculate the heat transfer requirement for a parallel or counter-parallel heat exchanger from the relation

q = UA ∆Tm

(10.56)

Consider an incremental area of heat exchanger surface area for a parallel flow heat exchanger as shown in Figure 10.8. The heat flow through an element of area dA may be written as

dq = − m h C ph dTh = m cC pcdT

(10.57)

The heat transfer could also be expressed as

dq = U(Th − Tc ) dA

From Equation (10.57)

dTh =

−dq m h C ph

(10.58)

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Heating and Cooling of Agro Products

dTc =



dq m cC pc

Thus,

 1 1  dTh − dTc = d(Th − Tc ) = −dq  +   m h C ph m cC pc 

(10.59)

Solving for dq from Equation (10.58) and substituting in Equation (10.59) gives



 1 d(Th − Tc ) 1  = −U  +  dA Th − Tc  m h C h m cC pc 

(10.60)

This differential equation may now be integrated from condition 1 to condition 2 in Figure 10.8 and this yields



ln

 1 (Th2 − Tc2 ) 1  = −UA  +  (Th1 − Tc1 )  m h C h m cC pc 

(10.61)

From Equation (10.57), we can write

m h C ph =

q Th1 − Th2

(10.62)



m cC pc =

q Tc1 − Tc2

(10.63)

Substituting these equations into Equation (10.61) gives

q = UA

(Th2 − Tc2 ) − (Th1 − Tc1 ) ln [(Th2 − Tc2 )/(Th1 − Tc1 ) ]

(10.64)

Comparing Equation (10.64) and Equation (10.56) we have

∆Tm =

(Th2 − Tc2 ) − (Th1 − Tc1 ) ln [(Th2 − Tc2 )/(Th1 − Tc1 ) ]

(10.65)

This temperature difference is called the log mean temperature difference Analysis for a cross-flow heat exchanger is more complicated owing to temperature variations across the flow. The results of this type of exchanger are available as correction factors and Equation (10.56) becomes

q = UA F∆Tm

where F is a factor to be obtained from the appropriate graph.

(10.66)

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Example 10.4 Water at the rate of 68 kg/min is heated from 35°C to 75°C by an oil having a specific heat of 1.9 kJ/kg-°C. The fluids are used in a counterflow double-pipe heat exchanger, and the oil enters the exchanger at 110°C and leaves at 75°C. The overall heat transfer coefficient is 320 W/m2 °C. Calculate the heat exchanger area.

Solution The total heat transfer is given by



q = mwCpw ∆Tw = 68 × 4180 × (75 − 35) = 11.37 MJ/min = 1.896 × 106 W and the log mean temperature is given by



∆Tm =

(110 − 75) − (75 − 35) = 37.44°C ln [(110 − 75)/(75 − 35)]

Since q = UA∆Tm , A=



1.895 × 105 = 15.82 m2 320 × 37.44

Example 10.5 Instead of the double-pipe heat exchanger of Example 10.4, it is preferred that a shell and tube exchanger be used with the water making one shell pass, the oil making two tube passes. Calculate the area required for this exchanger, assuming that the overall heat transfer coefficient is 320 W/m2 °C and the correction factor is F = 0.81.

Solution The total heat transfer is q = UAF ∆Tm , A=



1.895 × 105 = 19.53 m2 320 × 0.81× 37.44

Example 10.6 Alcohol at the rate of 0.20 kg/s is cooled from 75°C to 35°C in a counterflow heat exchanger. Cooling water enters the exchanger at 12°C and at the rate of 0.16 kg/s. The convection coefficient between the alcohol and the tube wall is 0.34 kW/m2·K and between the tube wall and the water is 0.225 kW/m2·K. The tubes may be assumed thin. Cp for the alcohol is 2.52 kJ/kg·K and for water is 4.187 kJ/kg·K. Calculate the capacity ratio, the effectiveness of the heat exchanger area.

Solution For the hot stream, alcohol



mhCph = 0.2 × 2.52 = 0.504 kJ/sK

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Heating and Cooling of Agro Products For the cold stream, water



mc Cpc =0.16×4.187=0.671 kJ/sK From Equation 10.49 the capacity ratio is



= C

mhCph 0.504 = = 0.75 mcCpc 0.671

An energy balance gives q = mwC pw ∆Tw = 0.2 × 2.50 × (75 − 35) = 0.16 × 4.18 × (Tc2 − 12) 20.15 = 0.67 × T c2 −8.05 Tc2 = 41.8°C





Effectiveness E is given by



E=

Th1 − Th2 75 − 35 = = 0.635 Th1 − Tc1 75 − 12

and the log mean temperature is given by



∆Tm =

(35 − 12) − (75 − 41.8) = 28°C ln [(35 − 12)/(75 − 41.8)]

Overall heat transfer coefficient is given by 1 1 1. 1 1 = + = + U halcohol hwater 0.34 0.225 U = 0.1355

Since q = UA ∆Tm,



A=

20.15 = 5.30 m2 0.1355 × 28

10.4 RADIATION When radiant energy strikes a surface, part of the radiation is reflected, part is absorbed, and part is transmitted as shown in Figure 10.9. Defining reflectivity ρ, absorbtivity α, and transitivity τ, we have

ρ + α + τ = 1

(10.67)

Most solids do not transmit any radiation and the transitivity is zero for such bodies. Then we have

ρ + α = 1

(10.68)

Two types of reflection phenomena are observed. If the angle of incident is equal to the angle of reflection, the reflection is called secular. On the contrary, if the incident radiation is distributed uniformly in all directions, it is called diffuse. These two types of reflection are shown in Figure 10.10.

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FIGURE 10.9  Radiation components.

FIGURE 10.10  (a) Specular and (b) diffuse radiation.

All materials have values of α and ρ between 0 and 1. However, It is useful and important to imagine a material for which α = 1 and ρ = 0. A body composed of this material is known as a black body; it absorbs all incident energy upon it and reflects none. For real materials, the highest value of α is around 0.97. Black-body radiation consists of emission over the entire range of wavelength and is not uniformly distributed. Then ebλ may be defined as the monochromatic emittance, the energy emitted per unit area at wavelength λ for a black body. In order to calculate the heat radiated by a black body, Max Planck formulated the concept of his quantum theory and derived the following expression for monochromatic thermal radiation emitted by a black body: e bλ =



C1λ −5 exp(C2 /λT) − 1

(10.69)

where λ is the wavelength, µm, T is temperature, K, C1 = 3.743 × 105 kW·µ4/m2, and C2 = 1.439 × 104 µk. The emittance power of a black body can be found by integrating the equation over the whole range of wavelengths. Thus

eb =

∫

∞

0

eλ dλ =

∫

∞

0

−5 1 ( C2 /λT )

Cλ

e

dλ = σT 4 −1

(10.70)

where σ is called the Stefan-Boltzman constant and has a value of 56.7 × 10−12 kW/m2K4. Radiation from real surfaces differs in many respects from black-body radiation. The emissive power of a real surface is always less than that of a black body at the same temperature. Most real materials exhibit some variation in ελ with wavelength. These are known as selective emitters. However, there is a second type of ideal surface known as gray surface where the emissivity, ε, is constant with wavelength. It must follow that for a gray body

e = ε eb = ε σ T4

(10.71)

For the purpose of heat transfer calculations, most real surfaces are regarded as gray surfaces.

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Heating and Cooling of Agro Products

10.4.1 Radiation Intensity and Shape Factor The intensity of black-body radiation, I is the radiation emitted per unit time and unit solid angle subtended at the surface and per unit area of emitting surface normal to the mean direction in space, and may expressed as:

I=

dE b (dA 2 /r 2 )dA1cosϕ

(10.72)

This is shown in Figure 10.11. dA2/r2 is the solid angle subtended by dA2. The surface dA1 is specified as diffuse, thus Lambert’s law states that I is constant in the hemispherical space above dA1. From the above definition of I, it thus follows that dEb/dA2 will have a maximum value at any given r when ϕ = 0 and it will be zero ϕ = 90°. In general

 dE b   dE b   dA  =  dA  cos φ 2 ϕ 2 n  

(10.73)

 dA  dE b = I  2 2  dA1cos φ  r 

(10.74)

From Equation (10.72)

and from Figure 10.12 it is seen that dA 2 = r dφ (rsin φ dθ) = r 2sinφ dφ dθ Hence

dE b = I dA1 sinφ cosφ dφ dθ

(10.75)

The total radiation passing through the hemispherical surface is φ= π /2 θ=2



E b = I dA1

∫ ∫

sin φ cos φ dφ dθ

φ= 0 θ= 0 φ= π /2



= 2π I dA1

∫

φ= 0

FIGURE 10.11  Illustration of intensity of radiation.

sin φ cos φ dφ = π I dA1

(10.76)

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FIGURE 10.12  Details from Figure 10.11.

Hence I=



e b σT 4 = π π

(10.77)

10.4.2 Radiation Exchange between Black Surfaces Consider the radiation exchange between two arbitrary disposed black surfaces of area A1 and A2, and at temperatures T1 and T2. Small element of surface dA1 and dA2 are considered. The distance between the two is r and the line joining the two centers makes angles of ϕ1 and ϕ2 to their normals. Then the solid angles are

dω1 =

dA 2 cosφ2 dA1 cosφ1 and dω2 = r2 r2

The radiant energy emitted by dA1 that strikes on dA2 is given by

 dA cosφ dE b(1−2) = I1 dA1 cosφ1  2 2 2 r 

  

(10.78)

Since both surfaces are black, this energy is absorbed by dA2. A similar quantity of energy is emitted by dA2. This can be expressed as

 dA cosφ  dE b(2−1) = I 2dA 2 cosφ2  1 2 1  r  

(10.79)

The net exchange is

dE b(1−2) − dE b(2−1) = dE b(12) =

dA1 dA 2 cosφ1 cosφ2 (I1 − I 2 ) r2

(10.80)

Equation 10.77 can be substituted into Equation 10.80 and this will result

dE b(12) =

σ dA1 dA 2 cosφ1 cosφ2 4 (T 1 − T2 4 ) πr 2

(10.81)

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Heating and Cooling of Agro Products

From Equation 10.78 the heat energy radiated by A1 that falls on A2 is given by dE b(1−2) =

∫∫ A1

A2

= σT1

4

cosφ1cosφ2dA1dA 2 . r2 cosφ1cosφ2dA1dA 2 π r2 A2



∫∫ A1

(10.82)

But total energy radiated by A1 is E b(1) = A1σ T 4



(10.83)

Hence the fraction of energy radiated by A1 that falls on A2 is F1−2, and it is known as the geometric configuration factor or shape factor of A1 with respect to A2. Thus, the energy radiated by A1 that falls on A2 is

E b(1−2) 1 I1 = E b(1) A1

∫∫ A1

A2

cosφ1 cosφ2 dA1dA 2 = F1−2 π r2

(10.84)

and E b(1− 2) = F1− 2A1σT14



(10.85)

Similarly, Equation 10.79 shows the total energy radiated by A2 that falls on A1 is expressed as

E b(2−1) = σT2 4

∫∫ A1

A2

cosφ1 cosφ2 dA1dA 2 π r2

(10.86)

and the total energy radiated by A2 is A2σT24 and hence

E b(2−1) 1 I2 = E b(2) A2

∫∫ A1

A2

cosφ1 cosφ2 dA1dA 2 = F2−1 π r2

(10.87)

and E b(2 −1) = F2 −1A 2σT2 4



(10.88)

The net energy exchange can be expressed in terms of either of the shape factor F1−2 or F2−1. Thus

E b(1− 2) = F1− 2A1 σ(T14 − T2 4 ) = F2 −1A 2 σ(T14 − T2 4 )

(10.89)

10.4.3 Heat Exchange by Radiation between Gray Surfaces The radiation from gray surfaces may be treated conveniently by using the network method of analysis. Radiosity is the sum of all radiation leaving a surface, emitted and reflected when no radiation is transmitted or

J = ε Eb + ρ G

(10.90)

J = ε E b + (1 − ε) G

(10.91)

Since τ = 0, ρ = 1−𝑎 = 1−ε so that

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Net energy loss of the surface is Q 1 = J −G = (E b − J) A (1 − ε)/ε



Q=

Eb − J Eb − J = 1− ε R A

(10.92)

Equation 10.92 can be considered to give net loss of energy from a gray surface. Here, the denominator is as the resistance to radiation heat transfer, the numerator as potential difference, and the heat flow as the current. Figure 10.13a gives the electrical analogy of the system. Now consider radiation energy exchange between two gray surfaces A1 and A2. Energy leaving A1 and reaching A2 is J1A1F12 Energy leaving A2 and reaching A1 is J2A2F21 Net energy exchange is Q1− 2 = J1A1F12 − J 2A 2F21 = (J1 − J 2 )A1F12 = (J1 − J 2 )A 2F21 =

J1 − J 2 J1 − J 2 J1 − J 2 = = 1 1 R A1F12 A 2F21

(10.93)

We have R, which signifies space resistance. An electrical analog of Equation 10.86 is shown in Figure 10.13b. Equations  10.88 and 10.89 establish the principles of electrical network to solve radiation exchange problems between gray bodies. We need only to connect a surface resistance to each surface and a space resistance between radiosity potentials. In this case, the heat transfer would be the overall potential difference divided by the sum of the resistances. The electrical analog of heat exchange by radiation between two gray surfaces is shown in Figure 10.14. The net flow of heat can be expressed as q=



E b1 − E b2 Q = A 1 − εf1 + 1 + 1 − ε2 ε1A1 A1F12 ε2A 2

(10.94)

When two surfaces are equal, and A = A1 = A2, Equation 10.87 gives q=



(a)

Q σ(T14 − T2 4 ) = 1 1 A + −1 ε1 ε2

(b)

FIGURE 10.13  (a) Resistances in radiation, (b) resistances in radiation.

(10.95)

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Heating and Cooling of Agro Products

FIGURE 10.14  Radiation network for two gray surfaces.

10.5 COOLING Cooling is a process of removal of heat and hence it is a process of reducing the temperature of the produce to below the ambient temperature. Precooling is no different from cooling and is the cooling prior to storage and processing as well. Household refrigerators operate at 4°C7°C. Chilling or cold storage uses slightly lower temperatures on the basis of the food to be refrigerated. Each fruit and vegetable has an optimum temperature for its shelf life. Undesirable changes in fruits and vegetables occur when the temperature is reduced below the optimum temperature. This is known as chilling injury. Chilling injury reduces greatly the quality of the product and shortens the shelf life. Chilling will preserve perishable foods for days or weeks. Cooling or precooling should address two questions. How much to cool (temperature limits)? And how fast should this be done (time limit)? In general, this limit is the lowest temperature to which it could be cooled without chilling or freezing injuries. Ideally, cooling should be carried out as rapidly as possible, but in practice as fast as feasible economically and technically. The  heat removal by change in temperature changes sensible heat of the produce or changes the state, i.e., solid to liquid (latent heat). Several methods are available for general cooling of products and these are room cooling, hydrocooling, vacuum cooling, etc. There are special techniques for cooling packaged materials and these are top icing, package icing, liquid nitrogen cooling, dry ice cooling, etc. Simple room cooling consists of an insulated room with appropriate refrigeration system, while forced-air cooling rooms are designed to promote faster cooling times. Hydrocooling makes use of water as a cooling medium since water is a better heat transfer medium than air. Package icing is used in the fields to cool the produce by using various forms of ice such as ice flakes or ice slurry. Vacuum cooling is another technique used to rapidly cool fresh, leafy vegetables. The produce is placed in the vacuum chamber and high vacuum is applied. Vacuum cooling works well with vegetables that have high moisture and a large surface area.

10.5.1 Cooling Rate The cooling rate is determined from Newton’s law of cooling, which states that the rate of change in temperature of a body surrounded by a medium at constant temperature is proportional to the difference in temperature between the body and the surrounding medium when the temperature difference is small. Mathematically,

dT = − k t (T − Te ) dt

(10.96)

Equation 10.96 can be written as T



∫

Ti

t

dT = k −dt T − Ta

(10.97)

T − Ta = e − k t T i −T a

(10.98)

∫ 0

On integration this equation yields

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Half Response Time The time of response of cooling may be compared to the time of radioactive decay and the time of one-half response is defined as the time necessary to obtain a temperature ratio of one-half. Likewise, one-fourth response time may be defined.

1. One-half response time: T − Ta = exp( − kt ) Ti − Ta





Since

T − Ta 1 = Ti − Ta 2 = t1/2



ln 2 0.931 = k k

(10.99)

2. One-fourth response time: T − Ta = exp( − kt) Ti − Ta



Since

T − Ta 1 = Ti − Ta 4 t1/4 =



ln 4 k

(10.100)

Thus, knowing k, t1/2 and t1/4 can be determined.

Factors affecting the cooling rate or half response time Half response time or cooling rate is a measure of the rate of cooling of a product under a given set of conditions. All product and process components that affect the rate of heat transfer will affect the rate of cooling, These are product-related factors and system-related factors. The product-related factors are product type, shape and size, thermophysical properties, and packing and packaging of the product. System-related factors are type of cooling medium, flow pattern, and product load. Example 10.7 In a hydrocooler operating at 2°C, carrots are cooled from 22°C to 10°C. Calculate how long it will take to reach 4.5°C. Assume a half cooling time of 5 min.

Solution From the Equation (10.99) we can write



k=

ln 2 0.693 = =0.138 min−1 5 t1/ 2

And from Equation 10.98 we can write



 T −Ta   4.5−2  t 45 = −ln  / k t = −ln 22−2  / 0.138 = 0.91 min T − T    i a

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Heating and Cooling of Agro Products

10.6 FREEZING Freezing is the unit operation in which the temperature of a food is reduced to below freezing point and a proportion of water undergoes a change state to form ice crystals. Freezing injury occurs when ice crystals are formed in the tissues. Tissues injured by freezing generally lose rigidity and become mushy after thawing. In freezing, a temperature of −18°C or below is used. Freezing will preserve the perishable foods for months or even years. In general, freezing equipment can be categorized into three classes: (a) air-blast freezers, (b) plate freezers, and (c) immersion freezes. An air blast freezer uses cold air velocities, and products of high density are frozen in large packages. In plate freezers, food products or packages are placed in direct contact with plates that are maintained at a desired freezing temperature. The  desired temperature is achieved by design of a refrigeration system. The  concept of immersion freezers involves bringing the product into direct contact with low-temperature refrigerant. Liquid nitrogen is probably the most widely used refrigerant for immersion freezers. Two other types of refrigerants that are used are CO2 and Freon 12.

10.6.1  Freezing Point Depression A solution freezes at a lower temperature than does the pure solvent. This phenomenon is called freezing-point depression. The freezing-point depression of a solution is a colligative property of the solution that is dependent upon the number of dissolved particles in the solution. The higher the solute concentration, the greater the freezing-point depression of the solution. Freezing-point depression is the decrease of the freezing point of a solvent on the addition of a nonvolatile solute. Examples include salt in water, alcohol in water, or the mixing of two solids such as impurities into a finely powdered drug. Utilizing the definition of chemical potential, the following expression is obtained

GA0s − GA0l = ln X A R gTA

(10.101)

which expresses the difference in molar-free energies between the liquid and solid phases. Since free energy can be related to enthalpy by the following expression:

∂ G h = ∂T A  TA  TA

(10.102)

Equation (10.101) can be expressed in terms of the differences in enthalpy of the liquid and solid phases or the latent heat of fusion as illustrated by the following equation:

hA0 l − hA0 s λ′ d ln X A = = 2 R gTA R gTA 2 dTA

(10.103)

Equation (10.103) relates latent heat of fusion with mole fraction and temperature. By integration of Equation (10.103), the following equation is obtained:

1  λ′  1 −   = ln X A R g  TA 0 TA 

(10.104)

Where TA0 is the freezing point of pure liquid (a) and XA is the mole fraction of water in solution. Equation  (10.104) is in a form acceptable for computing freezing-point depression; however, the

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expression is normally simplified for dilute solutions. For these solutions, the negative logarithm of the mole fraction of the solvent (−ln XA) can be approximated by the mole fraction of the solute (XB). By expressing the mole fraction of solute in terms of molality (m), the commonly used expression for freezing-point depression in dilute solutions is obtained: ∆Tf =



R gTA0 2 m L1000

(10.105)

Finally, the freezing-point depression is given by ∆Tf = K f × m w

where K f =

R g TAo

(10.106)

2

L1000

Kf is the molal freezing-point depression constant and m is the molality of the solute. Example 10.8 Compute the temperature at which ice formation begins in an ice cream mix for molality of 1.003.

Solution ∆Tf =



R gTA02 m 8.315 × (273.15)2 × 1.003 = 1.87 °C = L1000 333 × 1000

Hence the freezing point is 100−1.87 = 98.13°C.

Example 10.9 What is the freezing-point depression caused by adding 31.65 g of sodium chloride to 220.0 g of water? Kf for water is 1.86°C⋅kg⋅mol−1.

Solution The formula for freezing-point depression expression is ∆Tf = iKf × mw



where: ΔTf is the freezing-point depression i is the van’t Hoff factor Kf is the molal freezing-point depression constant for the solvent m is the molality of the solution The molality of the NaCl is calculated as





moles of NaCl = 31.65g NaCl ×

1mol NaCl = 0.5416 mol NaCl 158.44 g NaCl

mass of water = 220.0 g H2 O ×

1 kg H2 O = 0.220 kg H2 O 1000 g H2 O

Heating and Cooling of Agro Products



= m

271

0.5416 mol moles of NaCl = = 2.46 mol/kg kilograms of water 0.220 kg

The  van’t Hoff factor, i, is the number of moles of particles obtained when 1  mol of a solute dissolves. One mole of solid NaCl gives two moles of dissolved particles: 1 mol of Na+ ions and 1 mol of Cl− ions. Thus, for NaCl, i = 2. ΔTf is given by ∆Tf = iKf × m = 2 × 1.86 × 2.46 = 9.16°C



The freezing-point depression is 9.16°C.

