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T.C. Robber t s
Systems
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© 2002 by CRC Press LLC No claim to original U.S. Government works International Standard Book Number 1-56676-969-8
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Library of Congress Cataloging-in-Publication Data Robberts, Theunis C.
Food plant engineering systems / by Theunis C. Robberts. p. cm.
Includes bibliographical references and index. ISBN 1-56676-969-8 (alk. paper)
1. Food processing plants--Equipment and supplies. I. Title. TP373 .R55 2002
Dedication
To Marietjie for all the encouragement and to Marelize
Bruner for doing a splendid job with the artwork.
Preface
This book deals primarily with the ancillary equipment used in processing industries. Many books are dedicated to describing plants in specific industries or general plants in related industries. Most disregard the importance of ancillary equipment, such as motors, pumps, pipes, boilers, valves and controllers, necessary for the plant to operate. This book tries to fill the need for a text dealing not only with the bits and pieces that keep systems running, but also with how the peripheral pieces of a processing plant fit within a bigger picture.
A production line needs to be assembled to manufacture a product or range of products. Such a line will include many subsystems that must merge into an inte-grated system. All parts of the system are important, and the obvious parts receive much attention. Some parts, such as the peripherals or services including pumps, boilers, power transmission, water treatment, waste disposal, efficient lighting, and many others, may be completely forgotten. These parts of the operation are as important as the actual processing machines; without them, the system will collapse. Any plant must conform to the system* approach.
The information in this book is gathered from diverse sources and will be of particular value to plant operators, production managers, and students, who will gain understanding of the ancillary and peripheral components of the production line, environmental control systems, the building and the facilities, and the way in which various parts of a plant interact to increase overall production, even though individual parts may operate at less than their optimal levels. This will lead to a better under-standing of the requirements for true plant optimization.
Manufacturing systems can be broken down into tasks or activities. Some of these tasks will be performed in sequence, while others will be carried out simul-taneously. When you build a house, there is a sequence — foundation, followed by walls, floors and roof. Once the walls are done, the windows can be put in place while the furnace is installed in the basement. Any delay in a sequence of events will delay the project. A simultaneous activity that can be finished with time to spare will not delay the project if it is done within the allotted time. Once the slack or time available is at the point where a further delay will delay the project, all the slack has been used, and the activity becomes critical.
When a system is analyzed with the Critical Path Method (CPM) or Program Evaluation and Review Technique (PERT), one can identify the activities or opera-tions of the plant that have no slack. Keeping these critical elements optimally operational will ensure better balance in the system and optimized production. It is, however, a technique better taught in Operations or Project Management. Thus, this book will describe the features and characteristics of the service systems that allow
* Defined by Webster’s as “a regularly interacting or interdependent group of items forming a unified whole.”
for implementation during line upgrading or replacement. Just remember that slack can disappear rapidly in the case of an unforeseen breakdown.
For maximum sales and optimum profits, manufacturers try to produce a commodity at the lowest possible cost, as close as possible to what consumers believe they require. Good marketing strategies can convince consumers that they require the goods that you produce, such as cell phones, palm pilots, and many other successful products. The balance between value for money and profit should be such that it leaves the consumer with a sense of “delight” about the purchase. The price that a consumer is willing to pay has nothing to do with the input cost. The objective of the manufacturing system is to produce the desired or required product with the least manpower, equipment, energy, lost motion, and, therefore, cost. Systems may be simple or complex. In most systems, there are one or more subsystems, all supposedly working together in proper sequence and each con-tributing to the total effort.
The automobile is a good example of multiple subsystems working together in a system. The fuel subsystem consists of a fuel tank, a fuel line, a pump, and the injector (or carburetor), which mixes the gasoline with air in the proper proportion to create an ideal explosive vapor. The explosion provides the pressure to push a piston that is connected to the rest of the power train. The power train includes the engine, clutch, transmission, driveshaft, differential, axle, and wheel. The elements mentioned may have different components that make the subsystem operate properly, but everything must work together efficiently to accomplish the desired outcome.
Engineers and line and production managers, therefore, have the task of ensuring that all elements of the system are properly balanced to accomplish the job efficiently. If the driveshaft in an automobile was unable to deal with the torque, it would break and stop the operation, although all other elements would be in perfect working condition. On the other hand, if the element was too large, it could be unwieldy and inefficient. The same matter of judgment and proper design must be carried out on all of the machines, the building, and every part of the processing plant to obtain proper balance. The system components must work together to accomplish the required task, with each unit or subsystem functioning efficiently and dependably.
A system requires dependability of the operation. This is a function of continuous and preventative maintenance. Preventative maintenance requires data to predict when a component will fail. To optimize serviceability of electromechanical com-ponents prior to replacing them, predictive maintenance that will account for mea-sured vibration monitoring, operating dynamic analysis and infrared technology can be used. The primary focus of such a predictive maintenance system will be on the critical process systems. The operating system that produces the product that gen-erates revenue is more important than the auxiliary equipment.
Getting all the units to work together as a well-orchestrated system is what design and management of manufacturing are all about. This starts with subsystem optimization that will eventually be combined in total optimization, which focuses on the bits and pieces of equipment that keep the plant going. Most of these pieces of equipment are taken for granted and are overlooked when systems are designed or maintenance schedules are written.
Another objective of this book is to help the reader understand some of the terms frequently used regarding processing lines and give some of the principles underlying the terms. This book is written so that students and manufacturing personnel with only basic mathematical knowledge and a simple calculator will benefit from it. The examples were chosen to lead the reader to a full understanding of some of the mathematics that are expected in engineering texts.
Chapter 1 covers the language of processing systems, units and dimensions, and how to describe observations and operating parameters. Chapter 2 deals with general calculations including mass and energy balances. Chapter 3 addresses the properties of fluids, which leads to Chapter 4 about pumps and piping, and finally to a brief discussion of thermodynamics (Chapter 5).
Chapter 6 deals with electrical systems, including motors, starters, electrical heating, and lights. Chapters 7 and 8 deal with heating systems and steam generation, respectively. In Chapter 9, cooling and refrigeration systems are investigated. Water and waste and material handling systems are covered in Chapters 10 and 11.Chapter 12 concentrates on site, foundations, floors, walls, roofs, drains, insulation, piping, and electrical systems.
An overview of safety and EPA regulations is given in Chapter 13. The appendix contains a number of conversion tables and an introduction to mathematics.
The Author
Theunis C. Robberts, Ph.D., is the program director for the manufacturing program at the University of Minnesota, Crookston. He teaches Principles of Engineering, Product Development, Quality Systems, Cereal, Meat and Dairy Processing and Global Foods.
He received his undergraduate qualification in Food Technology at Pretoria Technikon, his Masters Dip. Tech. from Cape Technikon in South Africa and his Ph.D. from Warnborough University in England.
He did sorghum beer research at the National Food Research Institute of the CSIR in South Africa from 1970 through 1978. Since 1978, he has held various teaching positions at different polytechnics and universities. At the Technikon Wit-watersrand, he initiated and built a large pilot plant. An old building on campus was converted into a functional pilot plant facility. The processes include baking, milk, ice cream and cheese, meat products, beer making, and a canning line. The retort, exhaust box, steam cooker, and vacuum jam pot were built from his designs. The facility also has a walk-in freezer and refrigerator, a spray dryer, a short barrel extruder, powder mixing equipment and a form-and-fill packing line.
His consultation and product development work on an international level took him from frozen vegetables to anticariogenic beverages for toddlers. In one of the larger projects done for the government of Mozambique, he refit a bakery, changed product formulations, and made the quality systems operational. He holds a patent on a low-calorie chocolate beverage.
Dr. Robberts received the Ernest Oppenheimer Memorial Trust Award for senior researchers in 1992. He has published articles and presented lectures at several scientific meetings, including IFT, where he is a professional member.