10.7 HEATING Heating and cooling of food products are foremost processes in food processing plants. Most of the processed foods receive some type of treatments starting from the entry into the food processing plants to the arrival at consumers of the processed foods. Heat transfer is the basic unit operation in heating and cooling and has been discussed earlier. Some of the applications of heating processes such as evaporation, heat treatment, and pasteurization are discussed below. Evaporation: The concentration of solids fraction in a liquid food product is accomplished by evaporation of the moisture. This is accomplished by raising the product temperature to the boiling point and holding for the time required to achieve the desired concentration. Four basic components are required to accomplish the evaporation and these are an evaporation vessel, a heat source, a condenser, and a method of maintaining vacuum. Figure  10.15 shows the schematic diagram of the components of a simple evaporation system. Thermal treatment: Generally, higher temperatures above the optimum are lethal to bacteria, yeasts, molds, and bacterial spores as well. Most of the bacteria are destroyed at a temperature between 80°C and 100°C, but many bacterial spores are not killed at the boiling point of water at 100°C for 30 min of heating. To ensure the total destruction of microorganisms including spores, wet heat treatment at a temperature of 121°C is to be conducted for 15 min or longer. Hence all microorganisms may be totally killed or their number can be diminished to any desired value by increasing the temperature of foods holding these at the required high temperature for the desired length of time. Most enzymes are also inactivated if these are subjected to about 85°C and above. So heat treatment can effectively prevent both microbial and enzymatic spoilage. But at the same time, the chemical reactions are accelerated at elevated temperatures. To minimize the chemical deteriorations, food materials should be heated just sufficiently

FIGURE 10.15  Schematic diagram of the components of a simple evaporation system.

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in absence of oxygen so that microbial, enzymatic, and oxidative spoilage are prevented. The  well-established methods of food preservation by canning and pasteurization are based on the heat treatment principle. Pasteurization: Product is subjected to a temperature that kills a great many but not all of the organisms present; the process is called pasteurization. The heating not only kills many organisms but also greatly weakens and delays the development of those not killed, which is an important factor in the keeping of pasteurized products. The term pasteurization often applied to the preservation of fruit juices by heat. In this case, however, it is very probable that all organisms capable of growing in the liquid are destroyed by the heat and hence preservation is usually permanent. Pasteurization of milk refers to the process of heating of every particle of milk to at least 62.78°C for 30 min or 71.66°C for 15 s or any other time temperature combination that is equally efficient in approved and properly operated equipment. The main objectives of pasteurization of milk are to render milk safe for human consumption by destruction of all pathogenic microorganisms and to improve the keeping quality of milk by destruction of almost all spoilage organisms (85%–99%). Canning: The process of sealing foodstuffs hermetically in containers and sterilizing them by heat is known as canning. In 1804, Appert in France invented the process of sealing foods hermetically in containers and sterilizing them by heat. In honor of the inventor, canning is also known as appertizing. Fruits and vegetables are canned in the seasons when raw materials are plenty. Canned products are sold in off-seasons and give better returns to growers. During the canning process, spoilage organisms are destroyed within sealed containers by means of heat. The canning of fruits and vegetables consists of preparation of the food, filling, exhausting the container, sealing, thermal processing, and cooling the container and the contents.

10.7.1  Boiling-Point Elevation The boiling point is the temperature at which a liquid starts to boil and transforms to vapor. The liquid begins to boil at a particular temperature when its vapor pressure becomes equal to atmospheric pressure. The  presence of solute lowers the vapor pressure and makes it necessary to heat at a higher temperature to boil the solution. The boiling point of water (universal solvent) is known to be 100°C, which states that water starts to turn to a vapor state, and its vapor pressure becomes equal to atmospheric pressure at 100°C. Phase change: In  the evaporation process, the phase change that occurs is evaporation of water from the liquid state to the steam or vapor state. The  latent heat of vaporization is a well-known value for pure water and can be described as a function of pressure by the Clausius-Clapeyron equation, and minor variations in the Clausius-Clapeyron equation lead to the following expression:

ln P =

Lv + C R gTA

(10.107)

This clearly expresses the direct relationship between the latent heat of vaporization (Lv) and the vapor pressure (P). In  the case of food products that contain solids and other components that influence the latent heat of vaporization, an expression similar to the Equation (10.103) can be written as:



ln P′ =

L′v + C R g TA

(10.108)

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Heating and Cooling of Agro Products

Which relates the latent heat of vaporization for a fluid food product (L′v ) to the vapor pressure (P′) at the same temperature (TA). By combining Equations (10.107) and (10.108) at equal temperatures, the following equation is obtained: ln P′ =



L′v ln P + C′ Lv

(10.109)

Which establishes the log-log relationship between the pressures of pure water and the liquid food product. By plotting the logarithm of vapor pressure data for the food product against the logarithm of pressure of pure water at various temperatures, the relationship between the latent heat of vaporization of the fluid food product and the latent heat of vaporization of water is established. Information of this type is necessary to account for the change in latent heat as the concentration of the fluid food product occurs during evaporation. Boiling-point elevation: Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent. An expression that describes the extent of the boiling-point elevation can be derived by considering phase equilibrium and chemical potential of the two phases that exist in equilibrium. This derivation is obtained by utilizing the same procedure as is used to derive Equation (10.99), which described the freezing-point depression as a function of molality. The  derivation is almost identical with equation (10.100), which can be restated for the case of boiling-point elevation as follows: λv Rg



 1 1  −   = −ln X A  TA 0 TA 

(10.110)

where λv equals the latent heat of vaporization, TA0 is the boiling point of pure water and XA is the mole fraction of water in solution. By assuming that the boiling-point elevation is small and utilizes only the first term of the logarithmic expansion, equation (10.106) can be reduced to the following form: ∆Tb =



R gTA0 2 X b Lv

(10.111)

where XB is the mole fraction of the solute causing the boiling-point elevation. An additional modification can be made by introducing molality into equation (10.107) to obtain the following equation: ∆Tb =



R gTA0 2 m L v 1000

(10.112)

Finally the freezing-point depression is given by ∆Tb = K b × m

where K f =

R g TAo2 L v 10009

Kf is the molal boiling-point elevation constant and m is the molality of the solute.

(10.113)

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The boiling point elevation is directly proportional to the molality of the solute particles in the solution and it is given by: ∆Tb = K b × m w



(10.114)

where: ∆Tb is the Amount by which the boiling point is elevated Kb is the molal boiling-point elevation constant that is solvent dependent m is the molality (moles of solute particles per kg of solution) Boiling point of solution is given by Boiling point = boiling point of the solvent + ∆Tb T solution = Tsolvent + ∆T Example 10.10 Determine the temperature at which skim milk begins to boil. Composition of skim milk is approximately of 5.1% lactose.

Solution



= Molality

Lactose in1000 g 51 = = 0.1417 molecular weight of lactose 360

The boiling-point temperature depression is ∆Tf =



8.315 × (373.15)2 × 0.1417 = 0.073°C 2226 × 1000

Hence the boiling point is 100.073°C.

Example 10.11 The boiling point of a solvent is 70°C, the values for Kb and m are 0.5°C/m and 0.015 m. Determine the boiling point of the solution.

Solution The parameters given are



= m 0.015 = and Kb 0.5oC /m Water is the solvent; therefore Tsolvent = 70° C

The boiling-point formula is given by



Tsolution = Tsolvent + ∆T



∆Tb = Kb × mw

275

Heating and Cooling of Agro Products ∆Tb = 0.5 × 0,915 = 0.0075°C



Tsolution = 70 + 0.0075 = 70.0075°C

Example 10.12 Determine the boiling point of the solution whose m = 0.0285 m and Kb = 0.612°C/m. Water is the solvent here.

Solution The given parameters are M = 0.0285m and Kb = 0.612°C /m

Since water is solvent,

Tsolvent = 100°C

The boiling-point formula is



Tsolution = Tsolvent + ∆T



∆Tb = Kb × mw



∆Tb = 0.612 × 0.0285 = 0.017442°C Tsolution = 100 + 0.017442



= 100.017442°C



KEY TO SYMBOLS A C C1, C2 E Eb F12, F21 G Gr Gol Gos I J Lv Lv′ Nu P P′ Pr Q

surface area, m2 constant constant emissive power emissive power of black body shape factor global radiation per unit area per unit time Grashof number free energy in system of liquid phase at standard state free energy in a system of solid phase at standard state intensity of black-body radiation radiosity or total radiation leaving per unit area per unit time latent heat of vaporization, kJ/kg latent heat of vaporization of fluid food product Nusselt number vapor pressure of pure water vapor pressure of liquid food product Prandtl number quantity of heat

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R thermal resistance Rg universal gas constant Re Reynolds number T temperature, °C or K TA temperature, K Ta air temperature, °C Tab absolute temperature Tf final temperature of the mixture, °C Ti inside temperature, °C T0 outside temperature, °C h enthalpy hi inside convective heat transfer coefficient ho outside convective heat transfer coefficient hol enthalpy of liquid phase at standard state hos enthalpy of solid phase at standard state hc heat transfer coefficient, W/m2 K q quantity of heat per unit area r radial distance from the source ri inside radius ro outside radius x distance t time α absorptance 𝜌 reflectance τ transmittance ε emissivity σ Stefan-Boltzman constant λ wavelength ϴi angle of repose ϕ solid angle μi viscosity

EXERCISES 10.1 Calculate the heat flux through a plane slab 0.1 m thick for surface temperatures of 120°C and 30°C. Assume thermal conductivity = 2.0 × 10−4 K−1. 10.2 Freon at a mean bulk temperature of −10°C flows through a 20 mm bore pipe at velocity of 0.20 m/s. The Freon is heated by a constant wall heat flux from the pipe, and the surface temperature is 15°C above the mean fluid temperature. Calculate the length of pipe for heat transfer rate of 1.5 kW. At temperature 10°C: υ = 0.221 × 10−5, K = 72.7 × 10−6 kW/(mK), Pr = 4.0

N u = 0.023(Re)0.8Pr n

10.3 A tubular heater of the counterflow type is used to heat fuel oil of specific heat 3.14 kJ/kgK from 10°C to 26.7°C. Heat supplied by means of 1.51 kg/s of water enters the heater at 82°C. Determine the necessary surface area if the rate of heat transfer is 1.135 KW/m2K. 10.4 In a hydrocooler operating at 5°C, carrots are cooled from 25° to 10°C. Calculate the half response time of cooling for a cooling rate coefficient of 0.135 min−1 and time to reach 10°C.

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Heating and Cooling of Agro Products

10.5 Compute at which point ice formation begins of an ice cream mix of 10% butterfat, 12% solid-nonfat, 15% sucrose, and 0.22% stabilizer. Hints: Fraction solute = 0.15 + 0.12 × 0.545 = 0.2154 Fraction solvent = [1−(0.10 + 0.15 + 0.12 + 0.002)] = 0.6278





Fraction solute 0.2154 g solute = = 0.3431 Fraction solvent 0.6278 g solvent Molality =

g solute 0.3431 = 1.002 g solvent 0.342

BIBLIOGRAPHY Chakraverty, A. and De, D. S. 1980. Postharvest Technology of Cereals and Pulses. Oxford and IBH Publishing Co, New Delhi, India. Chakraverty, A. and Singh, R. P. 2016. Postharvest Technology and Food Process Engineering. CRC Press, Boca Raton, FL. Ditus, F. W. and Boelter, L. M. K. 1930. Heat Transfer in Automobile Radiators of the Tubular Type. Publications in Engineering. University of California, Berkeley, Vol. 2, 443. Fujii, T. and Imura, H. 1972. Natural convection heat transfer from plate with arbitrary inclination. Journal of Heat and Mass Transfer, 15: 750. Heldman, D. R. 1977. Food Process Engineering, 2nd edition. AVI Publishing Company, Westport, CT. Hollands, K. G. T., Unny, T. E., Raithby, G. D. and Konicek, L. 1976. Free convection heat transfer across inclined air layers. Journal of Heat Transfer. 98: 189. Holman, J. P. 1876. Heat Transfer. McGraw-Hill Kogakusha Ltd., Tokyo. Ozisik, M. N. 1972. Analysis of Heat and Mass Transfer. McGraw-Hill, New York. Ozisik, M. N. 1983. Heat Transfer: A Basic Approach. McGraw-Hill, New York. Sieder, E. N. and Tate, C. F. 1936. Heat transfer and pressure drop of liquids in tubes. Industrial Engineering Chemistry. 28: 1429–1435. Simonson, J. R. 1981. Engineering Heat Transfer. The Macmillan Press Ltd., London, UK. Sparrow, E. M., Ramsey, J. W. and Mass, E. A. 1979. Effect of finite width on heat transfer and fluid flow about an inclined rectangular plate. Journal of Heat Transfer. 101: 2.

11

Refrigeration and Cold Storage

This chapter presents principles of refrigeration, vapor compression refrigeration cycles, P-h charts, properties of refrigerants and applied psychrometric charts, computation of refrigerating effect, coefficient of performance, and compressor power requirement. It  also presents computation of sensible heat ratio, room sensible heat factor, bypass factor, cooling requirements, refrigeration requirements of air-conditioning and cold storage, design of air-conditioning of office buildings, and design of cold storage systems for potatoes and vegetables.

11.1 INTRODUCTION Refrigeration systems are used for cooling and dehumidifying purposes. Refrigeration is defined as the process of removing heat from a body that is below the temperature of its surroundings. Or refrigeration is defined as the transfer of heat from lower to higher temperature. There are two types of refrigeration: (a) absorption refrigeration and (b) vapor compression refrigeration. Agro products such as fresh fruits and vegetables are brought to the temperature for safe storage and transportation using refrigeration and air-conditioning. The primary function of cooling and precooling is lowering product temperature to respiration, transpiration, microbiological activity, enzymatic activity, and chemical activity, resulting in improved product quality and reduced heat load in storage. It is also important to know the temperature to be cooled and refrigerating capacity needed for storage and transportation. Temperate produce should be cooled to 0.2°C while tropical and subtropical produce should be cooled to 10°C–15°C. Cold storages are buildings designed for storage of commodities, particularly food materials of perishable nature in well-defined conditions of temperature and relative humidity. These well-defined temperatures and relative humidities for short-term and long-term storage of food materials, particularly fruits and vegetables, are prescribed by the researchers and professional organizations. Cold storages can be classified into (1) multipurpose cold storage systems for short- as well as long-term storage of different agro products and (2) specialized cold storage systems for single agro-product type. The heat removal resulting in a temperature change is called sensible heat of the product and is expressed as:

Q = mC p∆T

(11.1)

where: Q = heat removed, kJ m = mass of the product, kg Cp = specific heat of the product, kJ/kg°K ∆T = temperature difference, °C Another form of heat is called latent heat, which does not result in temperature change but causes the change of state and can be expressed as:

Q = m × L

(11.2)

where m = mass of the product that changes state, kg L = latent heat of fusion, kJ/kg 279

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11.2  VAPOR COMPRESSION REFRIGERATION CYCLE The liquid refrigerant in the receiver or supply tank is under high pressure. Because of the high pressure, the liquid is forced through the expansion valve into the region of low pressure by the compressor. The liquid refrigerant evaporates or boils to vapor in the evaporator. The heat required for evaporation comes from the surroundings and cooling results. The vapor moves at low p­ ressure through the vapor line to the compressor and is compressed to a high pressure as it passes to compressor. Here it returns to liquid state as the latent heat is transferred to the surroundings. The liquid then flows into the receiver. While designing a refrigeration system, it is customary to increase the capacity by a factor of 1.25 for safety; the refrigeration capacity is expressed in tons of refrigeration, and the temperature of the product should be reduced to the desired level within 75% of operating time.

11.3  PRESSURE–ENTHALPY (p-h) CHART A Pressure–Enthalpy (p-h) Chart is commonly used to analyze the refrigeration cycle. The condition of the refrigerant in any thermodynamic state can be represented as a point on the p-h chart. The condition of the refrigerant in any particular thermodynamic state can be located if any two of the properties are known, and the other properties of the refrigerant of that state can be determined directly from the chart to study the performance of the machines. The  p-h chart is divided into three areas separate from each other by the saturated liquid and saturated vapor lines. The region on the p-h chart to the left of the saturated liquid line is called the subcooled region, and the region on p-h chart to the right of the saturated vapor line is called the superheated region. The section between the saturated liquid line and the saturated vapor is the two-phase region. The horizontal lines extending across the chart are lines of constant pressure, and the vertical lines are lines of constant enthalpy. Thermodynamically, the process can be shown explicitly by the p-h  chart (Mollier chart). The vapor is drawn in the compressor cylinder during its suction stroke, and during the compression stroke the vapor is compressed isentropically to pressure p2 and temperature T3 and delivered out from the compressor. This point is represented by 3. State 3 shows the vapor in the superheated state. The  vapor at condition 3 passes into the condenser in which cooling water is supplied to remove heat from the vapor. The vapor is first cooled to the saturation temperature at pressure p2, and further removal of heat condenses it to liquid and removes its latent heat until point 4 is reached. Thus, in order to carry out this operation, the saturation temperature corresponding to pressure p2 should be sufficiently higher than the temperature of cooling water for efficient transfer. It may then be possible to even subcool the liquid vapor to a temperature below that at 4. The highpressure liquid is expanded through the throttle valve, and the liquid at 4 throttles to lower pressure p1; the condition achieved after a constant expansion is shown at 1. After ­throttling, the liquid partially evaporates at lower temperature T1 and lower pressure p1. Thus, after the throttle valve, the very wet vapor achieves a very low temperature. These vapors now pass the evaporator coils immersed in brine or to the chamber to be refrigerated. These absorb latent heat in brine in further evaporating itself. The vapor pressure may reach condition 2, i.e., dry saturated at p1. This completes the cycle, which is also called the simple saturated cycle. Heat extracted or net refrigerating effect is given by

R n = h 2 − h1

(11.3)

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Refrigeration and Cold Storage

And work done is given by W = h 3 − h 2



(11.4)

And coefficient of performance is given by COP =



R n h 2 − h1 = W h3 − h2

(11.5)

11.3.1 Unit of Refrigeration Refrigerating systems and their components are rated on the basis of ton. Refrigeration effect is compared with production of ice. A ton of refrigeration is defined as the refrigeration effect needed to melt down 1 ton of ice in 24 h and is given by



A ton of refrigeration =

336 × 1000 = 140,00 kJ/h 24

(11.6)

11.4 REFRIGERANTS A refrigerant is a cooling agent that absorbs heat during expansion and gives off heat during condensation. A broader definition may include secondary cooling agents such as brine or water. Major refrigerants are ammonia, Freon group, and methyl chloride.

11.4.1  Desirable Characteristics of Refrigerants

1. It should be nonpoisonous. 2. It should be noncorrosive. 3. It should be non-inflammable. 4. It should be nonexplosive. 5. It should be nontoxic. 6. It should be easy to detect the leaks. 7. It should be easy to locate the leaks. 8. It should have a low boiling point. 9. The refrigerant should be stable gas. 10. Moving parts in the refrigerant should be easy to lubricate. 11. Minimum difference between vaporizing and condensing pressure is desirable. 12. It should have a small relative displacement to obtain refrigerating effect. 13. It should have a well-balanced enthalpy for evaporation.

The standard comparison of refrigerants is based on an evaporating temperature of −15°C and a condensing temperature of +30°C. Also, the refrigerants are identified by numbers standardized by the American Society of Heating, Refrigerating and Air Conditioning Engineers (ASHRAE). Some common refrigerants in use are shown in Table 11.1.

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TABLE 11.1 Some Common Refrigerants Refrigerant No. R-11 R-12 R-22 R-500

Refrigerant Name

Chemical Formula of the Refrigerant

Trichloromonofluoromethane Dichlorodifluoromethane Monochlorodifluoromethane Azeotropic mixture of 73.8% (R-12) and 26.2% (R-152a) Azeotropic mixture of 48.8% (R-22) and 51.2% (R-115) Ammonia Tetrafluoromethane

R-502 R-717 R-134a

CCl3F CCl2F2 CHClF2

NH3 CH2FClF3

Example 11.1 A vapor compression cycle using Refrigerant 22 (R22) operates at a condensing temperature of 34°C and evaporative temperature of −10°C as shown in Figure 11.1. For a mass flow rate of the refrigerant of 0.45 kg/s, find the following:

1. The compressor power input 2. The refrigerating effect 3. The coefficient of performance (C.O.P)

Solution From the p-h diagram in Figure 11.1 and Tables of R22, we have



= h2 401 = .07 kJ / kg h3 427 kJ / kg

h4 = 241.84 kJ / kg



Compressor work W = m (h3 − h2 ) = 0.45 × ( 427 − 401.07 ) = 11.67 kW



Refrigerating capacity = m (h2 − h1 ) = 0.45 × ( 401.07 − 241.84 ) = 71.82



Coefficient of performance, COP = RE/W = 71.82/11.67 = 6.11

FIGURE 11.1  p-h diagram of the Example 11.1.

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FIGURE 11.2  p-h diagram of the Example 11.2.

Example 11.2 A  vapor compression cycle using R22 operates at a condensing temperature of 36°C and ­evaporative temperature of −16°C as shown in Figure 11.2. For the system capacity of 60 kW, find the following:

a. The mass flow rate b. The compressor power c. The refrigerating effect d. The coefficient of performance (C.O P)

Solution From the p-h diagram in Figure 11.2 and Tables of R22, we have



= h2 398.64 = kJ/kg h3 428 kJ/kg

h4 = 244.44 kJ/kg

Refrigerating capacity = m (h2 − h1 ) = m × (398.64 − 244.44 ) = 60 kW Therefore, m = 60/(398.64–244.44) = 0.39 kg/s



Compressor work W = m (h3 − h2 ) = 0.39 × ( 428 − 398.64 ) = 7.16 kW



Coefficient of performance, COP = RE/W = 60/7.16 = 8.38

Example 11.3 An ice plant operates in the ideal vapor compression cycle using refrigerant R134a. The refrigerant enters the compressor as saturated vapor at 0.15 MPa and leaves the condenser as saturated liquid at. 0.7  MPa. Water enters the machine at 30°C and leaves as ice at −5°C. For  an ice production rate of 10 kg per hour, determine the power input to the ice plant. The specific heat of ice and water are 2.1 and 4.18 kJ/kg°K, respectively, and the latent heat of fusion of ice is 344 kJ/kg.

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Solution Heat needed to be removed per kg of water to convert it into ice is Q = mcwater × ( t water − 0 ) + mLf + mcice × ( 0 − t ice )

Hence we have

Q = 10 × 4.18 × (30 − 0 ) + 10 × 344 + 10 × 2.10 × ( 0 − ( −5) ) = 4699 kJ



Therefore, the refrigerating capacity required is RE = 4699/3600 = 1.30 kW



From the p-h diagram in Tables of R134a, we have = h2 389.11 = kJ/kg h3 425 kJ/kg



h4 = 236 kJ/kg

Refrigerating capacity = m (h2 − h1 ) = m × (389.11− 2326 ) = 1.30 kW



Therefore, m = 1.30/(389.11–236) = 0.0084 kg/s Compressor work W = m (h3 − h2 ) = 0.0084 × ( 425 − 389.11) = 0.30 kW



Coefficient of performance, COP = RE/W = 1.30/0.30 = 4.31

Example 11.4 In a simple vapor compression cycle, the following are of the refrigerant R22 at various points: Compressor inlet:       h2 = 183.2 kJ/kg      v2 = 0.0767 m3/kg Compressor discharge:      h3 = 222.6 kJ/kg      v3 = 0.00164 m3/kg Condenser exit:         hf4 = 84.9 kJ/kg        v4 = 0.00083 m3/kg The  piston displacement volume for compressor is 1.5  liters per stroke and its volumetric efficiency is 80%. The speed of the compressor is 1600 rpm. Find:

1. Power rating of the compressor 2. Refrigerating effect

Solution From the p-h diagram Tables of R22, we have Piston displacement volume = 1.5 × 1000 × 10−6 m3/stroke = 0.0015 m3/revolution Power rating of the compressor is



Compressor discharge q = 0.0015 × 1600 × 0.8 = 1.92 m3 /min



Mass flow rate of compressor discharge m = Compressor discharge q / specific

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= volume 1.92/0.0767 = 25.03 kg3 /min Power rating of the compressor = m (h3 − h2 ) = 25.03 × ( 222.6 − 183.2) /60



= 16.44 kW



Therefore, the refrigerating capacity required is RE = m (h2 − h1 ) = 25.03 × (183.2 − 84.9) /60 = 41 kW

Example 11.5

A two-stage ammonia system designed for a fish processing plant uses flash gas removal and intercooling operation (Figure 11.3). The condensing temperature is 35°C. The saturation temperature of the high-temperature evaporator is 0°C and its capacity is 90 kW. The saturation temperature of the low-temperature evaporator is −4°C and its capacity is 170 of refrigeration. Draw the cycle on the p-h diagram and determine the following:

1. The rate of refrigeration compressed by the high-stage compressor 2. The power required by the compressors 3. The coefficient of performance

FIGURE 11.3  Two-stage ammonia system designed for a fish processing plant.

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FIGURE 11.4  The cycle on the p-h diagram of the two-stage ammonia system designed for a fish processing plant.

Solution The cycle on the p-h diagram of the two-stage ammonia system designed for a fish processing plant is shown in Figure 11.4. Given



= h1 1408.41 = kJ/kg h2 1655 kJ/kg



= h3 1461.70 = kJ/kg h4 1626 kJ/kg



= h5 366.07 = kJ/kg h6 m1 =

h= h= 200 kJ/kg 7 8

170 170 = = 0.1407 kg/s h1 − h 8 1408.41− 200



m = m= m2 = m8 1 7



m3 = m5 Heat and mass balance equations  about both the high-temperature evaporator and the intercooler are:



m5h5 + 90 + m2h2 = m3h3 + m7h7



m= m = 0.1407 kg /s 2 7



m5 × 366.07 + 90 + 0.1407 × 1655 = m3 × 1461.70 + 0.1407 × 200



m3 (1461.70 − 366.07) = 90 + 0.1407(1655 − 200)



m3 × 1095.63 = 294.72 m3 = 0.269 kg/s



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Refrigeration and Cold Storage Power rating of the compressors:



Low − stage power = 0.1407(h2 − h1) = 0.1407(1655 − 1408.41) = 34.69 kW



High − stage power = 0.269(h4 − h3 ) = 0.269(1626 − 1461.70) = 44.20 kW



Total power = 34.69 + 44.20 = 78.89 kW



COP =

Total cooling produced 90 + 170 = = 3.30 Win 78.89

11.5  CONSTRUCTION OF PSYCHROMETRIC CHART The psychrometric chart is the graphical representation of the thermodynamic properties of air on a chart. The psychrometric chart uses dry-bulb temperature as the abscissa and specific humidity or humidity as the ordinate, and the following moist air properties are plotted:

1. Vapor pressure 2. Relative humidity 3. Wet-bulb temperature 4. Specific volume 5. Enthalpy

Two skeleton psychrometric charts—ASHRAE and Carrier—are most widely used in the design of refrigeration and cold storage systems. The deviations between the two psychrometric charts are the ways enthalpy deviations are calculated and sensible heat ratios are constructed. The thermodynamic state of the moist are can be plotted on these charts if any two of these properties are known and other properties can be evaluated. Figure 11.5 shows a sample psychrometric chart displaying the different psychrometric properties available in this chart and which can be used for thermodynamic analyses.

FIGURE 11.5  Sample psychrometric chart.

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Example 11.6 Air at 27°C and 50% relative humidity is cooled and humidified to 10°C and 100% relative humidity. How much sensible heat and latent heat is removed from 500 liters/s of air? What is the refrigerating capacity needed? Show cooling and humidification process on a skeleton psychrometric chart.

Solution When the process is plotted on the psychrometric chart we can compute the sensible heat and latent heat from the psychrometric chart



Sensible heat removal = m × (ho − hc ) = ( 0.5/0.86 ) × ( 46.6 − 29.3) = 10.05 kW



Latent heat removal = m × m (ha − ho ) = ( 0.5/0.86 ) × (55.6 − 46.6 ) = 5.23 kW



Refrigerating capacity needed = 10.05 + 5.23 = 15.28 kW

Example 11.7 The sensible heat gain of a room is 4.8 kW and its latent heat gain is 1.4 kW. A conditioned air supply of 0.50 m3/s is to be delivered in the room. If the room is to be maintained at 25°C, find the relative humidity that will result in the conditioned room if the air supply is 17°C and 90% relative humidity.

Solution The sensible heat ratio is given by SHR =



4.8 = 0.77 4.8 + 1.4

Since the air entering inside the room is leaving the cooling coil, the air leaving the cooling coil is at 17°C and 90% relative humidity. So, the point L is located at 17°C and 90% relative humidity. A line parallel to the SHR line of 0.77 is drawn through L. The room condition will be the intersection of this line with the 60% relative humidity line and is indicated by O.

Example 11.8 The design conditions for a summer air-conditioning system for a small office building are as follows: Outside design conditions: 34°C, 25°C Inside design conditions: 25°C, 85% rh Room sensible heat: 75 kW Sensible heat from outside air: 26 kW Room latent heat: 26 kW Latent heat from outside air: 30 kW Outside air to meet ventilation requirement: 1.5 m3/s A direct expansion Freon—22 coil with bypass factor of 0.12 is used. Find the following:

1. Sensible heat ratio 2. The room apparatus dew point 3. The temperature of the air leaving the coil

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4. The coil apparatus dew point 5. Refrigeration capacity required

Solution

1. The sensible heat ratio is given by SHR =



75 = 0.743 75 + 26

2. The outside and inside room conditions are indicated at the points O and A, respectively. A line parallel to the SHR line of 0.77 is drawn through A and the point it intersects the saturation line is the room ADP temperature. It is indicated by R at ADP = 12.9°C. 3. The bypass factor (BPF) is defined as



BPF =

Tla − TADP Tr − TADP

Tla = Tr − (1− BPF)(Tr − TADP )



= 25 − (1− 0.12)(25 − 12.9) = 14.4°C



4. The temperature of the mixed air is indicated by the point C and is at 23.24°C. The line through the points C and L intersecting the saturation line at X is the coil apparatus dew point and is at 11.8°C. 5. The refrigerating capacity = total sensible heat load + total latent heat load = (75 + 26 ) + ( 26 + 30 ) = 157 kW

Example 11.9

The design conditions for a cold storage for storing 450 tons of vegetables are as follows: Outside design conditions: 36°C dry-bulb temperature, 25°C wet-bulb temperature Inside design conditions: 19°C dry-bulb temperature, 60% rh Total sensible heat gain: 21.72 kW Total latent heat gain: 26.06 kW Infiltrated air: 180 m3/h Fresh air supplied: 4500 m3/h If the air-conditioning is achieved by first cooling and dehumidifying and then heating, and the temperature of air entering the room is not to exceed 16°C, determine the following:

1. Amount of recirculated air if the recirculated air is mixed with fresh air before entering the cooling coil 2. Capacity of the cooling coil in tons of refrigeration and its bypass factor if dew point temperature of the coil is 6°C 3. Capacity of the cooling coil in kW

Solution The various points are plotted on the psychrometric chart (Figure 11.6). Point 1 indicates the outside air condition (36°C dry-bulb temperature, 25°C wet-bulb temperature) and point 2 indicates the inside design (19°C dry-bulb temperature, 60% rh). Point 3 is the mixed air condition of these airstreams, and point A is located by drawing vertical and horizontal lines through points 1 and 2, respectively.

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FIGURE 11.6  The cooling and dehumidifying and then heating process on the psychrometric chart. Room sensible heat factor is given by RSHF =



RSH 21.72 = = 0.454 RSH + RLH 21.72 + 26.06

Draw a line parallel to a-b through point 2 intersecting the vertical line at point 5 drawn through 16°C dry-bulb temperature, and point 5 represents the condition of the air entering the room. From the psychrometric chart, we have h5 = 33.1 kJ/kg                h2 = 39.2 kJ/kg                v1 = 0.9046 m3/kg Total mass of air passing through the room is given by ma =

=

Total sensible heat + Total latent heat Total room heat = Total heat removed (h2 − h5 ) 21.72 + 26.06 = 7.83 kg/s = 28,188 kg/h (39.3 − 33.1)

Mass of fresh air supplied from outside is given by



VF 4500 m = m= = = 4974.6 kg/h 1 F v1 0.9046 Mass of recirculated air is given by



m2 = mre = ma − mF = 28188 − 4974.6 = 23,213.4 kg/hm



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Refrigeration and Cold Storage Percentage of recirculated air is given by



mre % =

mre 23,213.4 × 100 = × 100 = 82.35% ma 28,188

Point 3 is located computing H3 from the following relation: m1 H3 − H2 = m2 H1 − H3



Draw a line joining the point 3 and APD = 16°C intersecting the constant specific volume line through point 5 at point 4. The lines 4–5 represent the heating of the air. From the psychrometric chart, we have



= h3 48.5 = kJ/kg h4 26.22 kJ/kg Cooling capacity of the coil is given by Qco =



ma (h3 − h4 ) 28,188(48.5 − 26.2) = = 44.9 tons 14,000 14,000

Heating capacity of the coil is given by



Qhe = ma (h5 − h4 ) =

28,188 (33.1 − 26.2) = 54.03 kW 3600

11.6  MOISTURE CONTROL AND STORAGE OF VEGETABLES CROPS Vegetable crops, in contrast to cereal crops, may require storage conditions that minimize moisture loss rather than drying. Vegetable crops such as onion require controlled drying or curing whereas potatoes, sugar beets, and other root crops require storage environment that minimizes moisture loss.

11.6.1  Potatoes Potato storage must be designed to satisfy at least three criteria: adequate volume, adequate strength, and environmental requirements. 11.6.1.1  Adequate Volume The required storage volume can be calculated from the total potato yield and bulk volume approximately 1.6  m3/t. For  example, 1000  tons of potato would require a storage volume of 1600  m3 (1.6 × 1000). 11.6.1.2  Adequate Strength Lateral wall pressure can be expressed as a fraction of fluid pressure. At any depth h from the top, the fluid pressure is defined as where: P = fluid pressure, kN w = specific weight, kN/m3 h = height, m

P = wh

(11.7)

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Thus, the lateral pressure PL due to the same depth of potato can be expressed as PL = k L wh

(11.8)



PL = 0.3 k L wh

(11.9)



( H/w ≤ 1.0 )

where k L = lateral pressure coefficient

1. For wide bins

where: H = height of the bin, m w = width of the bin, m 2. For narrow bins

PL = 0.2 k L wh



( H/w > 1.0 )

(11.10)

Total lateral force on the bin wall is obtained by integrating equation. For the bin, the lateral wall load per unit wall width is

PL =

∫

H

0

0.3 k L wh dh =0.15wH 2

(11.11)

For example, a 6 m high pile of potatoes (specific weight, w = 6.6 kN/m3) exerts a lateral force of 35.6 kN (0.15 × 6.6 × 62) per wall width. 11.6.1.3  Storage Environment Storage conditions around the stored potatoes affect both potential duration of the storage period and suitability of the tubers for desired end use. Environments that are suitable for one period storage and one end use are not necessarily suitable for another end use. 11.6.1.4  Suberization Period If the potatoes are to be stored for a period of greater than 1 month, a curing or healing period is recommended. The storage environment during suberization period should have a relative humidity above 90% (preferably 95%) to facilitate wound healing. The recommended temperatures are between 10.0°C and 15.6°C and the durations are 10 days to 2 months. 11.6.1.5  Short-Term Storage Potatoes placed in storage for a short period (up to 10 weeks) are stored at temperatures between 7.2°C and 10.0°C, and the recommended relative humidity is 90%. 11.6.1.6  Long-Term Storage Potatoes that are stored for a long period (over 10 weeks) should have temperatures between 4.4°C and 5.50°C. Seed potatoes may be stored at 3.3°C for long periods. Potatoes that are to be processed commercially should be stored at 7.2°C if they are fried and at 5.50°C if they are dehydrated. The recommended relative humidity is 90%.

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11.7  COOLING REQUIREMENT The cooling requirement is equal to heat gain during storage.

1. Field heat of the potatoes 2. Heat of respiration 3. Heat conducted through the walls and roofs 4. Heat from infiltration 5. Miscellaneous heat production Field heat is heat contained in potatoes that must be removed to lower tuber temperature to the desired level. The field heat of potatoes is calculated as

qF =



m C p (Ts − TF ) t

(11.12)

where: qF = field heat, kJ/h m = mass of the product, kg Ts = storage temperature, °C TF = field temperature, °C Cp = specific heat of potato, kJ/kg°C Consider 100 t (1000 × 100 kg) of potatoes at 15.0°C are placed into storage each day. They are to be cooled at 10.0°C in one day. The specific heat of potato is 3430 kJ/kg°C. The rate of field heat removed is

qF =

100,000 × 3430 × (10 − 15) = 19,800 W 24 × 60 × 60

Heat of respiration depends greatly on tuber temperature and time in storage as shown in Figure 11.7. For example, a 100,000 kg lot of potato that has been stored for one day and presently has a temperature of 10°C will generate heat of respiration of (100,000 × 0.055) 5500 W. Thus, heat of respiration must be determined separately for each lot of potatoes placed in storage due to their different periods of storage and temperature. Total cooling due to respiration is the sum of the respiration heats generated by all of the lots. Conduction transfer through walls and roofs of the storage depends on the temperature inside and outside of storage and the solar radiation. The  heat conducted through the walls and roofs is calculated using the following equation



qc = A

ETD Rk

where: qc = rate of heat conduction, W R k = thermal resistance, °C-m2/W A = surface area, m2 ETD = equivalent temperature difference, °C

(11.13)

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FIGURE 11.7  Heat of respiration for potatoes as a function of temperature and storage time.

Infiltration heat gain is the heat gain through the infiltration of outside air. Infiltration heat gain can be calculated from the following formula: qi =



m(h 0 − h i ) t

(11.14)

where: qi = rate of heat conduction, kW m = mass exchange of air, kg/s ho = enthalpy of the outside air, kJ/kg hi = enthalpy of the inside air, kJ/kg t = time, s The amount of air exchange between the inside and outside air is difficult to determine but may be estimated to be one-half of an air exchange per hour, and enthalpy is computed from the psychrometric chart. Miscellaneous heat gain is due to equipment, people working, lights, and other sources of heat inside the storage. These heats can be calculated by the estimates of their power consumption or by published data (ASHRAE). Example 11.10 A cold storage is designed to store 500 tons of potatoes for the following data: Outside design conditions: 35°C dry-bulb temperature, 28°C wet-bulb temperature Inside design conditions: 20°C dry-bulb temperature, 60% rh Water content in potatoes: 78% Loss of water content from potatoes: 0.01% per h People working inside the storage: 25 Fresh air supplied: 4400 m3/h Sensible heat gain through glass: 22,000 kJ/h Sensible heat gain through walls and ceilings: 45,000 kJ/h

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Refrigeration and Cold Storage Equipment and reaction heat of potatoes: 14,500 kJ/h Infiltrated air: 200 m3/h

If the air-conditioning is achieved by first cooling and dehumidifying and then heating, determine the following:

1. Amount of recirculated air in the cold storage if the recirculated air is mixed with fresh air before entering the cooling coil 2. Capacity of the cooling coil in tons of refrigeration and its bypass factor if apparatus dew point temperature of the coil is 5°C 3. Capacity of the cooling coil in kW

Temperature of the air entering the storage should not exceed 15°C.

Solution The outside design condition and the inside design condition on are plotted on the psychrometric chart. Point a indicates the outside air condition (35°C dry-bulb temperature, 28°C wet-bulb temperature), and point c indicates the inside design (20°C dry-bulb temperature, 60% rh). Point g is located by drawing vertical and horizontal lines through points a and c, respectively. From the psychrometric chart, we have = ha 90 = kJ/kg hc 43 = kJ/kg ha 90 = kJ/kg via 0.91 m3 /kg

Mass of infiltrated= air

200 = 219.8 kg/h 0.91

Sensible heat load from infiltrated air is given by



ISH = mia (hg − hc ) = 219.8(59 − 43) = 3517 kJ/h Latent heat load from infiltrated air is given by



ILH = mia (ha − hg ) = 219.8(90 − 59) = 6813 kJ/h Sensible and latent heat load from people working in the storage is given by



Qs = 25 × 120 = 3000 kJ/kg



Ql = 25 × 350 = 8750 kJ/h Water loss from potato is given by



Wl = 500 × 1000 × 0.78 ×

0.01 = 39 kg/h 100

Heat carried by the air as latent heat load from potato due to evaporation of moisture is given by



Qw = water loss × latent heat = 39 × 2473 = 96,447 kJ/h Total sensible heat (TSH) = Sensible heat through wall + Sensible heat through glass + Sensible heat due to infiltrated air + Sensible heat for human + Sensible heat for equipment and is given by

296



Agro-Product Processing Technology TSH = 45,000 + 22,000 + 3517 + 3000 + 14,500 = 88,017 kJ/h

Total latent heat (TLSH) = Latent heat due to infiltrated air + Latent heat for human + Evaporative load and is given by



TLH = 6813 + 8750 + 96447 = 1,12,2010 kJ/h Sensible heat ratio (SHR) is given by



SHR =

TSHT TSH + LSH

=

88,017 = 0.44 88,017 + 112,010

The various points are plotted on the psychrometric chart. Point a indicates the outside air condition (35°C dry-bulb temperature, 28°C wet-bulb temperature), and point b indicates the inside design (20°C dry-bulb temperature, 60% rh). Point c is the mixed-air condition of these airstreams. Draw a line parallel to the SHR line through point b intersecting the vertical line at point e drawn through 15°C dry-bulb temperature, and point e represents the condition of the air entering the storage. From the psychrometric chart, we have

= he = 32 kJ/kg hb

43 kJ/kg

v sa = 0.91 m3 /kg

Mass of the air passing through the storage is given by



msta =

TSH + TLH 88017 + 112010 = = 18,184 kg/h (hb − he ) 43 − 32

Mass of the fresh air is given by



Mass of freshair =

4400 4400 = = 4835 kg/h v sa 0.91

Mass of the recirculated air is given by



Mass of recirculated air = 18184 − 4835 = 13,339 kg/h Hence the percentage of the fresh air is given by



Percentage of fresh air =

4835 × 100 = 26% 18,184

and the percentage of the fresh air is given by



Percentage of recirculated air = 100 − 26 = 74% The mixing point c is located from the ratio of distance bc: ab = −0.26: 1 and this point is at 27°C dry-bulb temperature and 66% relative humidity. Also hc = 59 kJ/kg Now draw a line through point c and ADP = 5°C (point g) resulting the conditioner line; next draw a horizontal line through e, cutting the cg line at f. From the psychrometric chart, we have



hf = 27 kJ/k

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Refrigeration and Cold Storage Cooling capacity of the coil is given by Qco =



msta (hc − hf ) = 46.2 tons 3600 × 3.5

The bypass factor of the cooling coil is given by



= B

distance gf = 0.1 distance gc

Heating capacity of the coil is given by



Qhe =

msta (he − hf ) 18184 = (32 − 27) = 25.25 kW 3600 3600

EXERCISES 11.1 A  vapor compression cycle using Refrigerant 22 operates at a condensing temperature of 40°C and evaporative temperature of −9°C. For a mass flow rate of the refrigerant of 0.35 kg/s, find the following: a. The compressor power input b. The refrigerating effect c. The coefficient of performance (COP) 11.2 An ice plant operates in the ideal vapor compression cycle using Refrigerant R134a. The refrigerant enters the compressor as saturated vapor at 0.15 MPa and leaves the condenser as saturated liquid at. 0.7 MPa. Water enters the machine at 25°C and leaves as ice at −5°C. For an ice production rate of 25 kg per hour, determine the power input to the ice plant. The specific heat of ice and water are 2.1 and 4.18 kJ/kg-K, respectively, and the latent heat of fusion of ice is 344 kJ/kg. 11.3 The design conditions for a summer air-conditioning system for a small office building are as follows: Outside design conditions: 304°C, 25°C Inside design conditions: 21°C, 85% rh Room sensible heat: 70 kW Sensible heat from outside air: 20 kW Room latent heat: 25 kW Latent heat from outside air: 25 kW Outside air to meet ventilation requirement: 1.5 m3/s A direct expansion Freon—22 coil with bypass factor of 0.12 is used. Find the following: a. Sensible heat ratio b. The room apparatus dew point c. The temperature of the air leaving the coil d. The coil apparatus dew point e. Refrigeration capacity required 11.4 The design conditions for a cold storage for storing 450 tons of vegetables are as follows: Outside design conditions: 35°C dry-bulb temperature, 23°C wet-bulb temperature Inside design conditions: 20°C dry-bulb temperature, 60% rh Total sensible heat gain: 25 kW Total latent heat gain: 29 kW Infiltrated air: 175 m3/h

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Fresh air supplied: 4000 m3/h If the air-conditioning is achieved by first cooling and dehumidifying and then heating, and the air entering the room is not to exceed 15°C, determine the following: a. Amount of recirculated air if the recirculated air is mixed with fresh air before entering the cooling coil b. Capacity of the cooling coil in tons of refrigeration and its bypass factor if the dew point temperature of the coil is 5°C c. Capacity of the cooling coil in kW 11.5 A cold storage is designed to store 500 tons of potatoes for the following data: Outside design conditions: 38°C dry-bulb temperature, 28°C wet-bulb temperature Inside design conditions: 19°C dry-bulb temperature, 60% rh Water content in potatoes: 78% Loss water content from potatoes: 0.01% per h People working inside the storage: 20 Fresh air supplied: 4000 m3/h Sensible heat gain through glass: 22,000 kJ/h Sensible heat gain through walls and ceilings: 45,000 kJ/h Equipment and reaction heat of potatoes: 14,500 kJ/h Infiltrated air: 200 m3/h If the air-conditioning is achieved by first cooling and dehumidifying and then heating, determine the following: a. Amount of recirculated air in the cold storage if the recirculated air is mixed with fresh air before entering the cooling coil b. Capacity of the cooling coil in tons of refrigeration and its bypass factor if the apparatus dew point temperature of the coil is 5°C c. Capacity of the cooling coil in kW

BIBLIOGRAPHY Ameen, A. 2010. Refrigeration and Air-Conditioning. Prentice Hall of India. Arora, S. C. and Domkundwar, S. A. 2018. A  Course of Refrigeration and Air-Conditioning, 7th edition. Dhanpat Rai & Co., India. Ballaney, P. L. 2003. Refrigeration and Air-Conditioning, 4th edition. Khanna Publishers, India. Hall, C. W. 1980. Drying and Storage of Agricultural Crops. AVI Publishing Company, Westport, CT. Rajput, R. K. 2013. A Textbook of Refrigeration and Air-Conditioning, 3rd edition. S. K. Kataria & Sons, India. Ramaswamy, H. S. 2015. Post-harvest Technologies of Fruits  & Vegetables. DE Stech Publications Inc, Lancaster, PA. Severns, W. H. and Fellows, J. R. 1958. Air-Conditioning and Refrigeration. John Wiley & Sons, New York.