Contents
Chapter 1 Units and Dimensions
1.1 Introduction ... 1 1.2 History ... 1 1.3 Terms ... 2 1.3.1 Base Units ... 2 1.3.1.1 Definitions... 3 1.3.1.2 Supplementary Units ... 4 1.3.2 Derived Units ... 5 1.3.2.1 Definitions... 5 1.4 Dimensional Analysis ... 6
1.5 Units for Mechanical Systems... 8
1.6 Conversions and Dimensional Analysis ... 11
1.7 Problems... 12
Chapter 2 General Calculations 2.1 Introduction ... 13
2.2 Basic Principles... 13
2.2.1 Conservation of Mass ... 13
2.2.1.1 Mass Balances ... 13
2.3 Problems... 25
Chapter 3 Properties of Fluids 3.1 General Description ... 27
3.2 Physical Properties of Flow ... 27
3.2.1 Pressure ... 27 3.2.2 Density (ρ) ... 28 3.2.3 Specific Mass (w)... 28 3.2.4 Specific Gravity (SG)... 29 3.2.5 Bulk Modulus (k)... 29 3.2.6 Ideal Fluid ... 29 3.2.7 Viscosity ... 29
3.2.8 Temperature Effects on Fluid Flow ... 31
3.2.9 Shear-Thinning Fluids... 32
3.2.9.1 Bingham Plastics ... 32
3.2.9.2 Pseudoplastics ... 32
3.2.10 Shear-Thickening Materials... 33
3.3 Surface Tension and Capillary Action ... 33
3.3.2 Atmospheric Pressure ... 34
3.4 Hydrostatics... 35
3.4.1 Pressure Intensity ... 35
3.5 Basic Concept of Fluid Motion ... 37
3.5.1 Introduction ... 37
3.5.2 Conservation of Mass ... 38
3.6 Types of Flow ... 39
3.6.1 Turbulent and Laminar Flow ... 39
3.6.1.1 Reynolds Number ... 40
3.6.2 Rotational and Nonrotational Flow ... 42
3.6.3 Steady and Unsteady Flow ... 42
3.6.4 Uniform and Nonuniform Flow... 43
3.7 Conservation of Energy in Liquids... 43
3.7.1 Potential Head ... 43
3.7.2 Pressure Head... 43
3.7.3 Velocity Head ... 44
3.7.4 Energy Changes and Flow ... 44
3.7.5 Frictional Losses ... 46
3.7.5.1 Frictional Losses in Straight Pipes... 46
3.7.5.2 Fittings in Pipes ... 49
3.7.6 Mechanical Energy Requirement... 49
3.8 Problems... 50 Chapter 4 Pumps 4.1 Introduction ... 51 4.2 General Principles ... 51 4.3 Pump Efficiency ... 52 4.4 Centrifugal Pumps... 53 4.4.1 Operation ... 54
4.4.1.1 Net Positive Suction Head... 56
4.5 Suction Systems and Sump Design ... 56
4.6 Positive Displacement Pumps... 59
4.6.1 Rotary Pumps ... 61
4.6.1.1 External Gear and Lobe Pumps ... 61
4.6.1.1.1 External Gear Pumps... 61
4.6.1.1.2 Internal Gear Pumps... 62
4.6.1.1.3 Lobe Pumps ... 63
4.6.1.1.4 Peristaltic Pump... 63
4.6.1.1.5 Progressive Cavity Pump ... 65
4.7 Pipelines in the Industry ... 66
4.7.1 Piping Materials ... 66
4.7.1.1 Pipe Assembly ... 67
4.7.2 Stainless Steel Pipeline ... 67
4.7.3 Welding of Sanitary Pipe ... 67
4.8 Pipeline Construction ... 70
Chapter 5 Thermodynamics
5.1 Introduction ... 73
5.2 Energy of a System... 73
5.3 Characteristics of a System... 73
5.3.1 Energy of a System... 74
5.4 Gases and Vapors ... 75
5.4.1 Ideal Gas Equation... 76
5.4.2 Pressure Expressions... 77
5.4.2.1 The Perfect Gas Law ... 78
5.4.3 Mixtures of Gases ... 80
5.5 Problems... 83
Chapter 6 Electrical Systems 6.1 Introduction ... 85 6.2 Electricity Generation ... 85 6.3 Definitions ... 86 6.4 Energy Conversion ... 87 6.4.1 Electrical Energy... 88 6.5 Electric Motors... 88 6.5.1 Single-Phase Motors ... 88
6.5.1.1 Single-Phase Universal Motor... 88
6.5.1.2 Single-Phase Split-Phase Motor ... 88
6.5.1.3 Single-Phase Capacitor Motor... 89
6.5.2 Three-Phase Squirrel-Cage Induction Motors ... 89
6.6 Motor Management... 90 6.7 Power Transmission ... 91 6.7.1 Flat Belts ... 91 6.7.2 V-Belts... 92 6.7.3 Gears... 93 6.7.4 Calculation of Shaft RPM ... 93
6.7.5 Management of Drive Systems... 94
6.8 Control Equipment for Electrical Apparatus ... 94
6.9 Illumination of the Processing Facility ... 95
6.9.1 Light Intensity and Application ... 96
6.9.2 Types of Lamps... 96
Chapter 7 Heating Systems for Processing Plants 7.1 Heat Transfer... 99
7.2 Conductive Heat Transfer ... 99
7.2.1 Resistance to Heat Flow ... 101
7.2.2 Heat Flow Through Cylindrical Walls... 103
7.2.2.1 Steady-State Conduction through a Composite Cylinder 103 7.3 Convective Heat Transfer... 105
7.4 Heat Transfer by Radiation... 108
7.4.1 Stefan’s Law... 108
7.5 Steam Heating ... 110
7.5.1 Energy ... 110
7.5.2 Specific Heat Capacity... 110
7.5.3 Specific Latent Heat... 111
7.5.4 Energy Associated with Steam ... 111
7.6 Heat Pressure Diagram for Water... 111
7.6.1 Properties of Steam... 113 7.6.1.1 Saturated Liquid... 113 7.6.1.2 Saturated Vapor... 113 7.6.1.3 Wet Steam ... 114 7.6.1.4 Superheated Steam... 114 7.7 Steam Tables ... 115
7.8 Boiling Point of Water ... 116
7.8.1 Heating with Live Steam ... 116
7.8.2 Culinary Steam... 116
7.9 Hot Water Systems... 117
7.9.1 Energy Calculations for Water Heating... 117
7.10 Piping ... 119
7.10.1 Expansion of Pipes ... 119
7.11 Problems... 120
Chapter 8 Steam Generation 8.1 Introduction ... 123 8.2 Boilers ... 123 8.3 Fire-Tube Boilers ... 123 8.4 Water-Tube Boilers ... 124 8.5 Boiler Efficiency ... 125 8.5.1 Fuel-to-Steam Efficiency ... 126 8.5.1.1 Combustion Efficiency... 126 8.5.1.2 Thermal Efficiency ... 126 8.5.1.3 Stack Temperature ... 127 8.5.1.4 Radiation Loss ... 127 8.5.2 Boiler Blowdown ... 127 8.5.3 Water Treatment ... 128
8.6 Boiler Selection Criteria ... 128
8.6.1 Codes and Standards... 129
8.6.2 Steam Pressure ... 129 8.6.3 System Load... 129 8.6.3.1 Heating Load ... 130 8.6.3.2 Process Load... 130 8.6.3.3 Combination Load ... 130 8.6.3.4 Load Variations ... 130 8.6.3.5 Load Tracking... 131
8.6.4 Number of Boilers ... 131 8.6.4.1 Downtime Cost ... 132 8.6.5 Boiler Turndown ... 132 8.6.6 Fuels ... 132 8.6.6.1 Fuel Oil ... 132 8.6.6.2 Gaseous Fuels ... 133 8.6.6.3 Solid Fuels ... 134
8.6.6.3.1 Combustion of Solid Fuels... 134
8.6.6.4 Stack Emissions ... 135
8.6.6.4.1 Nitrogen Oxides... 135
8.6.6.4.2 Sulfur Oxides... 135
8.6.6.4.3 Carbon Monoxide... 135
8.6.7 Replacement Boilers ... 136
8.7 Boiler Fittings and Accessories ... 136
8.7.1 Pressure Gauge... 136
8.7.2 Steam Pressure Regulating Valve ... 136
8.7.3 Steam Trap ... 137 8.7.4 Water Columns... 137 8.7.5 Water Injection ... 138 8.7.6 Safety Valve... 138 8.7.7 Fusible Plug... 138 8.8 Steam Purity ... 140
Chapter 9 Refrigeration and Freezing 9.1 Overview ... 141
9.2 Natural Refrigeration ... 142
9.2.1 Natural Refrigeration to Subzero Temperatures... 143
9.3 Solid CO2... 143
9.4 Mechanical Refrigeration... 143
9.4.1 Capacity... 145
9.4.2 The Evaporating Coil... 145
9.4.2.1 Removal of Oil ... 145 9.4.3 Expansion Valve ... 146 9.4.4 Back-Pressure Regulator... 147 9.4.5 Compressors ... 148 9.4.6 The Condenser ... 148 9.4.7 Automatic Control... 148 9.4.8 Refrigerants ... 149 9.4.8.1 Ammonia... 150 9.4.8.2 Other Refrigerants ... 150 9.4.8.2.1 R-134a... 150 9.4.8.2.2 R-407 ... 153 9.4.8.2.3 R-32 ... 153 9.4.8.2.4 R-125 ... 153
9.5 Thermodynamics of Vapor Compression ... 153
9.6 Secondary Refrigeration ... 158
9.6.1 Congealing Tank System ... 158
9.6.2 Brine System ... 159
9.7 Management ... 162
9.8 Storage Rooms ... 163
9.8.1 Insulation ... 163
9.8.2 Prefabricated Polyurethane Panels for Cold Room Construction ... 164
9.8.3 Cold Room Doors and Devices ... 165
9.8.4 Management of Cold Rooms... 165
9.8.4.1 Defrosting ... 166
Chapter 10 Water and Waste Systems 10.1 Introduction ... 167 10.2 Water Quality ... 167 10.3 Potable Water ... 167 10.3.1 Water Safety... 168 10.3.1.1 Health Concerns... 168 10.3.1.1.1 Turbidity ... 169 10.3.1.1.2 Microbiological Concerns... 169 10.3.1.1.3 Radioactive Materials... 170 10.3.1.2 Aesthetic Concerns ... 170 10.4 Substances in Water ... 171 10.4.1 Hardness... 171 10.4.2 Iron... 172 10.4.3 Manganese ... 172 10.4.4 Nitrates... 172
10.4.5 Chlorides and Sulfates... 172
10.4.6 Gases ... 172
10.4.7 Silica ... 173
10.5 Treatment of Water Supplies ... 173
10.5.1 Treatment to Remove Turbidity ... 173
10.5.2 Softening Hard Water ... 174
10.5.2.1 Cold Lime Method ... 174
10.5.2.2 Ion Exchange ... 175
10.5.2.2.1 Base Exchange ... 175
10.5.3 Water Demineralizing ... 176
10.5.4 Filtering... 177
10.5.5 Care of Filters and Ion Exchangers ... 177
10.6 Treatment of Boiler Feedwater ... 178
10.6.1 Deaeration ... 179 10.7 Wastewater Treatment ... 179 10.7.1 Water Conservation... 180 10.7.2 Fluming Debris ... 181 10.8 Treatment Facilities... 181 10.8.1 Aerated Lagoon ... 181
10.8.2 Trickling Filters ... 182 10.8.3 Activated Sludge... 182 10.8.4 Anaerobic Treatment ... 182 10.9 Waste Disposal ... 182 10.9.1 Prevention of Waste ... 183 10.9.2 Solid Wastes... 183 10.9.3 Waste Incineration ... 184 10.9.3.1 Controlled Incineration ... 184 10.9.3.2 Incinerator Emissions ... 185 10.9.4 Landfill... 185 10.9.4.1 Landfill Methods... 185
10.9.4.2 Public Health and Environment... 186
Chapter 11 Materials Handling 11.1 Importance of Materials Handling... 187
11.2 Increasing Materials Handling Efficiency ... 188
11.2.1 Designing a Materials Handling System... 189
11.2.2 Kinds of Systems... 189
11.2.3 Analysis of a Materials Handling System ... 190
11.3 Materials Handling Equipment... 192
11.3.1 Conveyors ... 192 11.3.1.1 Chutes... 192 11.3.1.2 Roller Conveyors ... 194 11.3.1.3 Belt Conveyor ... 195 11.3.1.4 Slat Conveyors ... 195 11.3.1.5 Chain Conveyors... 196 11.3.1.6 Vibratory Conveyors ... 196 11.3.1.7 Screw Conveyors ... 196 11.3.1.8 Bucket Elevator... 196 11.3.2 Trucks... 196 11.3.3 Pallets ... 197 11.3.4 Bulk Handling... 197
11.4 Plant Layout for Materials Handling... 198
11.4.1 Space Arrangements ... 198
11.5 Efficient Use of Materials Handling Equipment ... 199
11.5.1 Maintenance... 199
Chapter 12 Manufacturing Plant Design 12.1 Building Design ... 201 12.2 Legal Aspects ... 201 12.2.1 Building Bylaws ... 201 12.2.2 Sanitation Codes ... 201 12.2.3 OSHA... 202 12.2.4 EPA ... 202
12.3 Expansion ... 202
12.4 Plant Location ... 203
12.5 Planning the Building on the Site ... 205
12.6 The Structure... 206 12.6.1 Foundations... 207 12.6.2 Supporting Structure... 207 12.6.3 Walls ... 208 12.6.4 Floors ... 209 12.6.4.1 Slope of Floors ... 209 12.6.4.2 Kinds of Floors ... 209 12.6.4.2.1 Concrete ... 209
12.6.4.2.2 Metal Plate Floor ... 209
12.6.4.2.3 Monolithic Floor Surfaces ... 210
12.6.4.2.4 Tile or Paver Floors ... 210
12.6.4.3 Drains ... 211 12.6.4.3.1 Floor Drains ... 211 12.6.4.3.2 Gutter Drains... 211 12.6.5 Windows ... 211 12.6.6 Doors... 212 12.6.7 Piping ... 212 12.6.8 Electrical ... 212 12.6.9 Ventilation ... 213 12.6.10 Hand-Cleaning Stations ... 213 12.7 Facilities Layout... 213 12.8 Summary ... 214 Chapter 13 Safety 13.1 Workplace Safety ... 215 13.2 Causes of Accidents ... 215 13.3 Safety... 217
13.3.1 Hazards in the Food Industry ... 217
13.3.2 Handling Materials ... 217
13.4 Accident Prevention ... 218
13.4.1 Color-Coding ... 218
13.4.2 Operating Instructions ... 218
13.4.3 Pressure Equipment ... 220
13.4.4 Grain Handling Equipment ... 220
13.4.5 Flammable Material... 220
13.5 General Plant Safety ... 220
13.5.1 Inspection of Equipment ... 221 13.5.2 Power Drives... 221 13.6 Fire Protection... 221 13.6.1 Fire Fighting ... 224 13.6.1.1 Portable Extinguishers ... 225 13.7 Noise Control ... 225
13.8 Occupational Safety and Health Administration ... 228
Appendix 1: Conversion Tables ... 231
Appendix 2 ... 239
A2.1 Some Help with Calculations ... 239
A2.2 The Plus/Minus Rule ... 239
A2.3 Fractions ... 239
A2.3.1 The Diagonal Rule... 239
A2.3.2 Do On Both Sides Rule... 240
A2.3.3 Adding Fractions ... 241
A2.3.4 Multiplying Fractions ... 242
A2.3.5 Dividing fractions ... 242
A2.4 Exponents ... 242
A2.5 Quadratic Equations ... 245
A2.5.1 Solution by Factorization ... 245
A2.5.2 Solution Using the Formula ... 245
A2.6 Simultaneous Equations ... 246
A2.7 Differentiation ... 248
A2.7.1 Rule 1... 248
A2.7.1.1 Repeated Differentiation... 249
A2.7.2 Rule 2... 249
A2.7.3 Rule 3... 249
A2.7.4 Rule 4... 251
A2.8 Applications of Differentiation ... 251
Appendix 3: Steam Tables... 253
1
Units and Dimensions
1.1 INTRODUCTION
Measurements have been used to describe quantities and sizes of artifacts from the earliest times. If one considers the size of the fish that got away, it becomes clear just how important it is to have a clear definition of size.