12

Separation

This chapter presents the contact equilibrium process—basic principles, gas absorption, extraction, and distillation; it also addresses applications of contact equilibrium in processing agro products in agro-processing plants such as carbonation of beverages, extraction of oil from soybeans, and sucrose from sugarcane and sugar beets. It also presents mechanical separation—filtration, sedimentation and centrifugation, and applications of mechanical separation in agro-processing plants such as cheese curd from water, and cream separation from skim milk.

12.1 INTRODUCTION Separation as a unit operation in agricultural and food processing plants is used in several unit operations and involves separation of selected or desired components from the agricultural and food products. Separation may be classified under two headings: (1) contact equilibrium process and (2) mechanical separation. Mass balance is the basic process in the contact equilibrium process and includes (a) gas absorption such as hydrogenation of oil and carbonation of beverages, (b) desorption or stripping such as off-flavor removal, (c) liquid-solid extraction such as extraction of oil from soybeans, (d) liquid extraction such as extraction of sucrose from sugarcane and beets, and (e) distillation used in manufacturing of alcoholic beverages. All of these processes involve the transfer of mass from one component or phase of the product to a secondary component or phase. In designing contact equilibrium processes, a fundamental understanding of these processes is essential. Mechanical separations usually refer to the following: (a) filtration such as separation of cheese curd from water, (b) sedimentation such as concentration of casein, and (c) centrifugation such as separation of cream from skim milk and physical forces used to accomplish the desired separation of the components. For example, in the case of filtration, forces are required to move fluid through the filter. In  sedimentation, gravitational forces are used to separate components. In  centrifugation, induced forces are applied and separation is achieved due to centrifugal forces that move fluid through the filter. In designing mechanical separation processes, a knowledge of these processes is required.

12.2  CONTACT EQUILIBRIUM PROCESS One of the basic principles involved in contact equilibrium is that an equilibrium is achieved between the two phases or two insoluble components of the mixture. Equilibrium condition can be achieved in any type of separation system including liquid-liquid, liquid-gas, liquid-solid, or gas-solid types.

12.2.1 Absorption One of the typical contact equilibrium processes is the absorption of gas in a liquid. Two important applications in food processing are the injection of hydrogen gas into an oil during the hydrogenation process and carbonation of beverages. A third process that needs attention is aeration of wastewater or liquid waste from processing plants. All of these processes involve the introduction of a gas phase into a liquid.

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The concept of equilibrium in a gas liquid system is very important in establishing the rate and extent of gas absorption. The concentration of gas absorbed in a liquid is directly proportional to the partial pressure of the same gas above the liquid. For situation in which the gas or gas mixture can be described by the ideal gas law and the liquid can be considered ideal, Raoult’s law will apply as given below: p = Px



(12.1)

where p = partial pressure of gas above liquid P = saturation vapor pressure x = mole fraction in solution For a situation in which liquid is considered nonideal and mole fraction values are small, Henry’s law is assumed to apply as follows: p = H c xsw



(12.2)

where Hc = Henry’s constant

12.2.2 Extraction The term extraction might be applied very broadly to describe all separation processes. In most of the applications in food industries this term refers to distinct phases: liquid-solid leaching and liquid-liquid extraction. Leaching refers to the removal of some soluble component from a solid by using liquid solvent. The leaching of sucrose from sugar beets utilizing hot water as a solvent and the removal of edible oils from soybeans are two examples of leaching related to food processing. 12.2.2.1  Rate of Extraction The rate at which extraction proceeds is a measure of the rate at which the solute is transferred from one phase to another phase. This rate depends on (a) particle size, (b) solvent, (c) temperature, and (d) agitation of the fluid. The rate of extraction can be described by the following equation

dM = K m (Cs − C) dt

(12.3)

where K m = overall mass transfer coefficient Cs = concentration of the solute at saturation condition C = concentration in extraction process M = mass of component being transferred in equilibrium contact Assuming the batch process in which the total volume of the solution is constant, the following equation can be derived:

dM = V′dC

(12.4)

This equation in the integral form can be expressed as

dC

s

where V′ = volume

Km

∫ C − C = ∫ V′ dt

(12.5)

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The solution of the equation in the following form illustrates best the extraction rate:



  K C = CS  1 − exp  − m  V′ 

  t    

(12.6)

12.2.2.2 Leaching The design of a system for leaching involves the determination of the amount of solvent required to produce a given amount of extract from a given quantity of solid material. Generally, this can be achieved by the use of material balance on the various components in the process. Three components must be considered in the leaching process and these are the solvent, the solute, and the insoluble solids. Initially, the solute is contained within the insoluble solids; during the leaching process, it distributes or diffuses into liquid solvent until the concentrations in the solvent and the insoluble solids are equal. This is referred to as equilibrium condition. Since there are three components involved in the leaching process, a three-component system represented by a right triangle is utilized to describe the system. As illustrated in Figure 12.1, each vortex of the triangle represents 100% concentration of one of the three components. In Figure 12.1, the concentration of solvent increases along the vertical axis while the concentration of the solute increases along the horizontal axis. Any location inside the triangle represents the mixture of the three components, and this triangle can be used to represent the process taking place inside triangle. Figure 12.2 shows the material balance of a single-stage leaching process, and the overall material balance yields the following equation 

F + S = E + W

where F = product feed in incoming streams S = solvent in the incoming streams E = solvent in the stream leaving the process W = product solids plus solvent and solute in the streams

FIGURE 12.1  Representation of three-component leaching process.

(12.7)

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FIGURE 12.2  Single-stage leaching process.

A material balance on the solute component yields the following equation X FC F + X SCS = X EC E + X WC W



(12.8)

where the solute component fraction includes the solute containing solvent component fraction (XSC) is nonexistent. In most cases, the solute fraction (XFC) of the feed is known and the solute fraction of the extract (XEC) leaving the process will be known. The solute fraction of the stream carrying the solids (XWC) away from the process may be known or may be considered insignificant. Similar mass balance equations can be written based on the fractions of the other two components. Several approaches are available for when designing a leaching process. Under typical conditions, when the streams entering the leaching process include the feed and pure solvent, the extract of the outlet of the leaching process will be predominantly solvent and the solute. In contrast, the solids carrying streams leaving the process will contain all three components involved in the process. Under these conditions and when the more common parameters are known, the material balance equations can be used to compute the fractions of solvent and the solids in the solid-carrying stream at the outlet of the leaching process. These equations are X WA =



S − EX EA W

(12.9)

X FBF W

(12.10)

X WB =



where XWA = solvent fraction of streams carrying solids XWB = insoluble solid fraction of streams carrying solids XEA = solvent fraction in the extract XFB = insoluble solid fraction in the feed An alternative approach utilizes the right-angle triangle system for determination of the unknown factors by referring to Figure  12.3. Point E in Figure  12.3 represents extract leaving the leaching process and assumes that the solid fraction is zero in this particular stream. The line passing through W and parallel to the hypotenuse of the right-angled triangle represents the solute concentration of the solid-carrying stream leaving the leaching process. The line from the right angle vortex of the triangle to E crosses the line describing the solid concentrations at W in the right angle. This point represents the concentrations of the three components in the solid-carrying stream leaving the leaching process. This process leads to the same type of results as material balance equations, assuming all of the same factors are given or specified. Example 12.1 A countercurrent extraction system is used to extract oil from 500 kg of soybeans per hour. The ­system is designed to extract oil from soybeans with 18% oil and provide 40% oil in the extract solution leaving the system at 400 kg per hour. If the weight of extract solution in the solids leaving the system is equal to 50% of the weight of the solids, compute the supply rate of the solvent.

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FIGURE 12.3  Right angle representation of extraction process.

Solution Product solids plus solvent and sludge in the extraction is given by W = 500 × 0.82 + 0.5 × 500 × 0.821 = 615 kg



Overall mass balance gives 500 + S = 400 + 615

Hence

S = 615+ 400−500



= 515 kg/h



12.2.3  Distillation The distillation process is used to separate two components of liquid solution and depends on the volatility of the two liquid phases. During distillation, both liquid components of the mixture are vaporized and the vapors produced will be different proportions of the components than the liquid. The proportion of vapor produced depends on the relative volatility of the two components, and the separation is done by condensation. The distillation process is designed on the basis of vapor-liquid equilibrium information. 12.2.3.1  Vapor-Liquid Equilibrium The distillation process involves the separation of liquid components having only slightly different volatilities and hence is important to understand the basic concept of this process. The vaporliquid equilibrium during distillation can be illustrated using a temperature-concentration diagram as shown in Figure 12.4. Heating a liquid from point A will increase the temperature until it reaches the lower curve at point B, and this curve is referred to as a boiling point curve. At  this point,

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FIGURE 12.4  Vaporization and condensation of components of liquid solution.

component A has mole fraction (xA) in the liquid while the vapor has mole fraction (yA) for component A as indicated by point C, shown in Figure 12.4. An increase in temperature above point B will decrease the mole fraction of component A in liquid, and this will also result in a decrease of mole fraction of component A in the vapor produced. During the process, the mole fraction (x A) will change from point B to E and the mole fraction (yA) will change from C to D. The final mole fraction in vapor is at point D while final mole fraction in liquid is at point E. These same changes will occur in reverse order if a vapor containing two components is reduced in temperature until the vapor phase no longer exists. The distillation involves the partial vaporization of the components in the liquid solution to produce enriched concentration in the vapor phase. Condensing the enriched vapor phase, a higher concentration of the primary component in the liquid will result. If the concentration is not enough as desired, the process should be repeated until the desired concentration is achieved. A second type of diagram considerably used in distillation process computations is the vaporliquid equilibrium diagram as shown in Figure  12.5. The  vapor-liquid equilibrium diagram represents the mole fraction of a component in vapor produced from the mole fractions of the same component in the liquid solution. Figure 12.5 can used to determine the effectiveness of a distillation process of the solution being separated. The relationship between the concentration of the specific components in the vapor and the composition of the same component in the liquid can be expressed in terms of a relative volatility (α). For a binary mixture, relative volatility (α) is the ratio of the volatility of component A = pA/xA to the volatility of component B = pB/xB and this results in the following equation:

α=

p A x B y A x B  y A   1 − x A  PA = = = p Bx A y Bx A  x A   1 − y A  PB

(12.11)

Equation 12.11 shows that the relative volatility (α) can be expressed as a ratio of vapor pressures of the two components.

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FIGURE 12.5  Vapor-liquid equilibrium diagram.

For ideal solutions, the relationship between partial pressure and mole fraction of a liquid solute can be expressed by Henry’s law, and the relationship between partial pressure and vapor pressure of a pure component in the liquid is expressed by Raoult’s law. For nonideal solutions of the type normally encountered during food processing, these equations are modified by introduction of an activity coefficient (γ). The modified Raoult’s law would be:

p = γXP

(12.12)

where the magnitude of the activity coefficient (γ) expresses the extent to which the system is nonideal. 12.2.3.2  Flash Vaporization Flash vaporization involves the vaporization of the volatile components from a liquid mixture and allows the vapor and liquid to come to equilibrium. The  process may be conducted as a batch operation or a continuous process, but in general the process involves the heating of the liquid in a heat exchanger followed by flashing of the liquid in a chamber. The chamber is at reduced pressure, which results in the flashing or vapor formation in an adiabatic process. It is usually designed to allow for separation of the liquid from the vapor, usually assuring that the liquid and the vapor leave at separate ports. The vapor moves to a condenser to produce enriched liquid which contains a higher concentration of the more volatile component of the original feed mixture. The design of the flash vaporization system is based on mass balance and equilibrium relationship between the liquid and the vapor. The  overall mass balance for flash vaporization can be expressed as: where F = feed rate D = vapor rate B = liquid rate

F = D + B

(12.13)

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Using the composition of the more volatile component of the mixture, a mass balance on that component can be expressed as:

Fx F = Dx D + Bx B

(12.14)

Using Equations 12.13 and 12.14, an expression for the ratio (B/D) can be derived as:

B xD − xF = D xF − xB

(12.15)

Most of the processes can be designed using Equation 12.15 and the operating line–equilibrium curve as shown in Figure 12.6. 12.2.3.3  Batch Distillation Batch distillation is conducted in a chamber equipped with a heating device and facilities for removal of vapor as it is produced. The vapor produced is removed immediately from the chamber and will have a high concentration of the most volatile component in the liquid mixture. As the distillation process continues and additional boiling and more vapor are produced, the concentration of the most volatile component in the liquid decreases with time. Batch distillation is the separation of the most volatile components in the liquid mixture according to the time at which the vapor may be separated. Consider that the original distillate (B) is changed by removal of a differential quantity (dB), and the original mole fraction (xB) is changed by removal of a volatile component with mole fraction (xD) to obtain a distillate with different composition. This change can be expressed as:

Bx B = x DdB + ( B − dB) ( x B − dx B )

(12.16)

If the second-order terms are neglected, Equation 12.16 can be expressed as:

dB dx B = B xD − xB

FIGURE 12.6  Operating line–equilibrium curve.

(12.17)

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Separation

By integration from the feed conditions (F and xF) to conditions existing at any time during distillation (B and xB), Equation 12.17 can be written as:

∫

F

B

dB F = ln = B B

∫

xF

xB

dx B xD − xB

(12.18)

In most situations, the integration of Equation 12.18 must be done by numerical techniques. 12.2.3.4 Fractionation Fractionation is carried out in a fractionation column and is referred to as rectification in many cases. It provides separation of a solution into components and also provides the desired level of purity. Figure 12.7 shows a schematic diagram of a fractional column. The fractional column involves two basic processes and at least two additional auxiliary processes. Two basic processes are rectification or enriching in the upper half of the column and stripping in the lower half of the column while the two additional auxiliary processes are involved in a reboiler at the bottom of the column and a condenser at the top of the column. The feed solution (F) is introduced at the middle of the column as shown in Figure 12.7. The feed moves downward through the stripping section, initially resulting in the formation of a vapor phase with the same component as the liquid phase, but with a higher concentration of the more volatile components. The liquid phase ( L) leaves the column at the bottom and enters the reboiler where

FIGURE 12.7  Schematic diagram of fractional column.

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heat is applied to produce a liquid (B) with high concentration of component with low volatility, and a vapor phase ( V ), which is returned to the stripping section of the column. Within the stripping section of the column, the vapor phase ( V ) establishes equilibrium with liquid moving downward and increases in concentration of the volatile components. The vapor phase V leaves the enriching section at the top of the column and enters into the condenser. In the condenser, heat is removed to produce liquid with a high concentration of volatile components. A portion of the condensate is removed as distillate (D) while the remaining portion is returned to the column as reflux liquid (L). The rates of liquid flow (L) and vapor flow (V) as well as the number of stages used to enhance contact between liquid and vapor phases will determine the extent of separation in the column. In designing the fractionation column, material balance must be conducted on both the enriching column and stripping column. A material in the upper section of the rectification or enriching column as shown in Figure 12.8 gives:

L n + D = Vn +1

(12.19)



L n x n + Dx D = Vn +1y n +1

(12.20)

where: xn = mole fraction of more volatile component in the liquid phase at n xD = mole fraction of more volatile component in the liquid phase at D yn+1 = mole fraction of the same component in the vapor phase Equations 12.19 and 12.20 are combined to obtain the equation for operating for enriching section and this gives:

y n +1 =

D L D L xn + xD = xn + xD V L+D L+D V

where the subscripts are dropped from liquid phase flow (L) since it does not change.

FIGURE 12.8  Material balance on rectification.

(12.21)

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Separation

FIGURE 12.9  Material balance on stripping section.

A material in the lower section of the stripping section of the column as shown in Figure 12.9 gives:

L m −1 = Vm + B

(12.22)



L m −1x m −1 = Vm y m + Bx B

(12.23)

Equations 12.22 and 12.23 can be used to obtain the equation for operating for the stripping section and this gives:

ym =

L B V + BL B x m −1 − x B = x m −1 − x B V V V V

(12.24)

where: V = constant flow rate for vapor phase in the stripping section The operating line for enriching has a slope of L/V and y-axis intercept of DxD/V while the operating line for stripping has a slope of L / V and y-axis intercept of −Bx B / V. The differences between both liquid-phase and vapor-phase flows in the enriching and stripping sections is directly related to feed rate and properties of the feed. Conditions developed from an analysis of the interception of these two operating lines lead to a definition of a ratio

i=

L−L F

(12.25)

and the slope of the i-line that connects the point of intersection with the diagonal. The slope will be

slope =

i i −1

which will change directly with properties of the feed (F).

(12.26)

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Example 12.2 A continuous distillation column is used to concentrate a 7% alcohol (by volume) feed to obtain brandy at 96% alcohol (by volume). The feed is introduced at a temperature below the boiling point such that i = 1.07 and the reflux ratio (Ln/V) is 0.8. If the feed rate is 15,000 gallons per hour and 0.047% alcohol (by volume) is lost with stillage, (a) determine the rate of brandy production and (b) develop the equation of operating line for the enriching section and stripping section. The mole fraction of feed, brandy, and stillage are 0.0228, 0.857, and 0.000125, respectively.

Solution Here 15,000 gal/h = 124,600 lbm / h = 6700 mole /h



Using Equation 12.13 we have 6700 = D + B

Using Equation 12.14 we have

67 × 0.0228 = D × 0.857 + B × 0.000125

Hence we have



B = 6523 mole /h

and D = 177 mole /h

The production rate of brandy is D = 177mole /h = 7225lbm /h



Using Equation 12.19 and reflux ratio of 0.8 we can write



Ln + D = Vn + 1 = Ln / 0.8



0.8Ln + 0.8 × 177 = Ln



142 = Ln



Ln = 708 mole /h = L and Vn+1 = 885 mole/h = V

Using Equation 12.20 we can write



yn + 1 =

708 177 xn + × 0.857 885 885

yn + 1 =0.8xn + 0.200 This equation is the equation for operating of the enriching section. Using Equation 12.25 and the value of i = 1.07, we can write



1.07=

L −L L −708 = 6700 F

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Separation



Lm−1 = 7877 mole /h = L



Using Equation 12.22 we can write



7877 = Vm + 6523



Vm = 1354 mole /h = V Using Equation 12.23 we can write



ym =

7877 6523 xnm−1 + × 0.000125 1354 1354 ym = 58xm−1 + 0.000602

This equation is the equation for operating of the stripping section.

12.2.3.5  Steam Distillation Steam distillation is used to separate the volatile components from relatively small amounts of nonvolatile components, and the boiling point of the mixture should be sufficiently low temperature to prevent the damage to the heat-sensitive product components. Steam distillation is commonly used to remove odors and flavors from oils and fats in agricultural and food processing industries. Normally, steam distillation is conducted for a mixture of water and insoluble liquid such as oils. The mixture is heated externally and the sum of the components is equal to the total pressure and can be expressed as:

p w + p 0 = P ′

(12.27)

where: pw = vapor pressure of water p0 = vapor pressure of insoluble component P′ = total pressure In this approach, distillation results in vaporization of insoluble component at a temperature below the normal boiling temperature. The vapor produced will contain both the components in proportion to the mole fractions of two components present. Steam distillation is possible without water being one of the components in the mixture by bubbling superheated steam or an insoluble gas through the mixture, and direct introduction of steam reduces the boiling point of the mixture. This reduction of pressure for a given amount of volatile component pressure can be expressed by the following relationship:

P′ − P w A M S = P M A wS

(12.28)

where: P = pressure of steam P′ = total pressure wA = mass of volatile component wS = mass of steam MS = molecular weight of steam MA = molecular weight of volatile component The right side of Equation 12.28 represents the ratio of the moles of volatile component to the moles of the steam. If the molecular weight of the volatile component is much greater than the steam, the vapor produced will contain large amount of volatile components.