When looking at the Eiffel Tower, the realization of just how many pieces of metal had to be precut and predrilled to make up that gigantic three-dimensional masterpiece is overwhelming. Even more important, the pieces had to fit in a specific spot. Doing the calculations of the forces involved and combining them with material strength and rigidity would be a huge undertaking with the aid of computers. How about doing it all on paper?
1.2 HISTORY
Measurements relate to the world as we observe it, and a standard needs to be chosen. In England, the inch was defined as the length of three grains of barley. Since nature is not consistent in the length of barley, the inch changed from year to year. The grain was the weight of a grain of pepper, and so forth. There is a rather amusing tale regarding the U.S. Standard railroad gauge* (the distance between the rails) of 4 ft, 8.5 in. Stephenson chose this exceedingly odd gauge that is used in many countries, because it was the standard axle length for wagons. The people who built the tramways used the same jigs and tools that were used for building wagons.
The ruts in the unpaved roads dictated the odd wheel spacing on wagons. If any other spacing was used, the wheels would break, because they did not fit into the ruts made by countless other wagons and carriages. Roman war chariots made the initial ruts, which everyone else had to match for fear of destroying their wagons. Since the chariots were made for or by Imperial Rome, they were all alike. The railroad gauge of 4 ft, 8.5 in. derives from the original specification for an Imperial Roman army war chariot. The Romans used this particular axle length to accom-modate the back ends of two warhorses. The railroad gauge is thus dictated through history by the size of the back ends of two Roman warhorses.
The space shuttle has two booster rockets of 12.17 ft in diameter attached to the sides of the main fuel tank. These are the solid rocket boosters made by Thiokol at a factory in Utah. The engineers who designed the boosters would have preferred to make them fatter, but they had to be shipped by train from the factory to the
* The URL for this information is http://www.straightdope.com/columns/000218.html.
2 Food Plant Engineering Systems
launch site. The railroad line to the factory runs through a tunnel in the mountains, and the boosters had to fit through that tunnel. The tunnel is only slightly wider than the rail cars that have dimensions dictated by the gauge. So, a major design feature of what is arguably the world’s most advanced transportation system was determined by the width of two horses’ behinds.
A problem of measurement is that someone has to choose a standard that others will use. The meter was chosen to be 1 × 10–7 of the distance from the
north pole to the equator along a longitudinal line running near Dunkirk, France. The units for volume and mass were derived from the meter. One cubic decimeter was chosen as the liter. The mass of water at 4˚C contained in one cubic centimeter was chosen as the gram. These original measures were replaced by standards that are much more accurate.
Because people, particularly engineers and scientists, need to communicate through numbers, it is important to define what numbers mean. In the same trend, to make sense of quantity, one must define accurately what measurements mean. In this book, the SI system is used because the preferred metric units are multiples of 1000, making conversion a simple matter of changing exponents by adding or subtracting three.
Some conversion tables are incorporated in Appendix 2. Please note the different notations that can be used. The solidus notation, i.e., m/s or g/kg or m/s/s, can be written in negative index notation as m s–1, g kg–1, m s–2. In the SI system, the
negative index notation is preferred.
1.3 TERMS
There is a lot of confusion regarding terms, units, and dimensions. For this discourse, we will limit dimensions to space, matter, time, and energy. Many people work in environments of infinite dimensions, but the understanding of those environments is outside the scope of this book. As a starting point, the term dimension will be defined as any measurable extent such as time, distance, mass, volume, and force.
Units are designations of the amounts of a specified dimension such as second, meter, liter, and newton. It is important to keep a sense of sanity within the confusing system of units (numbers) with different dimensions (measured extent) associated with them. The easiest way to do this is with dimensional analysis.
1.3.1 BASE UNITS
There are seven base units in the measurement system. The first four — mass, length, time, and temperature — are used by everyone, while the last three— electric current, luminous intensity, and amount of a substance — are used in technical and scientific environments.
Each of the base units is defined by a standard that gives an exact value for the unit. There are also two supplementary units for measuring angles (see Table 1.1).
1.3.1.1 Definitions
Most of the dimensions were selected in arbitrary fashion. The Fahrenheit scale was chosen such that the temperature for a normal human would be 100˚F; eventually the mistake was discovered, and we now have an average normal temperature of 98.6˚F. In the same way, we have many legends about the choice of the standard foot, inch, drams, crocks, grains, and bushels. Even in more modern times when the metric system was started, many standards were chosen in an arbitrary way that might have been founded in scientific misconception. In most cases, the older standards that relied on artifacts have been replaced by more scientific definitions. As our knowledge of science expands, new definitions will be formulated for standards.
Mass — Mass is a measure of matter that will balance against a standard mass. The mass of a body is therefore* independent of gravity — it is the same anywhere in the universe. The weight of a body is taken as a trans-lation from mass to force with a value of gravitational constant (g) as standard. Mass and weight can be the same if one applies the appropriate dimensional analysis corrections. (Mass is the only one of the seven basic measurement standards that is still defined in terms of a physical artifact, a century-old platinum-iridium cylinder with a defined mass of 1 kg kept at St. Sèvres in France. Scientists are working on a more accurate system to determine mass.)
Length — One meter is the length of a platinum-iridium bar kept at St. Sèvres, France. Scientists redefined the standard for length as a wavelength of orange light emitted by a discharge lamp containing pure krypton at a specific temperature. The length of this standard is 6.058
×
10–7 m.TABLE 1.1
Dimensions and Preferred Units Used in SI System
Dimension Unit Preferred Symbol
Mass Kilogram kg
Length Meter m
Time Second s
Absolute temperature Kelvin K
Temperature Degrees Celsius °C = K – 273
Electric current Ampere A
Luminous intensity Candela cd
Amount of a substance Mole mol
4 Food Plant Engineering Systems
Time* — Time is a definite portion of duration, for example, the hour, minute, or second. With the development of more sensitive measuring devices, the accuracy of the measurements was enhanced. The second was defined as a mean solar second equal to 1/86,400 of a mean solar day. This has been redefined as the time it takes for 9,192,631,770 electromagnetic radiation waves to be emitted from 133Cs.
Temperature — In layman’s terms, temperature can be described as the degree of hotness or coldness of a body relative to a standard. Temperature is proportional to the average kinetic energy in molecules or atoms in a substance. Three temperature scales are used: Kelvin (K), which is the SI-preferred scale, Celsius (˚C), and Fahrenheit (˚F), which is mainly used by laymen. The two fixed reference points are the melting point and boiling point of pure water at one atmosphere or 760 mm Hg pressure. In Kelvin, these points are 273.15 K and 373.15 K; in Celsius, these points are 0˚C and 100ºC. One Kelvin (K) is defined as 3.6609
×
10–3 of thethermody-namic temperature of the triple point of pure water (temperature at which ice, water, and water vapor are in equilibrium). Another frequently used reference point is the low temperature at which molecular motion will stop, 0 K or –273.15ºC.
Electric current — This is a measure of flow rate of electrons measured in ampere.
Luminous intensity — This is measured in candela. One candela is defined as the luminous intensity, in the perpendicular direction, of the surface of 1/600,000 m2 of a perfect radiator at 1772ºC under a pressure of one
atmosphere. In the older systems, this was referred to as candle power, and lightbulbs were frequently graded in candle power.
Amount of a substance — This is measured in mol. One mol is the amount of a substance of a system that contains as many elementary entities as there are atoms in 12
×
10–13 kg of 12C. In chemistry, we frequently referto mol per liter, or in preferred SI, as mol dm–3. 1.3.1.2 Supplementary Units
Circular measure — It is mathematically convenient to measure plane angles in dimensionless units. The natural measurement is length of circular arc divided by the radius of the circle. Since circumference of a circle with radius R is given by 2
π
R, 360˚ equal 2π
radians.Steradian — This is the measurement of solid angles, equal to the angle subtended at the center of a sphere of unit radius by unit area on the surface.
* Time is one of the most philosophical issues in theoretical physics. For some rather humorous interpretations of time, one can read the Diskworld books by Terry Pratchett, a physicist who decided to take scientific philosophy over the edge.
1.3.2 DERIVED UNITS
The clear definitions of the basic units allow for standardization of the dimen-sions. Mathematical combinations of standard base units are used to formulate derived units.
1.3.2.1 Definitions
Area — This is the product of two lengths describing the area in question (m2).
For rectangular units, it is given by width × length, and for circles, it is
π
r2.Land is frequently measured in hectare with 10,000 m2 equal to 1 hectare
equal to about 2.5 acres. Large land areas are measured in square kilometers.
Volume — This is the product of three lengths (m3) describing the space in
question.
1 m3 = 1 m × 1 m × 1 m = 103 mm × 103 mm × 103 mm = 109 mm3
= 106 cm3 = 106 ml = 103 liters
Density — The mass of a substance divided by its volume is the density.
(Remember to change the sign of the exponent when a unit is inverted, moved from below the line to above the line.)
Velocity — Distance traveled in a specific direction (m) divided by the time taken to cover the distance (s) is velocity.
or
Velocity is a vector because it is directional. Other vector quantities include force, weight, and momentum. Kinetic energy and speed have no direction associated with them and are scalar quantities. Other scalar quantities include mass, temperature, and energy.
Flow — This is measured in cubic meters per second (m3s–1) also referred to
as cumec.
Momentum — Linear momentum is the product of mass and velocity (kg ms–1 or Ns).
Acceleration — This is the rate of change of velocity of a body expressed in SI as (ms–2).
Gravitational constant — g = 9.81 ms–2.
Force — This is the product of mass (m) and acceleration (a).
ρ = kg = − m3 kgm 3 m s =V ms −1=V F=ma
6 Food Plant Engineering Systems
F, the resultant force on the body, is measured in Newtons; m is the mass in kg; and a is the ultimate acceleration in meters per second per second (N = 1 kgms–2). If a body of mass m starts from rest and reaches a velocity v in t
seconds as a result of force F acting on it, then the acceleration is v/t, and
is the rate of change in momentum.
Pressure — Force applied on a specific area (N m–2) is pressure. This is also
known as pascal (Pa). This is a very small unit, and the practical units are kilo pascal or bar. One bar is equal to approximately one atmosphere.
Frequency — This is defined as cycles per second = hertz = Hz, where 106
Hz = 1 megahertz = 1MHz.
1.4 DIMENSIONAL ANALYSIS
When working with numbers, it is important to remember that the number (unit) is associated with something (dimension). In a bag of fruit, there may be two apples and three peaches. Nobody will try to add the apples and peaches together and call them five apples. In the same way, we can only do calculations with numbers that belong together.
EXAMPLE 1.1
Convert 25,789 g to kilograms.
When we work with numbers that have different dimensions, they cannot be manipu-lated unless the dimensions are part of the manipulation. From basic definitions, we learn that density is mass/volume, so that
or in the more usual term,
F ma mv
t
= =
1bar=105Pa Nm
(
−2)
=0 1. MPa≅1atmosphere25789 25789 1 1 1000 25 789 g g kg g kg = = . density kg m = 3 density g cm = 3
EXAMPLE 1.2
It is important to give the density base because 1000 g = 1 kg and (100)3 cm3 = 1 m3.
When we convert a density of 1500 kg/m3 to g/cm3 or the frequently used g/ml, we
do it as follows:
EXAMPLE 1.3
Convert 20 kPa to kilograms per meter per second squared.
EXAMPLE 1.4
Calculate the mass of aluminum plate with the following dimensions: length = 1.5 m, width = 15 cm and thickness = 1.8 mm. The density of the aluminum = 2.7 g/cm3.
The dimensions can be canceled just as in calculations. The cm3 below the division
line will cancel the same term above the division line. The same calculation done in SI-preferred units would be as follows:
1500 1500 1000 100 100 100 1 500 000 1 000 000 1 5 1 5 3 3 3 kg m g cm cm cm g cm g cm g ml = × × × = = = , , , , . . / 20 20 1000 1 20000 20000 1 1 20 20000 1 1 20000 20000 2 2 2 1 2 2 kPa kPa Pa kPa Pa Pa Nm Pa kPa Nm kgms kgm s kg ms =
(
)
= =(
)
=(
)
= = − − − − − N /mass=volume×density
mass g volume cm density g
cm mass g cm cm cm g cm mass g cm g cm g
( )
=( )
× ( )
= × × ×( )
= × = 3 3 3 3 3 150 15 0 18 1 2 7 1 405 1 2 7 1 1093 5 . . . .mass kg volume m density kg
m
( )
=( )
3 × 3 mass kg m m m kg m( )
=1 5 ×0 15 ×0 0018 × 1 0 0027 0 0000001 3 . . . . .1.5 UNITS FOR MECHANICAL SYSTEMS
Work — Mechanical work (J) is force × distance moved in the direction of
the force. Work is measured in joule (J), where 1 J of work is done when a force of 1 N moves it to a point of application through 1 m.
EXAMPLE 1.5
The work to lift 1 kg of material a height of 3 m is as follows:
Power — The rate of work done on an object is power. One watt (1 W) is the
power expended when one joule (1 J) of work is performed in 1 s (W = J s–1).
Throughout the ages, the power unit on farms was a horse or an ox. When mechanical power units were introduced, it was natural that the power of the steam contraption was compared to that of a horse. Very few people know the kW output of the engine in their automobiles, but most will be able to tell you what it is in horsepower (hp). It is rather strange that we frequently use different measuring systems within one object. The size dimensions of the automobile are given in inches and cubic feet. The engine displacement is given in cc, which is metric and a way to denote cm3. One of the biggest wastes of time
and effort is in the continual conversion of units. One horsepower (hp) is equivalent to 550 ft lbf s–1. The conversion of horsepower to joule is as follows:
EXAMPLE 1.6
In conversion problems, dimensions and units become even more confusing. The conversion factor when 1 horsepower (hp) is converted to watt (W) is given as 1 hp = 745.7 W. The calculation with dimensional analysis will look like this:
mass kg m kg m kg
( )
= 0 000405 × = 1 2700 1 0935 3 3 . . J= × =N m kgm s2 −2 1 3× ×9 81. =29 43. J 1hp⇔745 7. W⇔745 7. Js−1 1hp 550 ftlb / s ftlb lb ft s ftlb lb s f f m 2 m f 2 = = × × = g g c cThe slight difference in value, 745.6 and not 745.7, is because limited decimals are used in the conversion values. In most cases, there are tables to work from. The conversions as shown in the example were done by someone and collected in tables.
Energy — The capacity to do work (J) is energy. Energy may be either
potential or kinetic. Energy can never be destroyed, but it may be changed from one form to another. More details regarding energy are given in
Chapter 2.
Potential energy — This is latent or stored energy that is possessed by a body,
due to its condition or position. For example, a 100 kg body of water in a position high above ground, such as in a tank with a free surface 10 m high, will have:
where m is mass flow rate, h is distance fallen in m, and g is acceleration due to gravity.
Potential energy is also released when a product such as coal, oil, or gas is burned. In this case, solar energy was stored by plants and changed into fossil fuel. Energy can appear in many forms, such as heat, electrical energy, chemical energy, and light. When energy is converted into different forms, there is some loss of utilizable energy.
Energy in = Total energy out + Stored energy + Energy lost Process design should minimize such nonproductive energy losses. The diesel engine delivers only about 35 to 50% useful energy from the fuel. Modern high-pressure steam generators are only about 40% efficient, and low-pressure steam boilers are usually only 70 to 80% efficient for heating purposes.
Kinetic energy — A body with mass m moving at velocity V has kinetic
energy: 1hp=745.62 Js−1=745.6W PE= mhg PE= × × = − = 100kg s 10 m 9.81 m s 9810 Js 9.81 kW 2 1 PE
If a body of mass m is lifted from ground level to a height of h, then the work done is the force mg multiplied by the distance moved h, which is
mgh. The body has potential energy = mgh.