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Example 12.3 The volatile component of an oil is removed by steam distillation at a pressure of 2.5 kg/cm2, and the molecular weight of the volatile component is 134. Compute the composition of the vapors leaving the distillation process. Assume total pressure to be 7.61 kg/cm2.

Solution We have from Equation 12.28 7.61− 2.5 w A × 18 = 2.5 134 × w S

This can be simplified in the form

w A 134 × 5.11 = 15.22 = wS 2.5×18



This shows that the volatile component is 15.22 times higher than steam. For a supply of steam at the rate of 2.55 kgm/min, the removal rate of volatile component is



w A =2.55×15.22 = 38.04 kgm /min

12.3  MECHANICAL SEPARATION PROCESS The separation process that depends primarily on physical forces is referred to as mechanical separation, and this is accomplished by filtration, sedimentation, and centrifugation. In the case of filtration, the force involved is that required to move the fluid through a filter medium. In sedimentation, the forces are those of gravity and the influence of gravitational forces on the different component of products. In centrifugation, an induced force is applied that results in separation of the product components due to centrifugal force.

12.3.1  Filtration The removal of solid from liquids is accomplished by somewhat different mechanisms than the removal of solid particles from air. In general, filtration process can be described by the manner in which the fluid is being filtered through the filter medium, the solids being deposited on the filter medium. Figure 12.10 shows a schematic diagram of the filtration process. The removal of solids build up on the filter medium, which results in an increase in the resistance to the flow as filtration progresses. In addition to filtration resistance, the filtration rate depends on the pressure drop across the filter medium, and the viscosity of the filtrate. The filtration rate can be expressed as

Rate of filtration =

Driving force Resistance

(12.29)

where the driving force is the pressure required to move the fluid through the filter media and the resistance is dependent on several factors. The overall resistance can be described by the expression

R = µ r′(L c + L)

(12.30)

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Separation

FIGURE 12.10  Illustration of filtration process.

where: Lc = thickness of the accumulated solids in the filter cake L = fictitious thickness of the filter material or medium µ = fluid viscosity r = specific resistance of filter cake Thickness Lc can be described by the following equation (Earle 1966) Le =



SV A

(12.31)

where: S = solid content of the fluid V = volume passing through the filter A = cross-sectional area Using Equations 12.30 and 12.31, the total resistance can be expressed as

 SV  R = µ r′  + L   A 

(12.32)

Combining Equations 12.29 and 12.32 following the equation for rate of filtration can be obtained dV = dt

A ∆P  SV  µ r′  + L  A 

(12.33)

Equation 12.33 is an expression that can be used to describe the filtration process and can be used for scale-up if converted to the appropriate forms. The filtration process may occur in two phases: (a) constant rate filtration normally occurring in the early stage of filtration, and (b) constant pressure filtration occurring in the final stage of filtration. Constant rate filtration can be described by following the integrated form of equation

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A ∆P  SV  µ r′  + L  A 

(12.34)

µ r′ SV 2 + L ∆V   A2t 

(12.35)

V = t Hence ∆P is given by

∆P =

The equation can be used to determine the pressure drop as a function of filtration rate. Equation 12.34 can be used in a different form if the thickness of the filter medium can be considered negligible. The following equation for pressure drop as a function of time can be derived ∆P =



µ r′V 2 A2t

(12.36)

In many situations, Equation 12.36 can be used to predict pressure drop requirements for a filter during the early stage of the process.

12.3.2  Sedimentation Separation of solids from fluid streams by gravitational or centrifugal forces is termed as sedimentation. In processing industries, the sedimentation is aimed to remove solid particles from either liquid or gas. 12.3.2.1  Sedimentation for Low Concentration Suspensions Particles in low concentration suspension settle at a rate of terminal velocity of the particle in the suspension, and this type of sedimentation is usually referred to as free settling. The terminal velocity of the particle can be established by examination of the forces acting on the particle. The force that resists the gravitational force is referred to as drag force, F D, and can be described by the following equation: FD = 3 πµ D u



(12.37)

where: u = relative velocity between the particle and fluid D = particle diameter µ = fluid viscosity This equation applies to the Reynolds number less than 0.2. For the Reynolds number greater than 0.2, the Coulson and Richardson equation  applies (Coulson and Richardson 1968). The  gravitational force FG is a function of particle volume and density difference along with gravitational acceleration as given by where: ρp = particle density ρf = fluid density gc = acceleration due to gravity

FG =

1 π D 3 ( ρ p − ρf ) g 6

(12.38)

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By setting Equation 12.37 equal to Equation 12.38, the following equation is obtained: ut =



D2 g ( ρ p − ρf ) 18 µ

(12.39)

From Equation 12.39, it is observed that the terminal velocity is directly related to the square of the particle diameter. In addition, the terminal velocity is dependent on the density of the particle and properties of the fluid. Equation 12.39 is the most common form of Stokes’ law and should be applied only for streamline flow and spherical particles. When the particle is not spherical, shape factor is used. 12.3.2.2  Sedimentation for High Concentration Suspensions When the concentration of solids in the suspension is sufficiently high, Stokes’ law is not applicable. When the particle size is between 6 and 10 microns, the particles fall at the same rate. This rate corresponds to the velocity predicted by Stokes’ law. The settling of particles results in very well-defined zones of solid concentrations. The rates at which solids settle in high concentration suspension have been investigated empirically and have been reviewed by Coulson and Richardson (1968). One approach is to modify Stokes’ law by introduction of the density and viscosity of the suspension in place of density and viscosity of the fluid. This results in the following equation: us =



KD2 g ( ρ p − ρf ) µ

(12.40)

where K = constant A second approach accounts for a void between suspension particles that allows movement of fluid to the upper part of the suspension column, and the expression for the particle settling velocity is given by up =



D2 g ( ρp − ρf ) f(e) 18µ

(12.41)

where f(e) = function of void in suspension Equation  12.41 is still a modification of Stokes’ law where the density of the suspension and viscosity of the fluid are utilized. Although there are some limitations to use these equations, these equations provide practical values for high concentration suspension of large particles. Example 12.4 A sedimentation tank is used to remove large solid particles from wastewater leaving the agroprocessing plant. The ratio of liquid mass to solids in the inlet to the tank is 9 kg liquid/kg solids and the inlet flow rate is 5  kg/min. The  sediment leaving the tank bottom should have 0.5  kg liquid/kg solids. If the concentration rate of solids in the water is 0.5 m/h, determine the area of sedimentation necessary.

Solution Using mass flow equation and the difference between inlet and outlet conditions on the wastewater, the following equation for upward velocity of liquid in the tank is obtained



uf =

(Ci −C0 ) ω ∆p

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where C0 = mass ratio of liquid to solid at the outlet Ci = mass ratio of liquid to solid at the inlet Based on the information provided, we have

C0 = 9 kg /s

ω = 5 kg /min ρ = 1000 kg /m3

Ci = 1 kg /s

−

Since upward liquid velocity should equal sedimentation velocity, the area becomes



A=

(9 − 1) × 5 = 48 m2 (0.50 / 60) × 1000

12.3.3 Centrifugation In many processes, the sedimentation does not work properly. In such situations, the separation can be accelerated through application of centrifugal force. The basic equation that describes the force acting on the particle moving in a circular path is given by

Fc =

m r ω2 gc

(12.42)

where ω = angular velocity m = mass, kg r = radius, m gc = acceleration due to gravity, m/s2 Since the angular velocity can be expressed as the tangential velocity of the particle and Equation 12.42 can be expressed as

Fc =

m u2 gc r

(12.43)

If the rotational speed is expressed in revolution per minute, Equation 12.43 becomes

Fc = 0.011′

m r n2 gc

(12.44)

n = rpm Equation 12.44 indicates that the centrifugal force acting on a particle is directly related to and dependent on the distance of the particle from the center of the rotation (r), the centrifugal speed of rotation (n), and the mass of the particle (m). For example, if a fluid containing particles of different densities is placed in a rotating bowl, the higher density particles will move to the outside of the bowl as a result of the greater centrifugal force acting on them. This will result in a movement of lower density particles toward the interior portion of the bowl. This is illustrated in Figure 12.11. This is the principle used in the separation of liquid food products with components of different densities. 12.3.3.1  Rate Separation The  rate of separation of materials of different densities can be expressed in terms of relative velocity between the two phases. An expression for this velocity is the same as the equation for

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Separation

FIGURE 12.11  Separation of fluids with different densities using centrifugal force.

terminal velocity given as Equation 12.39 in which the gravitational acceleration (g) is replaced by an acceleration parameter describing the influence of centrifugal force. This equation can be expressed as: 2



 2πn  a = r   60 

(12.45)

Substituting Equation 12.45 into Equation 12.39 yields

uc =

D 2 n 2 r ( ρ p − ρf ) 1640 µ

(12.46)

Equation 12.46 can be applied to any situation in which there are two different phases with different densities. 12.3.3.2  Liquid-Liquid Separation In the case of separation that involves two liquid phases, it is usually easier to describe the process in terms of the surface that separates the two phases during separation. The differential centrifugal force acting on an annulus of the liquid in the separation cylinder can be expressed as

dFc =

where dm = mass in the annulus of the liquid.

dm × rω2 gc

(12.47)

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The Equation 12.47 can be rewritten as dFc ρω2 r dr = dP = 2πr b gc



(12.48)

where dP = differential pressure across the annulus of the liquid b = height of the bowl By integration of Equation 12.48 between two different radii in the separation cylinder, the difference in pressure between these two locations can be computed from the following equation: P2 − P1 =



(

ρω2 r2 2 − r12 2gc

)

(12.49)

At some point in the cylinder, the pressure of one phase must be equal to the pressure in the other phase so that the expressions of this type given by Equation 12.49 can be written for each phase and can represent the radius of equal pressure in the following manner:

(

ρA ω2 rn 2 − r12



2gc

)

=

(

ρB ω2 rn − r22 2gc

)

(12.50)

Solving for the radius of equal pressures for two phases we can write  ρ r 2 − ρBr2 2  rn 2 =  A 1   ρA − ρB 



(12.51)

where −ρA = density of heavy liquid phase ρB = density of low density liquid phase − Equation 12.51 is a basic equation to be utilized in the design of the separation cylinder. The radius of equal pressure that represents two phases may be or are separated is dependent on two radii (r1 and r2). 12.3.3.3  Particle Gas Separation The separation of solid from a gas is a common operation in many food industries. Probably the most common is the separation of a spray-dried product from the airstream after the drying operation is completed. This is usually accomplished in a cyclonic separator. The basic equations presented earlier in Chapter 3 apply, and Equation 12.46 provides some indication of the rate at which separation of solid particles from an airstream can take place. It is obvious from this equation that the diameter of the particles must be known along with the density of the solid and the density of the airstream. Example 12.5 Design the inlet and discharge of a centrifugal separation of cream from whole milk. The density of skim milk is 1100 kg/m3. Provide the discharge conditions necessary if the skim milk outlet has a 2.50 cm radius and the cream outlet has a 5.0 cm radius. Compute the radius of the neutral zone. Cream density is 900 kg/m3.

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Solution Using Equation 12.51 the radius of the neutral zone can be computed as: 2

 5   2.5  1100 ×  − 900 ×     100   100  = 0.0109 rn2 = 1100 − 900

Hence rn is given by

R = 10 cm

EXERCISES 12.1 A countercurrent extraction system is used to extract oil from 1000 kg soybeans per hour. The system is designed to extract oil from soybeans with 20% oil and provide 45% oil in the extract solution, leaving the system at 750 kg per hour. If the weight of extract solution in the solids leaving the system is equal to 50% of weights of solids, compute the supply rate of the solvent. 12.2 A  continuous distillation column is used to concentrate a 7% alcohol (by volume) feed to obtain brandy at 96% alcohol (by volume). The  feed is introduced at a temperature below the boiling point such that i = 1.07 and the reflux ratio (Ln/V) is 0.8. If the feed rate is 15,000  liters per hour and 0.047% alcohol (by volume) is lost with stillage, (a) determine the rate of brandy production and (b) develop the equation of operating line for the enriching section and stripping section. The mole fraction of feed, brandy, and stillage are 0.0228, 0.857, and 0.000125, respectively. 12.3 The volatile component of an oil is removed by steam distillation at a pressure of 1.25 kg/ cm2, and the molecular weight of the volatile component is 134. Compute the composition of the vapors leaving the distillation process. 12.4 A sedimentation tank is used to remove large particles solids from wastewater leaving an agro-processing plant. The ratio of liquid mass to solids in the inlet to the tanks is 7 kg liquid/kg solids and the inlet flow rate is 3.5 kg/min. The sediment leaving the tank bottom should have 1 kg liquid/kg solids. If the concentration rate of solids in the water is 0.5 m/h, determine the area of sedimentation necessary. 12.5 Design the inlet and discharge of a centrifugal separation of cream from whole milk. The density of skim milk is 1150 kg/m3. Provide the discharge conditions necessary if the skim milk outlet has a 3 cm radius and the cream outlet has a 6.0 cm radius. Suggest the desirable radius for the inlet. Cream density is 900 kg/m3.

BIBLIOGRAPHY Coulson, J. M. and Richardson, J. F. 1968. Chemical Engineering, Vol. II. Pergamon Press, New York. Earle, R. L. 1966. Unit Operations in Food Processing. Pergamon Press, London, UK. Foust, A. S., Wenzel, L. A., Clump, C. W., Maus, L. and Andersen, L. B.1960. Principles of Unit Operations, 2nd edition, Wiley-India, New Delhi, India. Harriot, P. 1964. Process Control. Tata McGraw Hill Company LTD, New Delhi, India. Heldman, D. R. 1977. Food Process Engineering, 2nd edition, AVI Publishing Company, Inc., Connecticut. McCabe, W. L., Smith, J. C. and Harriott, P. 2013. Unit Operations of Chemical Engineering, 7th edition, McGraw-Hill Education (India) Private Limited. New Delhi, India. Sahay, K. M. and Singh, K. K. 1994. Unit Operations in Agricultural Processing, 1st edition, Vikas Publishing House, New Delhi, India.

13

Materials Handling and Conveying

This chapter presents concepts and principles of materials handling and classification of materials handling equipment. It also presents belt conveyors—details and design, chain conveyors, and chain conveyor design; screw conveyors—horizontal and inclined screw, bucket elevators, and gravity conveyors; pneumatic conveyors—advantages and disadvantages, and design; hydraulic conveyors— advantages and disadvantages, and design; and cranes, lifts, trucks and cart, and robotic handling system.

13.1 INTRODUCTION A materials handling system simply means loading, unloading, and moving of materials inside and outside of the processing plant, and the equipment used for materials handling are referred to as materials handling equipment. The efficiency of the processing plant is significantly affected by the efficiency of the handling of the materials from one unit operation to another. The importance of this movement doesn’t necessarily depend on its magnitudes. The delivery of a processed product from the processing plant is as important relatively as the movement of the product through a packaging plant. According to the American Material Handling Society, “Materials handling is the art and science involving the moving, packaging and storing of substances in any form.” Since materials handling is a system, it is important to understand the systems concept of materials handling. The term system means a complex combination of diverse components for a common plan or serving for a common purpose. In a processing plant, materials handling is a subsystem (component) of the production system. Materials handling can also be considered to be a system whose subsystems (components) are (i) design or method adopted, (ii) types of equipment used, (iii) operations of the materials handling such as packing/unpacking, movement, storage, etc., (iv) maintenance required for equipment, and mode of transportation of raw materials and processed products. The common objective of all of these components is to achieve the lowest-cost materials handling system. The systems concept based on systems thinking means that the materials handling system should consider the entire system, not any in isolation. This systems approach will ensure an overall cost-effective system for the processing plant. A well-designed materials handling system should be designed to achieve the following:

1. Improve the efficiency of the production system 2. Reduce labor cost 3. Minimize the damage of materials 4. Reduce overall cost by improving materials handling 5. Minimize accident 6. Maximize space utilization 7. Improve customer services 8. Increase efficiency and salability of the processing plant

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13.2  PRINCIPLES OF MATERIALS HANDLING The  principles of materials handling mean a set of prescribed rules established by authoritative practitioners. Stocker (1951) was the first to report the principles of materials handling and these have been further developed. A set of principles of materials handling are stated below:

1. Planning principle: All handling activities should be planned. 2. Systems principle: Integrate as many handling activities as possible, encompassing a full scope of operations such as production, inspection, packaging, receiving, storage, warehousing, shipping, and transportation. 3. Materials flow principle: Plan operations sequence and equipment arrangement to optimize material flow. 4. Simplification principle: Reduce, combine, or eliminate unnecessary movement and/or equipment. It increases efficiency of materials handling. 5. Gravity principle: Utilize gravity to move material whenever practicable. 6. Space utilization principle: Make optimum use of building volume. 7. Unit size flow principle: Increase quantity, size, and weight of loads handled. 8. Safety principle: Handling methods and handling equipment use must be safe. 9. Mechanization/automation principle: When appropriate, use mechanized or automatic materials handling equipment. 10. Equipment selection principle: Before selecting materials handling equipment, consider all aspects of materials handling, e.g., materials to be handled, moves to be made, and methods to be utilized. 11. Standardization principle: Materials handling methods and equipment should be standardized to the extent possible. 12. Flexibility principle: Use methods and equipment that can perform different tasks and applications. 13. Dead-weight principle: Reduce dead-weight movement. 14. Motion principle: Stoppage of mobile equipment should be minimum. 15. Idle time principle: Reduce idle or unproductive time of both materials handling equipment and manpower. 16. Maintenance principle: Do schedule maintenance and repair work of all materials handling equipment to minimize outage. 17. Obsolescence principle: Replace obsolete handling methods and equipment by more efficient methods or equipment to improve operations. 18. Control principle: Use materials handling equipment to improve production and inventory control and order handling. 19. Capacity principle: Use materials handling so that full production capacity can be achieved. 20. Performance principle: Select materials handling systems with higher efficiency measured in terms of expenses per unit load handled.

13.3  CLASSIFICATION OF MATERIALS HANDLING EQUIPMENT Materials handling equipment and systems used in processing plants are large and diverse in concept and design. It is, therefore, essential to classify the materials handling equipment into several categories for meaningful utilization of these equipment. Materials handling equipment may be classified as follows:

1. Belt conveyors 2. Chain conveyors 3. Screw conveyors

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4. Bucket elevators 5. Pneumatic conveyors 6. Hydraulic conveyors 7. Gravity conveyors 8. Cranes 9. Lift and carrying trucks and cart 10. Robotic handling system

13.4  BELT CONVEYORS The belt conveyor essentially consists of an endless belt operating between two or more pulleys, and the belt and its loads are supported by idlers. The installations may be a simple flat belt sliding over a long table or a heavy belt supported by antifriction bearings. Belt conveyors have high mechanical efficiency, little damage to the product being transported, high carrying capacity, and materials can be conveyed over a long distance. A properly designed belt has a long service life, but the initial cost is high. The factors to be considered for description of belt conveyors are the belt itself, drive, tension or take-up, idlers, and loading and unloading devices. The belt that supports and conveys the load is the essential and most important component of the belt conveyor system. The belt must be flexible enough to adjust the pulleys, wide enough to carry the load, and have enough strength to withstand the expected load and operating tension and a resistance surface. Stitched canvas, solid-oven, batala, and rubber belts are used. Stitched canvas and oven belts are usually impregnated with a waterproofing material. A rubber belt is made of canvas or oven material impregnated and vulcanized with rubber and covered with a rubber sheet. Batala belts are similar to rubber belts, but these are affected by temperature over 50°C. The belt drive should discharge the material at the head end pulley, and the pulley must be large enough to provide enough contact surface with the belt to ensure a positive drive. Additional contact surface may be provided using an idler pulley to provide more wrap contact of the belt. Pulley diameters must be large enough to keep from overflexing the belt. The drive pulley is connected to the drive motor through a suitable speed reduction gearbox and flexible shaft couplings. The drive of an inclined conveyor should include a braking device so that the downward movement is restricted in case of power failure. Take-up is necessary to maintain the desired tension in the belt to take into account the stretch of the belt and of contraction and expansion of the belt due to changes in moisture and temperature and also taking into account the invariable elongation of the belt with time due to the working of the belt under tension. These can be adjusted manually by adjusting the screw or automatically by attaching dead weight. The efficiency of the belt conveyor largely is dependent on idlers and the rollers used at certain spacing for supporting active as well as the return side of the belt, which are called idlers. Accurately made, rigidly installed, and well-maintained idlers are vital for smooth and efficient running of a belt conveyor. There are two types of idlers: flat belt idler and troughed belt idler (Figure 13.1). The flat belt idlers are used for granular materials and have an angle of repose not  less than 35°; the capacity of the troughed belt conveyors depends on the trough angle. Figure  13.2 shows the configurations of troughed belt conveyors for different trough angles. The troughed idlers with 20° trough are used for conveying all kinds of bulk materials, while troughed idlers with 30° and 45° trough are used for transporting small particle, lightweight material such as grain. The material can be fed onto the belt by hand or by a mechanism to provide continuous steady flow, and the material may be discharged over the end of the belt by a diagonal scraper by tilting one or more of the idler pulleys or by a tripper.

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(a)

(b)

(c)

FIGURE 13.1  (a) Flat belt idler, (b) Troughed belt idler for 20°, (c) Troughed belt idler for 30°.

(a)

(b)

(c)

FIGURE  13.2  (a) Troughing configuration for 20°, (b) Troughing configuration for 30°, (c) Troughing ­configuration for 45°.

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The major points to be considered in the selection and design of a conveyor belt are:

1. The capacity of the conveyor belt needed 2. Maximum belt capacity needed and selection of the belt 3. Selection of driving pulley 4. Motor power required 5. Selection of idlers and its spacing

13.4.1 The Capacity of the Conveyor Belt Needed Basically, this means what the capacity of the conveyor belt (tons/h) should be in terms of belt width and speed to convey the required quantity of the material by the belt conveyor system.

13.4.2  Belt Width Free-flowing material on a flat belt assumes the shape of an isosceles triangle (Figure  13.3a). The angle of dynamic repose ϕ may be considered to be equal to 0.35ϕ, where ϕ is the static angle of repose for the material. To avoid the spillage, the belt width B is taken to be at least 25% more than the base of the triangle b. Thus, b = 0.8B. As per Indian Standard b = 0.9B–0.05 m for B ≤ 2 m. Therefore, b = 0.8B is a more conservative estimate for B > 500 mm. The cross-sectional area F1 of the load on a flat belt as shown in Figure 13.3a is given by F1 =



bh 1 = ( 0.8B × 0.4Btanϕ ) = 0.16B2 tan(0.35ϕ) 2 2

(13.1)

where: b = base of the triangle, m h = height of the triangle, m B = width of belt, m ϕ = static angle of repose for the material Therefore, capacity Q1 of the belt conveyor is given by Q1 = 3600 F1 × V × γ = 576 B2 × V × γ × tan(0.35ϕ)



kg/h

where: γ = bulk density of material, kg/m3 V = velocity of belt, m/s

(a)

(b)

FIGURE 13.3  (a) Bulk load on flat belt conveyor, (b) Bulk load on troughed belt conveyor.

(13.2)

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For a troughed belt conveyor (Figure 13.3b), the cross-sectional area of the load F is given by F = F1 + F2



(13.3)

where: F1 = triangle area of the troughed belt conveyor, m2 F2 = trapezoidal area of the troughed belt conveyor, m2 The cross-sectional area F2 of the load on a troughed belt as shown in Figure 13.3b is given by F2 =



1 ( 0.4B + 0.8B) × 0.2B × tanλ = 0.12B2 × tanλ 2

(13.4)

The total cross-sectional area F of the load on a troughed belt as shown in Figure 13.3b is given by

F = 0.16B2 × tan(0.35ϕ) + 0.12B2 × tanλ = B2 0.16 × tan(0.35ϕ) + 0.12tanλ 

(13.5)

Therefore, capacity Q of the troughed belt conveyor is given by

Q = 3600 F × V × γ = B2 × V × γ × [576 × tan(0.35ϕ) + 432 × tanλ ]

kg/h

(13.6)

13.4.3  Belt Speed Recommended belt speed depends on the width of the belt, lump size factor of the bulk material, airborne factor, and abrasiveness factor; the speed factor is a summation of the lump size factor of the bulk material, airborne factor, and abrasiveness factor. Higher belt speeds may be considered under special design conditions only.

13.4.4  Belt Tension The motive force in a belt conveyor to drive the belt with load is transmitted to the belt by friction between the belt and the diving pulley, which is rotated by an electric motor (Figure  13.4). Considering no slip between the belt and pulley, the relationship between the tighter side tension and slack side tension is given by



FIGURE 13.4  Tensile force on belt.

T1 = eµα T2

(13.7)

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where: T1 = belt tension at tighter side, kg T2 = belt tension at slack side, kg α = wrap angle, radian µ = coefficient of friction between pulley and belt Effective pull Te = T1 − T2 in the belt against all the resistances is given by

Te = T1 − T2 = T2 (eµα − 1)

(13.8)

Te is the sum of all the resistive load against the motion of the belt carrying the load.

13.4.5  Selection of Belt Carcass The maximum force Temax often occurs at the starting up of a completely loaded conveyor belt from the rest. The ratio ζ between Temax and Te depends on the type of drive selected, which varies from 1.8–2.2 for direct on-line start of motor connected by a pin bus-type coupling, to a lower of 1.2 for delta starting of slip ring motor connected by flexible coupling or a three-phase squirrel cage connected with a fluid coupling with delayed chamber filling. Taking into consideration this maximum effective pull, Tlmax is computed from the following relationship:

 cµα  Tlmax = Teξ  µα   e −1 

(13.9)

Based on this maximum tensile force in the belt, the belt carcass should be selected from the manufacturer’s catalogs, having sufficient breaking strength to withstand this maximum tensile force.

13.4.6  Selection of Driving and Other Pulleys Large driving and tail end pulleys are generally fabricated from steel plate, and the diameter of pulley is selected based on the construction of the belt used. The recommended values of minimum pulley diameters are based on carcass thickness and fiber materials. The drive pulley may be covered with a layer of suitable material such as rubber, polyurethane, ceramics, etc.

13.4.7 Motor Power The power required to drive a pulley just for driving belt is given by



Pd =

Te × V 1000

where: Te = effective tension = (T1 − T2), Newton V2 = belt velocity, m/s Pd = driving power, kW

kW

(13.10)

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Considering the wrap resistance between the belt and the driving pulley, and the driving pulley bearing resistance, the actual power Pa is given by

Pa =



( R wd + R bd ) × V Te × V + 1000 1000

kW

(13.11)

where: Rwd = wrap resistance between belt and pulley, Newton Rbd = driving pulley bearing 1esistance, Newton Additional power requirements should be taken into account for each belt tripper and belt cleaner used with the conveyor. The final motor power PM is calculated based on the efficiency and is given by PM =



Pa η

(13.12)

The actual motor is chosen with a power rating of 15%–20% higher than the calculated power PM.

13.4.8  Selection of Idlers Depending on the type of belt conveyor, the carrying idlers can be troughed or straight, but the return idlers are generally always straight. The  major selection criteria are the roller diameters and the spacing of these idlers. The range of idler diameters to be selected depends on belt width, maximum belt speed, and types of materials to be conveyed. Empirical data have been found more applicable than rational data. Power for standard installations can be calculated from the following equations: Horsepower to drive empty conveyor is given by



Horse power to drive empty conveyor =

Belt speed, m/min ( A + B) × ( 3.281 × L ) × 0.3048 100

(13.13)

where: L = conveyor length, m A, B = constants Horsepower to convey material on level is given by

Horsepower to convey material on level = Capacity, tons/h ×

0.48 + 3.281× L 100

(13.14)

Horsepower to lift material is given by

Horsepower to lift material =

Lift, m tons/h × 1.015 × 0.3048 100

(13.15)

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13.5  CHAIN CONVEYORS Chain conveyor means a group of different types of conveyors and consists of endless chains driven by one set of sprockets at one end and supported by another set of sprockets at the other end. The materials are conveyed directly on the links of the chains or on specially designed elements attached to the chains. The load-carrying chain is supported by idle sprockets or guide ways. Belt conveyors are expensive, quiet, fast, mechanically efficient, and must be carefully engineered to ensure satisfactory performance, while the chain conveyors are not so expensive but may be noisy, are slow, are not mechanically efficient, and do not require specialized skill for design. Because of versatility in design and other advantages mentioned above, chain conveyors are well suited for a great variety of materials moving jobs, especially in agriculture where service is intermittent. Chain conveyors are classified as: (1) trolley chain conveyor, (2) scraper chain conveyor, and (3) apron chain conveyor.

13.5.1 Trolley Chain Conveyor Figure 13.5 shows a typical trolley conveyor. This type of conveyor consists of a series of trolleys suspended from an overhead endless track and propelled by an endless chain or cable with loads suspended from the trolleys. Different structural members can be used as a track for overhead trolley conveyors, which include I-beam, double angles, T-rails, steel bars, pipes, and fabricated sections. Overhead operations allow free floor space and no interference with equipment or traffic at the floor level. For this reason, trolley conveyors are also called overhead conveyors. This type of conveyor can be designed to make a sharp turn of 180° and is highly flexible. Also, this type of

FIGURE 13.5  A typical trolley conveyor.

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conveyor can be used for products of large unit size or for those that can be handled in boxes or baskets during processing. Meat products, fruits, and vegetables are examples of materials that can be handled in trolley chain conveyors.

13.5.2  Scraper Chain Conveyor Scraper chain conveyors used for granular, nonabrasive materials are simple, cheap, easily constructed, and may be operated at steep incline. However, power requirements are high and wear may be excessive. This type of conveyor is extensively used for moving raw products, beets, potatoes, and small grains in the processing plants. Scraper conveyors with attached flights are designed in a varieties of ways. The simplest of them is a single chain with flights operating in a wood or steel trough. Cross section of this type and other types are shown in Figure 13.6. Conveyors of this type are extensively used for moving the products of farming operations.

13.5.3 Apron Chain Conveyor If the flights in scraper conveyors are replaced by flat slats, steel plates or broads, this becomes a moving platform or an apron. This type of conveyor is used for conveying sacked materials and materials of large unit size. Design of chain conveyor The  designs of scraper conveyors are discussed here, but the designs of trolley and apron chain conveyors can be made in a similar manner. Flight height, length, and spacing depend on the expected duty of the conveyor. Flat flights are recommended for sacked materials, shallow flights for large-size materials such as ear corn and sugar beets. For  small grains and grains similar in size to small grains, flight height should be approximately 0.4 times the flight length and should be spaced approximately the length. Flight speeds vary 25–40 m/min. Low speeds should be used for materials of large granular size such as ear corn. Small granular materials such as small grains can be moved at higher speeds. Higher speeds may damage the product. Capacity should be provided by large-size flights rather than high speeds, and the capacity of a conveyor operating at an incline will be less than the capacity operating on the level.

FIGURE 13.6  Cross section of different types of flights (a) wooden conveyor, (b) steel pipe with a cylindrical flight, (c) wearing plates and (d) rollers.

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The theoretical power requirement for flight conveyors can be determined from the following equation:

Horsepower =

2vL c Wc Fc + Q(LFm + H) 4562

(13.16)

where: v = speed of conveyor, m/min Lc = horizontal projected length of conveyor, m Wc = weight of flights and chain, kg/m Fc = coefficient of friction for chains and flights Q = materials to be handled, kg/min L = horizontal projected length of loaded conveyor, m Fm = coefficient of friction for material H = height of lift, m The calculated horsepower must be adjusted to account for expected maximum load conditions, starting friction, loss in driving mechanization, variations in friction coefficients, type of power units, etc.

13.6  SCREW CONVEYORS A screw conveyor consists of a U-shaped trough in which a shaft with a continuous or interrupted spiral screw is rotated to push grain along the trough. The  screw shaft is supported by end and hanger bearings. Details of a screw chain conveyor are shown in Figure 13.7. Screw conveyors are used to handle finely divided powders; damp, sticky, heavy viscous materials; hot substances that may be chemically active; and granular materials of all types. Because of the simplicity, free from sharp recesses and cracks, and dustproof and ease in assembly, screw conveyors are used for moving food products such as powered milk and peanut butter. Although the screw conveyors are simple in design and relatively inexpensive, power requirement is high. The standard pitch screw has a pitch approximately equal to the diameter. The  screw conveyors are generally used to move granular products horizontally. However, the screw conveyor can be used to convey materials at any angle up to 90° from horizontal, but the capacity of the conveyor is reduced. The volumetric capacity of a screw conveyor depends on the screw diameter, inclination of the screw blade, speed of the blade, shaft diameter, and cross-sectional area of the loading. The theoretical capacity Q of the screw conveyor is given by

(

)

Capacity = 47.2 × D2 − d 2 × p × n

m 3 /h

(13.17)

where: Q = capacity of screw conveyor, m3/h D = screw diameter, m d = shaft diameter, m p = screw pitch, m n = rpm

FIGURE 13.7  Details of a screw chain conveyor: 1 screw diameter, 2 pitch of screw, and 3 screw length.

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Agro-Product Processing Technology

The theoretical power requirement for the screw conveyors for horizontal operation can be determined from the following equation: Horsepower =



QLWF 4562

(13.18)

where: Q = capacity of screw conveyor, m3/h L = conveyor length, m W = bulk material weight, kg/m3 F = material factor, 0.4 for paddy

13.6.1  Power Requirement for Screw Conveyors The driving power for a loaded screw conveyor can be determined from the following equation: P = PH + PN + Pl



(13.19)

where: PH = power necessary for conveying material, kW PN = driving power for the conveyor at no load, kW PI = power requirement for inclination of the conveyor, kW Power necessary for conveying the material is the product of mass flow rate m of the material; the length L of the material movement in the conveyor; and frictional coefficient λ, the progressive resistance coefficient; and is given by



PH =

m×L m×L×λ × λg = 3600 367

kW

(13.20)

where: m = mass flow rate, t/h L = length of material movement in the conveyor, m λ = progress resistance coefficient Power necessary for conveying the material at no load is low and is proportional to the screw diameter and total length of the screw. It is given by PH =



D×L 20

kW

(13.21)

where: D = nominal screw diameter, m L = length of screw, m Power necessary for inclination of the conveyor is the product of mass flow rate m of the material and the height to which the material is being conveyed and is given by



PI =

m×h m×h ×g = 3600 367

kW

(13.22)

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Materials Handling and Conveying

where: m = mass flow rate, t/h h = height, m Hence the total power requirement P is given by



P=

m × ( λL + h ) D × L + 20 367

kW

(13.23)

13.7  BUCKET CONVEYORS A bucket elevator consists of buckets attached to a chain or a belt that revolves around two pulleys, one at top and the other at the bottom. The vertical lift may range from a few meters to more than 50  m. Bucket elevators may be classified as (1) spaced bucket elevators and (2) continuous bucket elevators. Bucket elevators are efficient but more expensive. High efficiency results from the absence of frictional loss from sliding of materials on the housing. The bucket elevator’s capacity mainly depends on bucket size, conveying speed, bucket design and spacing, the way of loading and unloading, bucket, and the characteristics of the materials. Belt speed is the first critical factor to consider. Bucket elevators with belt carriers can be used at fairly high speed and the speed of the belt depends on the head pulley speed. If the belt speed is too low, the discharge of grains becomes more difficult; with too a high speed, the buckets are not fed well. In elevating of grain, the discharge from the bucket elevators is a combination of centrifugal and gravitational discharge. When a product turns around the pulley, it is influenced by two forces: centrifugal force and gravitational force. The  magnitude of the centrifugal force that is oriented outward is given by



cf =

WV 2 g× r



(13.24)

where: cf = centrifugal force W = weight of grains, kg/m V = velocity of product mass, m/s g = acceleration due to gravity, 9.81 m/s2 r = radius of the wheel plus one half of the projection of the bucket For optimum centrifugal discharge, cf and W must be equal at the point near the top of the travel and this gives



WV 2 g× r



(13.25)

V = g× r



(13.26)

W =

On simplification this gives

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Agro-Product Processing Technology

since we know 2πnr 60

(13.27)

V × 60 2πr

(13.28)

V=

We have

n=



n=



60 × g × r 29.9 = 2× π× r r



(13.29)

This equation gives the relationship between effective head wheel radius and its rpm for the most satisfactory discharge conditions. The discharge capacity of a bucket elevator Q in tons/h is given by



Q =

ρ C × F 3600 × V × × 1000 S 1000

tons/h



(13.30)

where: Q = capacity of bucket elevator, tons/h C = capacity of each bucket, l F = bucket filling factor, a constant V = elevator velocity, m/s S = bucket spacing, m ρ = material bulk density, kg/m3 The theoretical power requirement for the bucket elevator can be determined from the following equation:

Horsepower =

QHF 4562

(13.31)

where: Q = capacity of bucket elevator, kg/min H = lift of elevator, m F = factor, 1.5 for elevators loaded on the top side and 1.2 for elevators loaded on the bottom side. The  calculated horsepower must be adjusted to account for expected maximum load conditions, starting friction, loss in driving mechanization, variations in friction coefficients, type of power units, etc.

13.8  PNEUMATIC CONVEYORS The pneumatic elevator is used to move granular materials in a closed duct by a high-velocity stream of air. Pneumatic conveying is the process of conveying granular materials by floating the materials primarily in the air and then allowing them to flow to the destination through a closed pipe. Hence pneumatic conveying needs a source of air blowing or suction means for feeding product into the

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Materials Handling and Conveying

conveyor and ducts; also, a cyclone or receiving hopper for collection of the product and a feeder should feed the product at a specified rate in the pneumatic conveying system from the supply hopper. Energy is needed to transport the material from one place to another and is supplied by a mechanical method. Cyclone separators are most widely used to separate the product particles from air. A cyclone separator is a device that separates the bulk of product particles from the conveying airstream by centrifugal force. In some cyclone separators, a fabric filter is attached to remove the residual dust and fine product particles from the airstream. Advantages 1.  Mechanical simplicity means there’s only one moving part, fan, etc. 2. The initial cost is relatively low. 3. Conveying path can be random and may branch. 4. Conveying path can be changed easily. 5. A wide variety of materials can be conveyed. 6. The system is self cleaning. Disadvantages 1. High power is required. 2. The conveyed materials can be damaged. 3. Types of materials for pneumatic conveying are limited. 4. Friable or abrasive materials are not suitable for pneumatic conveying. 5. The movement of materials is unidirectional. 6. The length of pneumatic conveyors is limited. Design The selection of a fan or blower for a pneumatic conveyor is the most important aspect of a pneumatic system. Two important factors in the design of a pneumatic system are (1) supply pressure of air and (2) volumetric airflow rate. The supply air pressure depends on the pressure working length and the properties of the product to be conveyed. The volumetric flow rate depends on the necessary air velocity and pipe or duct size used in the system. Total pressure drop and horsepower requirement can be computed based on the material loading ratio, which essentially adjusts the system pressure and horsepower by adding material loading factor. The material loading factor is the ratio of the weight flow of the material to the weight flow of air. For any given material, there is a minimum transport velocity required to convey the material. Therefore, the airflow rate will depend on the size of the pipe or duct. For any given system, several different pipe or duct sizes may be used, but only one will be the most economical. Typically, material loading of 2:1–1:1 or less is acceptable for standard industrial fans, and material loadings of 6:1 may be used for pressure blowers. Once pipe has been selected, the airflow rate can be calculated based on the minimum transport velocity of the material being loaded. Table 13.1 lists some agro products and the corresponding TABLE 13.1 Common Conveying Velocities Material Dry vegetable pulp Salt Corn Flour

Velocity, m/min

Material

Velocity, m/min

1200 1600 1700 1000

Wheat Oat Sugar Cotton

1800 1400 1700 1200

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Agro-Product Processing Technology

conveying velocities. In general, materials up to approximately 800 kg/m3 can be conveyed with an air velocity of 1500 m/min. The steps to correct the system pressure for material loading are given below:

1. Find the material loading ratio, R. 2. Find the system pressure loss due to acceleration of the material Pac = R × Pvc



(13.32)

where: Pac = system pressure loss due to acceleration, m R = material loading ratio Pvc = velocity pressure of standard air, m



3. Find the system pressure loss due to lifting the material

P lf = R ×

L 8.30



(13.33)

where: Plf = system pressure loss due to lifting material, cm water L = lifting height, m

4. Find the system pressure loss due to friction Pfr = f × R ×

H 8.30



(13.34)

where: Pfr = system pressure loss due to friction, cm water f = friction coefficient

5. Find the system pressure loss due to elbows (90°) Pel = π × f × R × Pvc



(13.35)

where: Pel = system pressure loss due to elbow, cm water Add this loss for each elbow.



6. Add all losses to the system pressure requirement for airflow through the dust. The horsepower can be adjusted by adding the material loading factor to one. Then multiply the fan horsepower by this number to get the horsepower with material loading ratio. BHP = (1 + R ) × BHPstd



where BHP is the brake horsepower. This method gives reasonable estimates only but not an accurate one.

(13.36)

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Materials Handling and Conveying

13.9  HYDRAULIC CONVEYORS Moving bulk materials through pipes or channels in a stream of water is called hydraulic conveying. The hydraulic conveyors are used to move bulk materials in a closed pipes and channels by a highvelocity stream of water. The mixture of water and materials is called pulp and a pump is used to convey the pulp under pressure. This type of conveying system is widely used in food industries for conveying pulp and fruit and vegetable juices. Advantages 1. High capacity of materials can be conveyed over a long distance. 2. It is relatively simple and the running cost is low. 3. Conveying can be combined with other processes. 4. There is flexibility in selection and modification. 5. It is safe and easily controllable. 6. Maintenance costs are low. Disadvantages 1. Materials that can be handled are limited. 2. It cannot be used for cold conditions when water freezes. 3. It cannot be used in increased humidity when operating in closed conditions. 4. Disposal or recirculation of water is often difficult and or/costly. 5. Some materials degrade due to attrition. 6. Choking of pipes particularly at bends and fittings occurs. 7. Crushing and mixing cause added power-intensive operation. Design The design of a hydraulic conveying system requires knowledge of computation of system pressures and the horsepower requirement. The equation for pump work based on Bernoulli’s equation is



Pa α V2 P α V2 + g Z a + a a + η Wp = b + g Z b + b b + h f ρ 2 2 ρ

(13.37)

where: P = system pressure, m Z = height above datum level, m V = velocity of fluid, m/s η = overall efficiency of pump Wp = pump work per unit mass of fluid, m.kg/kg hf = friction head, m This equation can be used to calculate pump work from different system pressures. The system pressure in the pipeline is made up of the pressure required to overcome the resistance to the motion of the pulp in the pipeline and the velocity required to impart velocity V to the flowing mass.



Pvc = L

λ V 2 V 2  Lλ  V 2 × + = + 1 × d p 2g 2g  d p  2g

(13.38)

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Agro-Product Processing Technology

where: Pvc = velocity pressure, m λ = hydraulic loss factor dp = pipe diameter, m V = pulp velocity, m/s L = length of the pipe, m Value of λ may be taken as 0.04–0.045 for normal velocity and pulp concentration. The equation for computation of power P in kW is given by



P=

m × Wp 1000

(13.39)

where: m = mass flow rate, kg/s This equation can be used to calculate electric motor power from pump work and mass flow rate of the fluid.

13.10  GRAVITY CONVEYORS Gravity conveyors consist of a series of rollers or wheels set at level or slightly inclined to materials in boxes in particular. The materials move by gravity or by hand at level condition.

13.11 CRANES Cranes in materials handling are used for lifting or lowering a load by hook and moving it horizontally in which the hoisting mechanism is an integral part of the cranes. The cranes may be fixed or mobile. A crane thus essentially consists of a steel structure, hoisting mechanism, load handling attachment, and drive and controls. Cranes are not used extensively in agricultural processing, but the cranes are used for isolated cases.