If the mass is released and starts to fall, the potential energy changes to kinetic energy. When it has fallen the distance h and has a velocity V, then:
Kinetic energy for a rotating body is more technical. However, if the mass can be assumed to be concentrated at a point in the radius of rotation, the moment of inertia can be used.
EXAMPLE 1.7
An object with a mass of 1000 kg moving with a velocity of 15 ms–2 at the average
radius would be found to have the following kinetic energy:
Kinetic energy is of importance in the movement of all objects in the food processing industry.
Centrifugal force — An object moving around in a circle at a constant tip
speed is constantly changing its velocity because of changing direction. The object is subjected to acceleration of rω2 or v
t2/r. Newton’s law says
that the body will continue in a straight line unless acted upon by a resultant external force called centrifugal force.
where m = mass, r = radius in m, ω = angular velocity = 2πN radians per second. KE= mV 2 2 mgh= mV 2 2 KE= mV 2 2 g KE= × × 1000 15 2 9 81 2 . KE= 11476 89. J CF mr r = ω2 = 2 mvt
A satellite orbiting Earth is continuously falling toward Earth, but the curvature of Earth allows the satellite to fall exactly as much toward Earth as the Earth’s curvature falls away from the satellite. The two bodies, therefore, stay an equal distance apart.
Centrifugal force is used in separators for the removal of particular matter or separation of immiscible liquids of different densities. The efficiency of a separator can be defined as the number of gravitational forces, where the number of g forces is equal to centrifugal force divided by gravita-tional force.
Torque — This is a twisting effect or moment exerted by a force acting at a
distance on a body. It is equal to the force multiplied by the perpendicular distance between the line of action of the force and the center of rotation at which it is exerted. Torque (T) is a scalar quantity measured by the product of the turning force (F) times the radius (r).
1.6 CONVERSIONS AND DIMENSIONAL ANALYSIS
A viscosity table lists viscosity in units of lbm/(ft s). Determine the appropriate SI
unit, and calculate the conversion factor. The original units have mass (lbm), distance
(ft), and time (s). In SI, the corresponding units should be kg, m and s.
But, viscosity is also given in Pa s:
Pa s = (N/m2) s = kg × m/s2× s/m2 = kg/ms CF= mrω =rω 2 2 mg g T F r T Nm = × = kg / ms lb / fts conversion kg / ms lb fts 1 kg 2.2046 lb 3.281ft 1m 1s 1s kg / ms lb / fts 1.48866 kg / ms =
(
)
× = × × × = × m m m m1.7 PROBLEMS
1. Set up dimensional equations and determine the appropriate conversion factor to use for each of the following:
a. ft2/h = (mm2/s) × conversion factor
b. Hundredweight/acre = kg/ha × conversion factor c. BTU/ft3 ˚F = kJ/m3K × conversion factor
2. Make the following conversions, using standard values for the relationship of length and mass units.
a. 23 miles/hr to m/s b. 35 gal/min to m3/s
c. 35 lb/in2 to kg/m2
d. 65 lb/in2 to bar
e. 8500 ft-lbf to joules and to kilowatt-hours
13
General Calculations
2.1 INTRODUCTION
Many people dread what they will find when they hear words like mathematics, algebra, and calculations. This is true for most college students and is unfortunate. Most mathematical problems have an associated reading part, and many students say that translating words into numerical equations is the most difficult aspect of mathematics. Calculations are manipulations of equations into which numbers are plugged and are then worked out on a calculator. Calculations are easy if you know all of the associated rules, most of which are included in Appendix 2. If you are unsure of your grasp of the things that make mathematics easy, look in the appendix and place a sticky note there so that you can refer to it as often as necessary.
2.2 BASIC PRINCIPLES
Many processes involve adding (x + y = z) and taking away substances (a – b = c) or fractions (2x – 1/
2x = 11/2x). Similarly, energy can be added or removed in the
processes that constitute the system. In many of the processes, the change in physical quantities can be accompanied by chemical reactions and separation processes.
2.2.1 CONSERVATIONOF MASS
The basis of mass balances is the conservation of mass. The total material input is equal to the total material output plus any accumulation of material in the equipment. This relationship will apply to the total process, individual components and any chemical reactions that took place. In the case of a continuous process, the calcu-lations are simplified when the process reaches a steady state condition. In this case, the accumulation can be regarded as a static quantity and, therefore, ignored so that input is equivalent to output.
In smaller batch-scale operations, total input is normally removed at some stage so that input is equal to output plus waste (another form of accumulation). The waste is normally a small quantity compared to the large amount of accumulation that can be found in some processes.
2.2.1.1 Mass Balances
Look at the simultaneous equations in Appendix 2. Mass balances are simultaneous equations that can be solved by thinking and using logic. Logic cannot be taught — it can merely be developed, and once developed, will be there for all decisions
14 Food Plant Engineering Systems
in life. Mass balances are essentially easy and are good for helping us gain confidence in calculation skills.
To make any headway in a mass balance, the system or part of the system must be defined. It might be necessary to look at different subsystems prior to formulating the total mass balance. An easy method of breaking the system into identifiable sections is to look at the big picture regarding specific components. Putting a boundary around the system will allow us to identify what goes in and what comes out of the bounded system.
For example, when sieving wheat, the amount of wheat that enters the system (feed) must be equal to the amount of wheat that exits the system (product), plus the oversized and undersized materials (undesired materials) that were separated, plus the dust (undesired material) that was removed plus any bits and pieces that got stuck in the screens (accumulation). Any material stream that enters or exits the system must be part of the calculation, including any air or vapor transfer. Streams can divide, recirculate, or feed into different parts of the system. Keeping track of the mass fractions in the streams is essential to the solution of a problem.
Once the system boundary has been chosen, a basis for the calculation must be chosen. It is possible to use the flow of a specific line per minute, hour, or day. If the flow rate does not change, it is easier to use the flow per time as a volume or mass without considering time. However, it is much simpler to use a quantity of material, such as 100 kg, and convert this into flow per time at the end of the calculation. It is easiest if the compound chosen as the base goes through the system unchanged in mass. In a dehydration process, the water content in the product and the vapor in the air continuously change. All that stays the same is the solid content of the material. Therefore, it is best to use the solids as the base.
Before starting, try to visualize and interpret the problem. Ponder the problem, and then start following the steps. After practicing, the visualization process will be quick and natural.
If the following steps are used, calculations become much easier:
• Read the problem at least twice before doing anything else. Follow all the steps listed below, and read the problem again while checking the diagram and mass fractions to ensure that everything makes sense. • Draw a simple diagram of the process, and indicate everything that goes
in and comes out.
• If a complex system is used, show the boundaries of each subsystem and calculate each subsystem individually.
• Designate symbols for each quantity that needs to be calculated. • Give the mass fraction (percentage) of each quantity that enters or exits
the system.
• Use simultaneous equations to solve the problem (see Appendix 2).
Be sure to use a method that will work for you. Try to visualize the operation as clearly as possible, breaking it down into a logical sequence of events.
EXAMPLE 2.1
A simple example of a logical sequence of events that has nothing to do with mass balances is found in the manual for “Do an Appendectomy for Idiots.”
1. Is the patient on the table? If yes, go to 2; if no, place the patient on the table and go to 1. (Please note the feedback to ensure that the condition is true. Do not jump to the second item after you have given a command. Make sure the command has been properly executed!)
2. Can you see the patient’s navel? If yes, go to 3; if no, turn the patient over and go to 2.
3. Can you see the lower left abdomen? If yes, go to 4; if no, get your glasses and go to 3.
4. Cut the abdomen. Can you see guts? If yes, go to 5; if no, go to 4.
Find the appendix, rip it out, cut it off, put it in a bottle of alcohol to make a paperweight, and sew up the abdomen after counting swabs and hardware, etc. A simple old mass balance is certainly easier than an appendectomy, so get started.
EXAMPLE 2.2
An aqueous solution contains 25% (m/m) soluble solids. After evaporation, the solution lost half of the original mass (Figure 2.1). What is the concentration of soluble solids in the concentrated solution?