13.12  LIFT AND CARRYING TRUCKS AND CART A lift consists of a box-type cage or car that moves vertically up and down through the designed opening in the floors called the shaft of the lift. The movement of the car is guided by guide rails vertically in the shaft. The car is suspended from the top and moves up and down by a hoisting mechanism at the top of the shaft. Lifts are used for vertical movement of materials and people between different floors in multistoried factories or buildings. Grain crops after harvesting and threshing are bagged and then transported to procurement centers, milling complexes, and to the buyers. Grains are mainly transported by animal carts, trucks, and tractor trolleys.

13.13  ROBOTIC HANDLING SYSTEM In a discrete mass manufacturing process, the feeding of raw materials or jobs into a machine for processing and removal of the finished products are an important activity and are known as materials handling in the workplace. Materials handling in the workplace is defined as the handling of the materials after they have been delivered for use at the workplace and before they are again picked up by some conventional handling process and equipment for the next operation. Materials handling

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Materials Handling and Conveying

in the workplace is not only monotonous and fatiguing, but it also involves personal safety of the operators. Moreover, the handling time saved directly increases the productivity of the manufacturing process. It is in this context that industrial robots have been increasingly used in materials handling in the workplace. According to the Robotic Institute of America, a robot is a programmable, multifunctional manipulator designed to move material, parts, or specialized devices through variable programmed motions for the performance of a variety of tasks. Materials handling by robots are of two types and these are (1) materials transfer and (2) machine loading and uploading. Materials transfer is the task of moving a part or job from one location to another within the workplace where robots are applied for this task whereas the robot-centered machine cell is the most common example of machine loading and or/ uploading. Both of these two types are applicable to the different phases of materials handling in the workplace and are finding applications in industrial processing and manufacturing. Robotic food manufacturing is a rising trend in the food industry. Example 13.1 A flat belt conveyor is used to supply 2000 boxes of agro products per hour, and the size of box is 220 × 180 × 100 mm. Calculate the belt width and the speed of the belt. Assume clearance between the boxes to be 200 mm.

Solution The 220 mm side of the box is placed along the width of the flat belt and clearance on either side of the box is 100 mm. Hence the belt width is given by Belt width  =  220  mm side of the box  +  clearance on either side of the box  =  220  + 100 × 2 = 420 mm. The maximum velocity of the flat belt is given by

V=



2000 × (180 + 200 ) = 12.67 m/min 60 × 1000

Example 13.2 A troughed belt with a 20° surcharge is used to convey 15 tons of wheat per hour. What width of belt should be used if the maximum operating speed is 150 m/min? Assume the belt incline to be 15° and density of wheat to be 600 kg/m3.

Solution Here, Q = 15 tons/h, V = 150 m/min, γ = 600 kg/m3, ϕ = 20°, and ʎ = 15°. Substituting these values in the following relationship

B=



Q V × γ × [576 × tan7.5 + 432 × tan15]

=

45 × 1000 150 × 600 × [576 × tan7.5 + 432 × tan15]

=

15000 = 0.93 m 17242.37



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Agro-Product Processing Technology

Example 13.3 A troughed belt 0.60 m wide is connected to a set of three pulleys so that the belt inclines 18° to transport paddy weighing 600 kg/m3. The belt has a clear margin of 5 cm and the surcharge angle of the paddy is 20°. a. Calculate the capacity of the belt for a belt speed of 30 m/min. b. Calculate the width of the belt for a half-capacity belt.

Solution Here, B = 0.9 m, V = 200 m/min, γ = 600 kg/m3, ϕ = 20°, and ʎ = 18°. Substituting these values in Equation (13.6), the belt capacity is given by Q = B2 × V × γ × [576 × tan(0.35ϕ) + 432 × tan λ ]



n(0.35 × 20) + 432 × tan18] = (0.6)2 × 0.50 × 0.60 × [576 × tan = 22.796 tons / h Belt width calculated for Q = 11.398 tons/h from the following relationship

B=



=

11.398 0.5 × 0.6 × [576 × tan7.0 + 432 × tan18] 11.3980 = 0.425m 63.24



Example 13.4 The effective tension of a conveyor belt is 100 Newton, the wrap angle is 220°, and the coefficient of friction is 0.45. Calculate the maximum tensile force to select the belt carcass from the manufacturer’s data. Assume the ratio between Temax and Te to be 2.0.

Solution Here, Te = 1000 Newton, ζ = 2.0, µ = 0.45, and α = 220°. α in radian is given by α=



π × 220 = 3.83 180

Substituting these values in the following relationship  cµα  Tlmax = Te ξ  µα   e − 1



 e.45×3.83   5.61  = 100 × 2.0  0.45×3.83  = 200   = 243.34 Newton e 1 −  4.61  

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Materials Handling and Conveying

Example 13.5 A  troughed belt of cross-sectional area 0.00089  m2 and length 10  m has a capacity 60  m3/h. Calculate the belt speed and the effective tension. Assume density to be 600 kg/m3.

Solution Here, Q = 60 m3/h, L = 10 m, and the cross-sectional area = 0.0089 m2. The belt speed is given by belt speed =

capacity 60 = cross-sectional area × 60 0.0089 × 60

= 112.34 m/min





Horsepower to drive empty conveyor belt speed,m/min ( A + B) × (3.281L) × 100 0.3048

Horsepower =

112.34 ( 0.2 + 0.00140 ) × ( 3.281× 10 ) × 0.3048 100

=

= 24.35





Horsepower to convey material on level



Horsepower = Q ×

0.48 + 0.01L 60 × 600 0.48 + 0.01× 10 = × 100 1000 100

= 0.1764 Horsepower to lift the material is given by



Horsepower =

lift,m t/h 10 36 × 1.015 × = × 1.015 × 0.3048 1000 0.3048 1000

= 1.1988 Therefore, the total horsepower requirement is



Horsepower = 24.35 + 0.18 + 1.20 = 25.73 The effective tension is given by



Te =

kW 25.73 × 0.745 = belt speed (112.34/60 )

= 10.24

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Agro-Product Processing Technology

Example 13.6 Design a steel chain conveyor with grain moving in the open top to operate at a 30° incline and elevate shelled corn to a height of 5 m at a rate of 40 tons/h. Assume a flight speed of 30 m per min and the density to be 600 kg/m3.

Solution Here, the cross-sectional area is B × 0.4B m2, and density of shelled corn ρ = 600 kg/m. The width of the flight is given by B=



=

Q 0.4 × relative capacity × flight speed 40, 000/(600 × 60) = 0.41 m 0.4 × 0.55 × 30

The theoretical length of the conveyor is given by

= L



5 = 10 m sin30

The load tension is given by





T = weight of material per m × (LmFm + H)

Here, the cross-sectional area = B × 0.4 B m2, length = 1 m, and density = 600 kg/m3, which gives the weight of wheat per m as

Weight per m = B × 0.4B × 600 = B2 × 0.4 × 600



= 0.412 × 0.4 × 600 = 40.34 kg/m



The load creates a tension in the chain and is given by T = weight per m × (LmFm + H)



(

)

= 40.35 × 12 × cos300 + 5 = 621.08 k g



Example 13.7 A screw conveyor is of 0.30 m of diameter and the pitch is mounted on a shaft of 0.04 m diameter; it operates at 150  rpm. The  wheat grain of density of 850  kg/m3 is conveyed. Assume actual capacity of the screw conveyor to be 50% of the theoretical capacity. Determine the actual capacity of the conveyor and horsepower requirement of the motor for a screw length 8 m if the horsepower material factor for wheat is 0.40.

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Materials Handling and Conveying

Solution Here, D = p = 0.30 m and d = 0.04 m. The theoretical capacity of the screw conveyor Q is given by

(

)

Capacity = 47.2 × D2 − d2 × p × n m3 /h

=

47.2 × 0.302 − 0.042 × 0.30 × 150 60

(

)



= 3.129 m 3 /min Hence the actual capacity of the screw conveyor is given by Actual capacity = 0.50 × 3.129 = 1.56 m3 /min



The theoretical power requirement for the screw conveyors is given by: Horsepower =



=

QLWF 4562 1.56 × 8 × 850 × 0.4 = 093 hp 4562



Hence the capacity of the motor to drive screw conveyor is 1 hp.

Example 13.8 A 5 m long screw conveyor with 0.30 m of diameter and 0.28 m of screw pitch is mounted on a shaft of 0.15 m diameter and operates at 300 rpm. The wheat grain of density of 770 kg/m3 is transported using this screw conveyor. Assume actual capacity of the screw conveyor to be 50% of the theoretical capacity. Determine the actual capacity of the conveyor and horsepower requirement of the motor for the screw conveyor if the power transmission efficiency is 50%. Assume material factor for wheat to be 0.40.

Solution Here, D = 0.30 m, p = 0.28 m, and d = 0.15 m. The theoretical capacity of the screw conveyor Q is given by

(

)

Capacity = 47.2 × D2 − d2 × p × n m3/h



=

47.2 × 0.302 − 0.152 × 0.28 × 300 60

(

)

= 4.453 m 3 /min Hence the actual capacity of the screw conveyor is given by



Actual capacity = 0.50 × 4.453 = 2.22 m3/min

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Agro-Product Processing Technology

The theoretical power requirement for the screw conveyors is given by: Horsepower =



=

QLWF 4562 2.22 × 5 × 770 × 0.4 = 0.76 hp 4562



The capacity of the motor to drive the screw conveyor is given by: = Actual power requirement



horsepower 0.76 = = 1.52 hp efficiency 0.50

Hence the capacity of the motor to drive the screw conveyor is 1.5 hp.

Example 13.9 A bucket elevator is designed for lifting maize grain vertically to a height of 25 m. Each unit of the buckets has a volume of 3.50 × 10 −3 m3 and the buckets are placed 0.40 m apart in a belt. The center-to-center distance between the head and foot wheel is 25.30 m. The diameter of the head pulley is 0.30 m and it rotates on a shaft of 0.05 m diameter. The weight of the belt and empty buckets per m is 5 kg. The density of maize is 750 kg/m3. Determine the speed and power requirement to lift 4 tons of maize per hour. Assume degree of lifting to be 80% and coefficient of friction to be 0.4 at axle of the pulley.

Solution Here, the capacity of the elevator is Q  =  4000  kg/h and density of maize is ρ  =  750  kg/m3. The length reserved for each bucket = 0.4 m. The volume of maize per min is given by



Q =

4000 = 0.089 m3 /min 750 × 60

and the volume of maize per bucket is



Volume of each bucket = 3.50 × 10−3 × 0.8 = 2.8 × 10−3 m3



The number of buckets lifted per minute is given by Number of buckets lifted per min =



=

Volume of maize lifted per min Volume of each bucket 0.089 = 31.79 2.8 × 10 −3



The linear speed of the belt is given by



Speed of the belt = Number of belts per min × length reserved for each bucket = 31.79 × 0.4 = 12.72

m/min



345

Materials Handling and Conveying The weight of the belt and pulley is given by



Weight of the belt and pulley = 2 × one way length of belt and pulley × weight per m = 2 × 25.30 × 5 = 253 kg The weight of maize on tension side is given by Weight of maize =

=

lifting height × unit bucket volume × den of maize unit bucket length 25 × 2.8 × 10 −3 × 750 = 131.25 kg 0 .4



Therefore, the total load is given by



Total load = 253 + 131.25 = 384.25 kg and the frictional resistance is given by



Frictional resistance = 384.25 × 0.4 = 153.7 kg The power lost in the friction of the head pulley is given by



Power lost in friction =

frictional resistance × actual friction height 60 × 60

The actual friction height covered per min is given by



Actual frictional height = belt velocity ×

shaft diameter 0.05 m/min = 12.7 × 0.30 pulley diameter

Therefore, power lost in the friction of head pulley is given by



Power lost in friction =

153.7 × 12.7 × 60 × 75

0.05 0.30 = 0.072 hp

The power required for lifting the maize is given by



Power lifting the maize =

4000 × 25 = 0.37 hp 60 × 60 × 75

Therefore, the total power requirement is given by



Total power requirement = 0.072 + 0.37 = 0.442 hp Hence a 0.5 hp electric motor is required.



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Agro-Product Processing Technology

Example 13.10 A pump draws mango pulp of specific gravity 1.84 through a steel pipe of 75 mm diameter from a storage tank and discharges through a steel pipe of 50 mm diameter into an overhead tank at a height of 15 m. The velocity of mango pulp in the suction line is 0.914 m/s and the frictional losses of the entire piping system are 15 m. The efficiency of the pump is 60%. Determine the developed pressure by the pump and the horsepower of the motor.

Solution The equation for pump work is Pa P α V2 α V2 + g Z a + a a + η Wp = b + g Zb + b b + hf 2 2 ρ ρ



Here, Pa = P since the pressure at the top of the storage and the discharge end is atmospheric. The kinetic energy factor α is taken to be 1. The equation above can be expressed as:

η Wp =



(

)

Vb 2 − Va 2 g + hf ( Zb − Z a ) + gc 2g



Here, Zb – Za = 15 m and the cross-sectional area of the 75 and 50 mm steel are given by



A75 =

π (0.075)2 = 4.42 × 10 −3 m3 4

A 50 =

π (0.050)2 = 1,96 × 10 −3 m3 4

and

The velocity in 50 mm pipe is

Vb =



0.914 × 4.42 = 2.10 m 4 × 1,96 × 10 −3



Then



0.60 × Wp = 15

(

)

2.102 − 0.9142 g + +3 gc 2g c

= 18 +

( 2.10

2

− 0.914

2 × 9.81

2



) = 18.18

and since 1 kgf = 9.81 N and 1 Joule = 1 Nm



Wp =

18.18 × 9.81 = 297.24 Nm/kg or Joule/kg 0.60



347

Materials Handling and Conveying The pressure developed by the pump is given by



 V 2 − Vb 2  Pb − Pa = specific gravity × 0.1×  a + (Zb − Z a ) + hf  2g  



Hence the pressure developed by the pump is





 2.062 − 0.9142  + 15 + 15  = 5.55 kg/cm2 Pb − Pa = 1.84 × 0.1×  2 × 9.81   The mass flow rate of mango pulp is m = 4.42 × 10 −3 × 0.914 × 1.84 × 1000 kg/s = 7.43 kg/s





and the power is given by P=



m × Wp 7.43 × 297.24 = = 2.20 kW 1000 1000

The power developed by the mango pulp is P = 2.20 × 0.60 = 1.32 kW

Example 13.11

A pneumatic conveyor must lift maize 15 m and move it horizontally 25 m with one elbow at 90° and it lifts 0.03 m3/min. The velocity to transport the grain is 30 m/s, and the material loading factor is 0.25. Standard brake horsepower is 112. Determine the total pressure head and the horsepower of the motor. Assume the grain to surface frictional coefficient to be 0.045 and frictional coefficient for elbow is 4.50.

Solution The steps to correct the system pressure for material loading are given below: The velocity pressure of standard air is



Pvc =

V2 30 2 = = 5.53 cm water 2g × 8.30 2 × 9.81× 8.30

Hence the system pressure loss due to acceleration of the material



Pac = R × Pvc = 0.25 × 5.52 = 1.38 cm water



and the system pressure loss due to lifting the material



Plf = R ×

L 15 = 0.25 × = 0.45 cm water 8.30 8.30

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The system pressure loss due to friction Pfr = f × R ×



25 H = 0.045 × 0.25 × = 0.03 cm water 8.30 8.30

and the system pressure loss due to elbows (90°) Pel = π × f ∗ R × Pvc = π × 4.5 × 0.25 × 5.52 = 19.51 cm water



System pressure loss is the summation of all the losses P and is given by P = Pvc + Plf + Pfr + Pel = 1.38 + 0.45 + 0.03 + 19.51





= 21.37 cm water

The horsepower can be adjusted by adding the material loading factor to one. Then multiply the fan horsepower by this number to get the horsepower with material loading and this gives. BHP = (1+ R ) × BHPstd



= (1+ 0.25) × 112 = 140 BHP



EXERCISES 13.1 A flat belt conveyor running at 15 m/min is used to supply 1000 boxes of apples in a processing plant, and the size of each box is 220 × 180 × 120 mm. Calculate the number of boxes supplied per hour. Assume clearance between the boxes to be 150 mm. 13.2 A troughed belt 0.90 m wide is connected to a set of three pulleys so that the belt incline is at 200 to transport paddy weighing 620 kg/m3. The belt has a clear margin of 5 cm and the surcharge angle of the paddy is 18°. a. Calculate the capacity of the belt for a belt speed of 115 m/min. b. Calculate the width of the belt for a half-capacity belt. 13.3 A troughed belt with a 20° surcharge is to convey 35 tons of wheat per hour. What width of belt should be used if the maximum operating speed is 110 m/min? Assume the belt incline to be 15°. 13.4 The effective tension of a conveyor belt is 100 Newton, the wrap angle 220°, and the coefficient of friction is 0.45. Calculate the maximum tensile force to select the belt carcass from the manufacturer’s data. Assume the ratio between Temax and Te to be 1.5. 13.5 A troughed belt with a 20° surcharge is to convey 45 tons of wheat per hour. What width of belt should be used if the maximum operating speed is 15° m/min? Assume the belt incline to be 15°. 13.6 Design a steel chain conveyor with grain moving in the open top that operates at a 30° incline and elevates shelled corn to a height of 7.5 m at a rate of 50 tons/h. Assume a flight speed of 30 m per min, and the capacity is 40 t/h. 13.7 A screw conveyor is of 0.35 m of diameter and the pitch is mounted on a shaft of 0.04 m diameter; it operates at 200 rpm. The wheat grain of density of 850 kg/m3 is conveyed. Assume actual capacity of the screw conveyor to be 50% of the theoretical capacity. Determine the actual capacity of the conveyor and horsepower requirement of the motor for a screw length 10 m if the horsepower material factor for wheat is 0.40.

Materials Handling and Conveying

349

13.8

A 5 m long screw conveyor with 0.30 m of diameter and 0.28 m of screw pitch is mounted on a shaft of 0.15 m diameter and operates at 350 rpm. The wheat grain of density of 770 kg/m3 is transported using this screw conveyor. Assume actual capacity of the screw conveyor to be 55% of the theoretical capacity. Determine the actual capacity of the conveyor and horsepower requirement of the motor for the screw conveyor if the power transmission efficiency is 50%. Assume material factor for wheat to be 0.40. 13.9 A  bucket elevator is designed for lifting maize grain vertically to a height of 15  m. Each unit of the buckets has a volume of 1.50  ×  10 −3  m3 and the buckets are placed 0.40 m apart in a belt. The center-to-center distance between the head and foot wheel is 12.50 m. The diameter of the head pulley is 0.30 m and rotates on a shaft of 0.05 m diameter. The weight of the belt and empty buckets per m is 5 kg. The density of maize is 750 kg/m3. Determine the speed and power requirement to lift 1.5 tons of maize per hour. Assume the degree of lifting to be 80% and the coefficient of friction to be 0.4 at axle of the pulley. 13.10 A pump draws orange pulp of specific gravity 1.24 through a steel pipe of 75 mm diameter and discharges through a steel pipe of 50 mm diameter into an overhead tank at a height of 15 m. The velocity of mango pulp in the suction line is 1.0 m/s and the frictional losses of the entire piping system is 5 m. The efficiency of the pump is 65%. Determine the developed pressure by the pump and the horsepower of the motor. 13.11 A pneumatic conveyor must lift wheat 10 m and move it horizontally 20 m at the rate of 15 m/s. It has one elbow at 90° and lifts 0.045 m3/min. Determine the pressure drop across the fan and air horsepower. Standard brake horsepower is 112. Determine the total pressure head and the horsepower of the motor. Assume the grain to surface frictional coefficient to be 0.045, and the frictional coefficient for the elbow is 4.50.

BIBLIOGRAPHY Bolz, H. A. and Hagemann, G. E. 1958. Materials Handling Handbook. Roland Press, New York. Henderson, S. M. and Perry, R. L. 1976. Agricultural Process Engineering, 3rd edition, AVI Publishing Company, Connecticut. Ray, S. 2017. Introduction to Materials Handling, 2nd edition, New Age International, New Delhi, India. Sahay, K. M. and Singh, K. K. 1994. Unit Operations in Agricultural Processing, 1st edition, Vikas Publishing House, New Delhi, India. Stocker, H. E. 1951. Materials Handling. Prentice Hall.

14

Process Dynamics and Control

This  chapter presents an introduction to process control; dynamic behavior of first-order and ­second-order processes such as transfer function; second-order processes such as transient responses and performance indices; and frequency analysis such as Bode plots and Nyquist plots, frequency response of controllers, controller design, and process modeling and control. It also presents the use of control theory to design process control systems to speed up production and quality of processed agro products.

14.1 INTRODUCTION To a great extent, the art of automatic control systems dominated our modern way of life. The automatic toaster, the alarm clock, the thermostat, and the microwave oven for warming food have all influenced the current way of our life. Even the applications of automatic control systems extend from nuclear work and guided missiles to agro-product processing plants. Although it is expected that in the near future technological development will make it possible to travel to outer space and other planets, the processing engineers are slow to use control theory to design process control systems to speed up the production and quality of processed agro products on this planet. Thus, it is imperative that processing engineers have an understanding of both theory and practice of process control. The two main aspects of this chapter are process dynamics and process control. The term process dynamics refers to the dynamic behavior of the process, but the primary objective of process control is to maintain the desired conditions of the process safely and efficiently while satisfying the environmental and product quality requirements. To achieve these goals, we need to use feedback control systems.

14.2  FEEDBACK CONTROL SYSTEMS Systems may be defined as a combination of components or parts for a common goal. They may be classified as open-loop systems and feedback systems. In open-loop systems, the input is independent of output; in the case of feedback systems, the input is influenced by the output. Figure 14.1 shows a simple feedback control system.

14.3  BLOCK DIAGRAMS The first step in the analysis of control systems is to draw a physical diagram showing the flows of the different interacting components influencing the performance of the system, and this diagram is called a block diagram. For control systems, it is helpful to use a block diagram to represent the functional relationships among the parts of the systems. Each part is represented by a block showing its input from another block and the output to the other block. The mathematical input-output relationship is called transfer function, and the blocks are connected by arrows showing the direction of the information flow. A block diagram of a simple process control system is shown in Figure 14.2. Blocks may be rearranged to simplify the algebraic manipulations of the transfer functions. Block diagrams of a complex system may have multiple feedback loops as well as several inputs. Figure 14.3 shows a double-loop block diagram of a simple process control system, and Figure 14.4 shows its simplified single-loop form of the block diagram.