You can use any mass as long as the final mass is half of the original mass and the amount of water vapor removed is the same as the final mass.
Mass balance
Soluble solids balance
FIGURE 2.1 2kg=1kg vapor
(
)
+1kg concentrate(
)
2 0 25 1 1 0 0 5 kg× = kg× + kg× = . . X X EVAPORATOR Wet Material 25% Sol Solids 2 kg Product X % Sol Solids 1 kg Vapor 0 % Sol Solids 1 kg16 Food Plant Engineering Systems
The concentrate will contain 50% soluble solids. In the original 2 kg, there were 500 g soluble solids and 1500 g water for a final mass of 2000 g. To lose half the mass, there are 500 g soluble solids in 500 g water, or 50% soluble solids. This is a simple problem, but one in which you can check your answer with simple logic. Checking the relative amounts of all the compounds is the most important part of mass balances.
EXAMPLE 2.3
A company produces a new Asian-style condiment that will contain 2% (m/v) salt and 6% (m/v) sugar in the final product. The salt and sugar are added as solutions that are fed into the main pipeline. If the salt solution contains 18% (m/v) salt, and the sugar solution contains 60% (m/v) sugar (see Figure 2.2), how much of each of the concen-trates must be added to the condiment stream if it flows at 100 l per minute? Since the flow is given in liters/minute, liters can be used.
Overall mass balance
Salt balance
Sugar balance
Solving for two unknowns, we have the required two equations:
FIGURE 2.2
Condiment Mix ( )C + Salt ( ) A + Sugar ( ) B = Product ( )P
100×0.0+ ×A 0.18 0.0 + ×B =
(
100+ +A B)
×0.02 100×0.0 + ×A 0 0. + ×B 0.6 =(
100+ +A B)
×0.06 0 18 2 0 02 0 02 0 16 0 02 2 1 . . . . . ( ) A A B A B = + + − = →Multiply Equation (1) with 27 to ensure that the B values will be the same. Adding the two equations will eliminate B, and we can solve for A. That value is then used to solve for B.
Adding Equation (2) to Equation (3) gives:
Substituting in Equation (2) gives:
The process requires that 14.1 l/min of the salt solution and 12.68 l/min of the sugar solution be added. The new flow rate will be 126.8 l/min. It should contain 6% sugar — 126.8 times 0.06 is 7.61 kg of sugar and 2.54 kg salt. How much sugar is in 12.7 l of a 60% solution? There are 60 kg per 100 l — 0.6 times 12.68 kg in 12.68 l. This is equal to 7.61 kg. Calculating the salt in the same way, we get 2.54 kg salt in 14.1 l solution. The answer is verified. (Remember that answers can vary slightly because of rounding up or down.)
EXAMPLE 2.4
This example will expand the scope of our mass balance calculations. The same principles are presented in a deluxe version.
A winery is making a new fruit-flavored blended mix. The final product contains 25% clear peach juice. The wine portion can be blended from A, B, or C. The alcohol and sugar levels for the wine and the blend are given in Table 2.1.
Use 100 kg as a basis for the blending operation. With that, one can easily convert to larger quantities. 0 60 6 0 06 0 06 0 06 0 54 6 2 . . . . . ( ) B A B A B = + + − + = → 27 0 16 27 0 02 27 2 4 32 0 54 54 3 × − × = × − = →
( )
. . . . A B A B 4 26 60 60 4 26 14 1 . . . A A = = = − × + = = + = = 0 06 14 1 0 54 6 0 54 6 0 846 6 846 0 54 12 68 . . . . . . . . B B B18 Food Plant Engineering Systems
Total mass balance
Alcohol balance
Sugar balance
There are four unknowns in the problem but only three equations, therefore, the problem cannot be solved. Read the problem again. We need to use 25% peach juice; therefore, J is equal to 25 kg, and the three equations can be rewritten using 25 in place of J.
Total mass balance
Alcohol balance
Sugar balance
Multiply the total mass balance with 0.143, and subtract the answer from the alcohol balance to remove A. Multiply the total mass balance with 0.037, and subtract the answer from the sugar balance, again eliminating A. This gives two equations with two unknowns.
TABLE 2.1
Alcohol and Sugar Levels for the Wine and the Blend
Ingredient Alcohol % (m/m) Sugar % (m/m)
Wine A (A) 14.3 3.7 Wine B (B) 16.9 2.4 Wine C (C) 15.3 5.4 Peach Juice (J) 0 13 Blend (P) 12 5.5 A+ + + = 100B C J 0 143. A+0 169. B+0 153. C+0 0. J=100×0 12. 0 037. A+0 024. B+0 054. C+0 13. J=100×0 055. A+ + =B C 100−25=75 0 143. A+0 169. B+0 153. C=12 0 037. A+0 024. B+0 054. C=5 5. −25×0 13. =2 25.
Multiplying the second equation with two and adding it to the first equation will eliminate B.
Substitution of B in one of the formulas will solve for B.
Substituting B and C in the total mass balance gives A.
The blend will be made up of 22.82 kg A, 47.07 kg B, 5.11 kg C, and 25 kg peach juice. To check the answers, use the alcohol balance.
The sugar balance can be verified the same way. To use Solver in Microsoft® Excel, set up the problem as shown in Table 2.2.
Using the amount of product to produce and forcing the target to a set value of 100 kg, requires us to make sure that the quantities of ingredients produced are not negative. In mathematics, negative numbers are useful. In manufacturing situations, it means that you have to remove material from the product, which is not easy. Other constraints include the amounts of alcohol and sugar in the final product. This simple problem can be done in about 4 min on Solver (see Table 2.3 for the answer report). If you have Microsoft Office and Excel, and if Solver is not available in the Tools menu, then just go to Tools and find Add-Ins — Solver will be there.
EXAMPLE 2.5
One ton of sliced carrots is dehydrated in a concurrent flow dehydrator from 85% (m/m) to 20% (m/m) moisture content. Air with moisture content of 0.012 kg water per kg dry air enters the dehydrator at a rate of 500 kg dry air per kg dry solid (see
Figure 2.3). What is the moisture content of the air leaving the system?
0 026 0 01 1 275 0 013 0 017 0 525 . . . . . . B C B C + = − + = − − + = − = = 0 026 0 034 1 05 0 044 0 225 5 11 . . . . . . B C C C kg 0 026 0 01 5 11 1 275 47 07 . . . . . B B + × = = kg A= 22 82. kg 0 143 22 82 0 169 47 07 0 153 5 11 12 3 263 7 955 0 782 12 . . . . . . . × + × + × = + + =
Food Plant Engineering Systems
TABLE 2.2
Problem Setup in Excel Solver
A B C D E F
1 A B C D P
2 Alcohol 14.3 16.9 15.3 0 12 3 Sugar 3.7 2.4 5.4 13 5.5 4 Amount 22.8146847135268 47.0716789228462 5.113636363627 25 = SUM(B4:E4) 5 Alcohol = B4*B2/100 = C4*C2/100 = D4*D2/100 = E4*E2/100 = SUM(B5:E5) 6 Sugar = B4*B3/100 = C4*C3/100 = D4*D3/100 = E4*E3/100 = SUM(B6:E6)
If the solids entering contain 85% (m/m) water, then the system contains 85 g water and 15 g solids in 100 g material. The amount of water entering the dehydrator with 1000 kg is as follows:
The amount of solids is 150 kg. But, the final solids amount also contains 20% (m/m) moisture. Therefore, the final mass is:
TABLE 2.3
Answer Report for the Excel Solver Solution Target Cell (Value of)
Cell Name Original Value Final Value
F4 Amount P 100.000 100.000
Adjustable Cells
Cell Name Original Value Final Value
B4 Amount A 22.815 22.815
C4 Amount B 47.072 47.072
D4 Amount C 5.114 5.114
Constraints
Cell Name Cell Value Formula
F6 Sugar P 5.500 F6 = F3
F5 Alcohol P 12.000 F5 = F2
B4 Amount A 22.815 B4> = 0
C4 Amount B 47.072 C4> = 0
D4 Amount C 5.114 D4> = 0
Note: The cell references refer to those in Table 2.2.
FIGURE 2.3
1000×0 85. =850kg
150
22 Food Plant Engineering Systems
The amount of water removed per hour from each ton is:
The amount of air entering the system is:
The moisture in the incoming air is:
The total moisture leaving the system is:
Moisture content of the air leaving the system is:
kg moisture/kg dry air
There are other ways of dealing with the same problem.
The moisture entering the system with a kg of dry material is given by the following:
The moisture leaving the system with the “dry” product is:
The moisture entering the system with the air is:
Since water in must equal water out, the water uptake by the air is:
850kg−37 5. kg=812 5. kg 150kg 500kg 75000 kg kg × = 75000kg 0 012kg 900 kg kg × . = 812 5. kg+900kg=1712 5. kg 1725 5 75000 0 023 . . kg kg = 0 85 0 15 1 5 67 . . × kg= . kg 0 20 0 80 1 0 25 . . × kg= . kg 500×0 012. kg=6 0. kg 5 67. kg+6kg−0 25. kg=11 42. kg
Thus, the moisture content of the air leaving is as follows:
kg moisture/kg dry air
There are many ways to tackle most problems. Be cautious regarding the percentage composition of substances. The percentage can be given on a dry mass basis or a wet mass basis producing different results.
If 20 g of a pure substance is mixed with 40 g water, the total mass is 60 g. On a wet mass basis, the moisture content will be:
And on a dry mass basis, the moisture content of the wet mixture will be:
This example is silly, but it shows the effect of the base used for calculations.
EXAMPLE 2.6
We use crystallization as a way to purify some soluble chemical compounds. If we dissolve 100 kg of refined sugar that contains 94% (m/m) sucrose, 3% (m/m) water, and 3% soluble solids that will not crystallize in 25 kg water at 80˚C, we should have an almost-saturated solution.
Calculate the percent sucrose in the solution.
Total sucrose = 94 kg Total water = 3 + 20 kg Total mass = 100 + 25 kg
This solution will be saturated at 78% sucrose. If we cool the solution to 20˚C, where a saturated solution contains 67% (m/m) sucrose, how much sucrose will crystallize?
If we assume that the inert soluble solid will not compete with the sucrose for water molecules, the amount of sucrose that will stay in solution can be calculated.
11 42 500 0 023 . . kg kg = 40 60×100= . %66 67 40 20×100=200% 100 0 94 100 25 100 75 2 × + × =
(
)
. . % m m/ 125×0 67. =83 75. kg24 Food Plant Engineering Systems
The difference between this amount and the total amount that went in is the amount that will crystallize.
After centrifugation, the crystals contain 10% of the supernatant. What is the mass of crystals and other liquor that will exit the centrifuge?
What will the final mass of sugar be after the crystals are dried to 0% moisture?
First, one should calculate the amount of moisture, sugar, and soluble solids in the adhering moisture.
Sugar content
Water and soluble solids
In the original 125 kg, there were 3 kg soluble solids. After removal of 10.25 kg pure sucrose, there were 3 kg soluble solids in 114.75 kg total soluble phase. The concen-tration of the soluble solid is as follows:
The amount of the soluble solid in the crystals is:
This amount of soluble solid will adhere to the recrystallized sucrose. What will the percentage contamination be after drying?
The solids amount is:
The amount of soluble solid in the sucrose is now:
94−83 75. =10 25. kg 10 25 0 9 11 39 . . = . kg 1 24. ×0 67. =0 83. kg 1 24. −0 83. =0 41. kg 3 114 75. ×100=2 61. % 1 24. ×0 0261. =0 032. kg 10 25. +0 032. =10 282. kg
that can crystallize are purified.
2.3 PROBLEMS
1. What is the final salt concentration if 20 kg of a solution containing 21% (m/m) salt are mixed with 10 kg of a solution containing 12% (m/m) salt? 2. How much water is removed when 100 kg of a solution containing 11% (m/m) total solids have to be concentrated to produce a liquid containing 35% (m/m) total solids?
3. While making ketchup, the tomato concentrate flows through a pipeline at 50 kg/min. At what rate must a 23% (m/m) solution of salt be added to ensure 3% salt in the final product?
4. A brewer’s spent grain contains 80% (m/m) moisture. When the grain has lost half of its weight, what is the moisture content?
5. Calculate the quantity of dry air required to dry 50 tons of food from 19 to 11% moisture. The air enters with a moisture content of 0.003 kg water/kg dry air and exits with a moisture content of 0.009 kg water/kg dry air.
6. Orange juice concentrate 3:1 is reconstituted by mixing one volume of concentrate with three volumes of water. Consumers indicated that this tastes artificial; therefore, manufacturing is producing a new 4:1 concen-trate. Before packaging, this will be blended with fresh juice to produce a 3:1 concentrate. If you want to produce 1000 kg of the new product, how much 4:1 concentrate must be prepared, and how much juice must be kept in reserve for dilution purposes?
7. To produce sausage formulations, packers use lean beef, pork belly, and soy concentrate. The manufacturer uses 3% soy in final sausage mass. Water is added to the formulation as ice. The final sausage must contain 15% protein, 60% moisture, and 25% fat.
The ingredients contain the following:
How much of each ingredient should be combined to make 600 kg of sausage emulsion?
Ingredient Protein (%) Moisture (%) Fat (%)
Lean beef 20 67 13
Pork belly 10 40 50
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Properties of Fluids
3.1 GENERAL DESCRIPTION
A fluid is a substance that cannot maintain its own shape and has no rigidity. It will flow and alter its shape to fill the container that retains it. It differs from a solid in that it suffers deformation due to shear stress, however small the shear stress may be. The only criterion is that sufficient time should elapse for the deformation to take place. In this sense, a fluid is shapeless.
Fluids may be divided into liquids and gases. A liquid is only slightly compress-ible, and there is a free surface when it isplaced in an open vessel. On the other hand, a gas always expands to fill its container. A vapor is a gas that is near the liquid state. It is easier to think of fluids as compressible and incompressible.
Water will be the liquid used as an example. It is a well-known commodity that can be used as an example in many problems. Water may contain up to 3% of air in solution, which at subatmospheric pressures, tends to be released. Provision must be made for this when designing pumps, valves, pipelines, etc.
3.2 PHYSICAL PROPERTIES OF FLOW
The principal physical properties of fluids are described in the following sections.
3.2.1 PRESSURE
Pressure is force acting on a specific area. If the force is measured in newtons and the area in square meters, then the pressure is in pascal (Pa).
The pressure caused by a column of liquid acts equally in all directions and depends upon the depth of and density of the liquid. It is thus defined as force per unit area, and it is expressed as weight-force and area or the height of a column of liquid that will produce a similar pressure at its base.
3
Pa m = N2 Pressure Pressure ( ) ( ) ( ) ( ) Pa Force kg head h density Pa kg m kg m = × × = × × N N ρ 3Pressure can also be expressed as the column of liquid (h) times the density of the liquid (ρ) multiplied by the gravitational constant (g).
EXAMPLE 3.1
A tank is filled with water to a depth of 2.5 m. What is the pressure at the bottom of the tank due to the water alone?
At any point below the surface of a liquid, the pressure is equal to pressure on the surface Ps plus the pressure of the column of liquid above it to give:
EXAMPLE 3.2
What is the total pressure at the bottom of the tank if the surface of the water in the tank is at atmospheric pressure? The normal value for atmospheric pressure is 101.3 kPa.
3.2.2 DENSITY (ρρρρ)
The density of a fluid is its mass per unit volume. In the SI system, it is expressed as kg/m3. Water is at its maximum density of 1000 kg/m3 at 4˚C. There is a slight
decrease in density with increasing temperature, but for practical purposes, the density of water is 1000 kg/m3. Relative density is the ratio of the density of a liquid
to that of water.
3.2.3 SPECIFIC MASS (W)
The specific mass of a fluid is its mass per unit volume. In the SI system, it is expressed in N/m3. At normal temperatures, water is 9810 N/m3 or 9.81 kN/m3
(approximately 10 kN/m3 for ease of calculation). Pressure Pressure Pa Pa density kg m gravity m s head m