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FIGURE 14.1  Simple feedback control system.

FIGURE 14.2  Block diagram of a simple process control system.

FIGURE 14.3  Block diagram of a multi-loop simple process control system.

FIGURE 14.4  Block diagram of a process control system reduced to a single loop.

14.4  DYNAMIC BEHAVIOR OF FIRST- AND SECOND-ORDER SYSTEMS 14.4.1  First-Order System A thermometer is a first-order system, and its response can be described by a first-order differential equation. Consider a mercury thermometer suddenly immersed in a water bath of temperature T1. Assume the heat capacity of the glass to be negligible and the mercury is at a uniform temperature T2. The system is shown in Figure 14.5. The transient heat balance is given by

Heat in − Heat out = Heat accumulated

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FIGURE 14.5  Thermometer, a first-order system.



U oc A(T1 − T2 ) = mC pm

dT2 dT =C 2 dt dt

(14.1)

where: m is the mass of mercury Cpm is the specific heat of mercury Uoc is the overall heat transfer coefficient A is the surface area of the mercury thermometer t is the time Rearranging Equation 14.1 we have



mC pm dT2 + T2 = T1 U oc A dt

(14.2)

dT2 + T2 = T1 dt

(14.3)

τ

The term τ = (m × Cpm /UA) has unit of time and is called the time constant of the system. The time constant is a product of a resistance to heat flow, 1/UA or R, and a capacity of heat storage, m × Cpm or C. In an electrical system, the time constant is a product of resistance and capacitance. In firstorder systems, the time constant can be obtained directly from the resistance and capacity of the system without writing the material balance equation. With zero initial condition, the Laplace transform of the Equation 14.3 gives

τsT2 + T2 = T1

(14.4)

This Equation 14.4 is expressed as the ratio of output to input that gives the transfer function of the system and is given by

T2 1 = T1 τs + 1

14.4.2  Step Input Let us consider that the input is suddenly increased from 0 to U at t = 0.

(14.5)

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For T1 = U we can write Laplace transform of U as T1 ( s ) =



U s

and we can also write T2 =



U s( τs + 1)

The actual temperature T2 is given by  U   1 −1  T2 = −1   = U  s(τT + 1)  T s( 1) τ +    



(14.6)

and the solution of Equation (14.6) using the table of Laplace transform is T2 = U(1 − e − t / τ )



(14.7)

14.4.3 Ramp Input The response for a ramp input T1 = bt or T1 = b/s2 is given by   U  − t/τ 1  T2 = −1  2  = bτ  e + τ − 1  s ( 1) τ + T    



(14.8)

When the transient term is negligible, the output lags the input by a time equal to the time constant T2 ≅ bt − bτ = b( t − τ)



(14.9)

14.4.4  Sinusoidal Input Now consider the response of the first-order system to a sinusoidal input: For T1 = Tmax sin ωt or T1 = Tmaxω/(s2+ω2)

  T 1  Tmax −1  1 T2 = ℑ −1  2 max 2  = ω   (1 + s2 /ω2 )(τs + 1)  (s ) ( s 1) ω τ + +    

(14.10)

The inverse transform gives the following solution

T2 =

Tmax ω

 Tmax ω2 − t/τ ω sin (ωt − ϕ)  e +   2 2 (1 + τ2 /ω2 )1/2 ′   1 + τ ω

(14.11)

where Φ = tan−1τω Once the transient response dies out, the output becomes a sinusoidal response with an amplitude of Tmax/(1 + τ2ω2)1/2 and a phase angle of Φ°, and this steady-state response is called the frequency response. A complete solution of Equation (14.10) is required to know how many cycles are needed to die out the transient response before the response becomes a true sine wave.

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14.4.5  Second-Order System A manometer is a second-order system, and its response can be described by a second-order differential equation. Mass suspended from a spring is a classic example of a second-order system. But the manometer is considered here because of the familiarity of the process engineers with a manometer. Consider the U-tube manometer as shown in Figure 14.6. Assume the frictional pressure drop is proportional to the velocity of the fluid in the manometer, and assume that all the fluid accelerate uniformly. The force balance is given by

 ALρ d 2 h g  dh = A  P − 2hp  − RA gc dt 2 g dt c  

(14.12)

where: A is the cross-sectional area ρ is the liquid density P is the applied pressure R is the frictional resistance h is the half of the liquid level difference t is the time With laminar flow, the resistance is given by the Hagen-Poiseuille equation:

∆P 32Vµ = 2 Lc D gc

or R =

32Lµ 2ρg

(14.13)

Substituting into Equation 14.12 and rearranging gives

L d 2 h 16Lµ dh Pg + + h = c = hi 2 2 2gc dt 2ρg ρg D dt

FIGURE 14.6  U-tube Manometer, a second-order system.

(14.14)

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Equation 14.14 may be written in the standard form

1 d 2 h 2ξ dh + + h = h i ωn 2 dt 2 ωn dt

(14.15)

where: ωn is the natural frequency, rad/sec ξ is the damping coefficient The system response may be underdamped (ξ  1). With a damping coefficient less than 1, the system output overshoots the final value and oscillates before coming to equilibrium. The solution of the system for underdamped condition ξ  1 is given by

h 1 = 1+ ( τ a e − t / τa − τ b e − t / τ b ) τ b − τa hi



(14.18)

14.5  THE LAPLACE TRANSFORM The  Laplace transform is a mathematical tool that provides a systematic and relatively simple method of solving linear differential equations. Differential equations are transformed into linear algebraic equations  by replacing time t by a variable s in a complex plane. Solving these algebraic equations and performing inverse transformation gives the solution of the original equation. The Laplace transform of function f(t) is defined as:

F(s) =  f (t )  =

∫

∞

f (t ) est dt



(14.19)

0

where F(s) is the function of the complex variable s and £ denotes the transformation process. The inverse Laplace transform is written as

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f (t ) =



1 2πj

∫

σ + j∞

σ − j∞

F(s) est ds

(14.20)

Each term in Equation (14.19) is transformed separately.  f1 (t ) + f2 (t )  = F1 (s) + F2 (s)



(14.21)

When a time function is multiplied by a constant, the transform is multiplied by the same constant.   kf1 (t )  = k F(s)



(14.22)

Table 14.1 gives some important transforms. The Laplace transform of the first derivative is ℑ f ′(t )  = s F(s) − f0



(14.23)

where f0 is the initial value of f(t) For second- or higher-order derivatives, the highest power of s in the transformed equation corresponds to the order of the derivative.  f ′′(t )  = s2 F(s) − sf0 − f0′



(14.24)

TABLE 14.1 Some Important Laplace Transforms f(t) Step function, u(t) e − at sin ωt cos ωt e − at f(t) tn d (k) f(t) k) f ( ) (t) = dt (k)

∫

F(s) 1 s 1 s+a ω s 2 + ω2 F(s + a) 1 s+a n1 s n+1 s k F(s) − s k −1f(0 + ) − s k −2 f ′(0 + .) − ....f (k −1) (0 + )

t

− f(t)dt

Impulse function δ(t)

F(s) + s

∫

0

−∝

f(t)

s

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In most of the process control problems, the initial value of the function and the initial values of the derivatives are zero. The transform of an integral has s in the denominator is zero.  ℑ−1 [ f ( t ) ] = ℑ  



t



∫ f (t) dt  = 0

F(s) s

(14.25)

The Laplace transform of a function with a delay L is expressed as ℑ [ f ( t − L) ] = e − Ls F(s)



(14.26)

where ℑ [ f ( t − L) ] = 0 for t < L

Example 14.1

Transform the following equation and determine transfer function Y(s)/X(s).

Solution

7

d2y dy +3 + 5y = 10 x dt 2 dx

Taking Laplace transform, we can write



7s2 Y(s)−s y 0 −y′0  +3 s Y(s)−y 0  +5 Y(s)=10 X(s) For initial values of y0 = 0 and y0′ = 0 we have the following transfer function: Y(s) 10 = 2 X(s) 7 s + 3 s + 5



14.6  TRANSFER FUNCTION A block in the block diagram represents a function or dynamic characteristics of the component. The ratio of the transform of the output to the transform of the input is called the transfer function. The transfer function of a system or element represents the relationship describing the dynamics of the system under consideration. Figure  14.7 shows a block diagram of a system with a negative feedback control. The  equations describing this system in terms of transfer variables are:

C ( s ) = G ( s ) .E ( s )

(14.27)

B ( s ) = H ( s ) .C ( s )

(14.28)

E ( s ) = R ( s ) − B ( s )

(14.29)

= R ( s ) − H ( s ) . C ( s )

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FIGURE 14.7  Block diagram of system with a negative feedback.



C (s ) = G (s ) R (s ) − G (s ) H (s ) C (s )

(14.30)

Combining these equations, the overall closed-loop transfer function is given by

C(s) G(s) = R(s) 1 + G(s)H(s)

(14.31)

The open-loop function, independent of where the loop is opened, is G ( s ) H ( s ). The characteristic equation of the closed-loop system is obtained from the denominator of the control ratio:

1 + G(s)H(s) = 0

(14.32)

14.6.1 Routh–Hurwitz Stability Criterion Consider the system represented by the transfer function:

C(s) b0 + b1s + b2s2 + b3s3 +  + b ms m = R(s) a 0 + a1s + a 2s2 + a 3s3 +  + a ns n

(14.33)



(a 0 + a1s + a 2s2 + a 3s3 +  + a nsn )C(s) = (b0 + b1s + b2s2 + b3 s3 +  + b msm ) R(s)

(14.34)

To find the transient response of this system, the complementary function is determined by putting R(s) = 0, and since C(s) ≠ 0,

(a 0 + a1s + a 2s2 + a 3s3 +  + a nsn = 0

(14.35)

This equation, the denominator of the closed-loop transfer function equation is zero, is called the characteristics equation. The  Routh-Hurwitz stability criterion states that the system whose characteristic equation  is stable if and only if all the elements in the first column of the array formed by the coefficients in the below manner have the same algebraic sign. If the elements in the first column are not all of the same sign, then the number of the changes of sign in that column equals the number of roots with positive real parts. Consider the following equation:

a nsn + a n −1sn −1 + a n −2sn −2 +

+ a1s1 + a 0 = 0



(14.36)

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The procedure for constructing the Routh-Hurwitz array is as follows: Step 1: Arrange the coefficients of the given equation in two rows in the following manner: sn s n −1



a n a n −2 a n −4 . . . a n −1 a n −3 a n −5 . . .

Step 2: Form the third row from the first and second rows, the fourth row from the second and fifth, and continue the procedure of forming a new row from the two preceding rows until only zeros are obtained. In general, an array of n+1 rows result with the last two rows each containing a single element. sn s n −1 s n −2 s n −3 .

an a n −1 b n −1 c n −1 .

a n −2 a n −3 b n −3 c n −3 .

a n−4 a n −5 b n −5 c n −5 .

.

.

.

.

.

.

.

.

s



0

h n −1



where:

b n −1 =



(a n −1 )(a n −2 ) − a n (a n −3 ) −1 a n a n −2 = a n −1 a n −1 a n −1 a n −3

b n −3 =



1 an

a n −4

a n −1 a n −1 a n −5



and

c n −1 =



1 a n −1

a n −3

b n −1 b n −1

b n −3



Example 14.2 Apply the Routh-Hurwitz stability criterion to the following equation to find the stability of the system:



s4 + 7 s3 + 17 s2 + 17 s1 + 6 = 0

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Solution Following the procedure outlined above, we can find an array of coefficients as follows: 1st row: 2nd row:

1 7

17 17 6

3rd row:

14.58

4th row:

14.12

5th row:

6

6

Since all the coefficients in the first column are of the same sign (positive), the given equation has no root with positive real parts and the equation represents a stable system.

14.7  TRANSIENT RESPONSE System transient or time performance is the response of prime importance for control systems. Initially, it is important to determine whether the system is stable or not utilizing the Routh-Hurwitz stability criterion. If the system is stable, the response to a specific input signal will provide several measures of performance. Since the actual input to the system is usually unknown, a standard test signal is normally used. The standard test input signals commonly used are (1) step input, (2) ramp input, and (3) parabolic input. The equations representing these signals are given in Table 14.2. The ramp input is the integral of the step input, and parabolic is simply the integral of ramp input. The unit step input is the easiest to generate and evaluate and is usually chosen for performance tests.

14.7.1  Performance of Second-Order System Let us consider a single-loop second-order system and determine its response for a unit step unit input. Utilizing the generalized notation we have Y(s) =



ωn 2 R(s) s + 2ζωns + ωn 2 2



(14.37)

With a unit step input, we can write Y(s) =



ωn 2 , s(s +2ζωns + ωn2 )

(14.38)

2

For which the transient output as obtained using the Laplace transform is 1 y( t ) = 1 − e − ζωn t sin(ωnβt + θ) β



(14.39)

TABLE 14.2 Test Input Signals Test Signal Step Ramp Parabolic

R(s) r(t) = 1, t > 0 = 0, t  0 = 0, t  0 = 0, t  0) in the numerator of a transfer function causes the sinusoidal output of the process to lead the input (ϕ > 0); hence, a left half-plane zero is often referred to as a process lead. Now consider the amplitude ratio and phase angle. Substituting s = jω into G(s) = 1 + τs we can write G(jω) = jωτ + 1

(14.55)



AR = G(jω) = ω2τ2 + 1

(14.56)



ϕ = ∠G(jω) = tan −1(ωτ)

(14.57)

from which we can write

Therefore, a process zero contributes a positive phase angle that varies between 0° and +90°. The output signal amplitude becomes very large at high frequencies, which is a physical impossibility.

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Consequently, a process zero is always found in combination with one or more poles. The order of the numerator of the process transfer function must be less than or equal to the order of the denominator. 14.8.1.5  Time Delay e − jωθ The  time delay is the remaining important process element to be analyzed. Frequency response characteristics of the time delay can be obtained by substituting s = jω: G(jω) = e − jωθ



(14.58)

and this can be expressed by substituting Euler’s identity G(jω) = cos ωθ − jsinωθ

(14.59)



AR = G(jω) = cos2θ + sin 2θ

(14.60)



 sinθ  φ = ∠G(jω) = tan −1  −   cosθ 

(14.61)



φ = −ωθ

(14.62)

We can write

Since ω is expressed in radians/time, the phase angle in degrees is −180ωθ/π. Figure  14.10 shows the Bode plot for the time delay. The  phase angle approaches −∞ as ω becomes large. This unbounded phase lag is an important attribute of a time delay and is detrimental to closedloop system stability.

FIGURE 14.10  Bode plot for time delay e −θ s .

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14.8.1.6  Bode Stability Criterion Before stating the Bode stability criterion, we should introduce two important definitions:

1. Critical frequency ωc is defined as the value of ω for which ϕOL(ω) = −180°. This frequency is also referred to as the phase crossover frequency. 2. Gain crossover frequency ωg is defined as the value of ω for which AROL(ω) = 1.

Assume that open-loop frequency response has only a single critical frequency ωc and a single gain crossover frequency ωg. Then the closed-loop system is stable if AROL(ω)  1. The gain margin provides a measure of relative stability. Now consider phase margin of the Bode plot. The angle ϕg denotes the phase angle at the gain crossover frequency ωg where a = 1. The phase margin is defined as:

γ = 180 + ϕg

(14.64)

The  phase margin also provides a measure of relative stability. Figure  14.11 illustrates the gain margin and phase margin for the Bode plot. In general, a well-tuned controller should have a gain margin between 1.7 and 4.0 and a phase margin between 30° and 45°.

FIGURE 14.11  Illustration of gain margin and phase margin for Bode plot.

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14.8.2 Nyquist Criterion A Nyquist plot is an alternative representation of frequency response information and is an Argand diagram of the complex function G(s) H(s) in which frequency ω appears as an implicit parameter. A Nyquist diagram of G(s) H(s) can be constructed directly from G(jω) H(jw) and ∠G(jω) H(jω). On the other hand, a Nyquist plot is more compact and sufficient for many important analyses—for example, determining the system stability. Most of the recent interest in a Nyquist plot has been in connection with designing multi-loop controllers and for sensitivity analysis. Consider the following transfer function

G(s) H(s) =

c(s − z1 ) (s − z 2 ) (s − z n ) (s − p1 ) (s − p2 ) (s − p m )

(14.65)

where: z is the zero of transfer function p is the pole of transfer function c is the constant. The Nyquist method is to let the Laplace variable s take the value jω and to plot a locus of the transfer function G(jω) H(jω) on an Argand diagram for ω ranging from −∞ to + ∞. It can be shown that the number of clockwise encirclements of the origin made by the G(jω) H(jω) locus (N) is equal to the number of zeros with positive real parts (Z+) minus the number of poles of G(s) H(s) with positive real parts (P+), i.e.,

N = Z + − P+

(14.66)

Likewise, the number of clockwise encirclements of the origin made by the 1 + G(jω) H(jω) locus is equal to the number of zeros of 1 + G(s) H(s) with positive real parts minus the number of poles of 1 + G(s) H(s) G(s) H(s). The latter result may also be obtained by plotting instead G(jω) H(jω) but observing encirclement of the −1, 0 point, and this is the procedure adopted in the Nyquist approach (Figure 14.12). For stability, there must be no zeros of 1 + G(s) H(s) with positive real parts (Z+ = 0). The rule becomes: For a closed-loop system to be stable, the locus of the open-loop frequency response function G(jω) H(jω), when plotted on the complex plane, shall encircle the −1, 0 point in an anticlockwise direction a number of times equal to the number of poles of G(s) H(s) with positive real parts. If there are no poles with positive real parts, then the rule simplifies to the requirement that the G(jω) H(jω) locus must not encircle the −1, 0 point.

FIGURE 14.12  Plots of 1 + G(jω) H(jω) and G(jω) H(jω).

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14.8.2.1  Gain Margin and Phase Margin The gain margin on a polar plot is given directly by –a where a is the value of G(ω)H(ω) when the curve cuts the negative X(ω) axis, since the margin is referred to as a gain of 1. The phase margin is found by drawing a circle of unit radius about the origin and drawing the radius from the center to the point where G(jω)H(jω) cuts the unit circle. The phase margin is then the angle between this radius and the negative X(ω) axis, positive below the axis, negative above.

14.9  FREQUENCY RESPONSE OF CONTROLLERS In order to use frequency response analysis to design control systems, the frequency-related characteristics of feedback controllers must be known for the most widely used forms of the PID controller.

14.9.1  Proportional Controller Consider a proportional controller with positive gain G c (s) = K c



(14.67)

In this case, G c (jω) = K c is independent of ω. Therefore,

AR = K c

(14.68)

ϕ = 0

(14.69)

and

14.9.2  Proportional–Integral Controller Consider a proportional integral controller has the following transfer function: 1    τIs + 1  G c (s) = K c 1 +  = K c  τ s  s τ I    I 



(14.70)

Substituting s = jω we have

 1  j  G c (jω) = K c 1 +  = K c 1 − τ j ω ωτ I Ι   

  

(14.71)

Thus, the amplitude ratio and phase angle are

AR = G c (jω) = K c 1 +

(ωτI )2 + 1 1 = Kc 2 (ωτI ) ωτI

(14.72)



φ = ∠Gc (jω) = tan −1 ( − 1/ωτI ) = tan −1 (ωτI ) − 90

(14.73)

On the basis of Equations  14.56 and 14.57, the integral action dominates at low frequencies. As ω → 0, AR → ∞, and ϕ → 90°. At high frequencies, AR = Kc and ϕ = 0°, neither is a function of ω in this region.

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14.9.3 Ideal Proportional–Derivative Controller The ideal proportional–derivative controller is rarely used in actual control systems, but it is a component of PID control at high frequencies. Its transfer function is G c (s) = K c (1 + τDs)



(14.74)

The frequency response characteristics are similar to those of an LHP zero:

AR = K c (ωτD )2 + 1

(14.75)



φ = tan −1 (ωτD )

(14.76)

14.9.4  Proportional–Derivative Controller with Filter The PD controller is most often realized by the transfer function  (τ s + 1  G c (s) = K c  D   ατD + 1 



(14.77)

where α has a value in the range of 0.05–0.2. The frequency response for this controller is AR = K c



(ωτD )2 + 1 (αωτD )2 + 1

(14.78)

φ = tan −1 (ωτD ) − tan −1 (αωτD )

(14.79)

The pole in Equation 14.77 bounds the high-frequency asymptote of the AR.

14.9.5  Parallel PID Controller The parallel PID controller can be developed both in parallel and series forms, and either of these versions exhibits features of both the PI and PD controllers. The simpler version is the following parallel form:

 1 + τ I s + τ I τ Ds 2    1 Gc (s) = K c  1 + + τ Ds  = K c   τ Is  τ Is   

(14.80)

Substituting s = jω and rearranging gives

    1 1 Gc (s) = K c  1 + + jωτD  = K c 1 + j  ωτD − ωτI jωτI    

  

(14.81)

The controller settings are Kc = 2, τf = 10 min, τD = 4 min, and α = 0.1. The phase angle ranges from −90° (ω →0) to + 90° (ω → ∞). By adjusting the values of τf and τD, we can change the shape and location of the notch in the AR curve. Decreasing τf and increasing τD narrows the notch, whereas the opposite changes broaden it. The center of the notch is located at ω = 1/ τI τD and AR = Kc. Varying Kc merely moves the

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amplitude ratio curve up or down, without affecting the width of the notch. Generally, integral time τf should be larger than τD, and typically τI = 4τD.

14.9.6  Series PID Controller The simplest version of the series PID controller is  τ s +1  G c (s) = K c  I  ( τDs + 1)  τIs 



(14.82)

This controller transfer function can be interpreted as the product of the transfer functions for PI and PD controllers. Since the transfer function (Equation 14.82) is physically unreliable and amplifies high-frequency noise, we consider a more practical version that includes a derivative filter.

14.9.7  Series PID Controller with a Derivative Filter The series PID controller with a derivative is given by  τ s + 1   τ Ds + 1  G c (s) = K c  I    τIs   ατDs + 1 



(14.83)

where 0.05