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AN INTRODUCTION TO FOODDEHYDRATION AND DRYING

Donald G. Mercer, Ph.D., P.Eng.

Disclaimer

The author assumes no responsibility or liability for any problems of any mannerencountered through the application of the principles discussed herein.

All “Examples” and “Case Studies” are based on generic or hypothetical cases and donot represent any specific or proprietary processes in current or past use. Such

“Examples” and “Case Studies” are intended for instructional purposes only.

© Donald G. Mercer 2007Reproduction in whole or in part by any means,electronic or otherwise, is forbidden except with

the expressed permission of the author.

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An Introduction to Food Dehydration and Drying Outline: Page i

AN INTRODUCTION TO FOOD DEHYDRATION AND DRYING

OUTLINE

CHAPTER 1: GETTING STARTED

1.1 Learning Objectives

1.2 Background1.2.1 Definition of “Drying”1.2.2 Historical Development

1.3 Reasons for Drying Foods1.3.1 Spoilage Reduction1.3.2 Enhanced Storage Life1.3.3 Changed Storage Conditions1.3.4 Weight Reduction1.3.5 Increased Convenience1.3.6 Changed Properties

1.4 Methods of Water Removal and Dehydration1.4.1 Traditional Methods1.4.2 Non-Traditional Methods

1.5 Factors Influencing Drying1.5.1 Introduction1.5.2 Product Attributes

1.5.2.1 Particle Size1.5.2.2 Particle Shape1.5.2.3 Composition, Structure, and Porosity1.5.2.4 Moisture Content1.5.2.5 Surface Characteristics1.5.2.6 Specific Surface Area1.5.2.7 Specific Heat Capacity1.5.2.8 Seasonal Variation1.5.2.9 Cultivar or Varietal Differences

1.5.3 Drying Equipment Attributes1.5.3.1 Type of Dryer1.5.3.2 Dryer Design Features1.5.3.3 Air Temperature1.5.3.4 Retention Times1.5.3.5 Relative Humidity1.5.3.6 Volumetric Air Flowrate1.5.3.7 Linear Air Velocity1.5.3.8 Air Flow Patterns

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An Introduction to Food Dehydration and Drying Outline: Page ii

1.5.3.9 Seasonal and Daily Variations

1.6 Effects of Drying on Products1.6.1 Introduction1.6.2 Nutritional Degradation1.6.3 Loss of Structural Integrity1.6.4 Reduction in Functionality1.6.5 Flavour and Aroma Changes1.6.6 Colour Changes1.6.7 Case Hardening1.6.8 Leaching of Soluble Constituents

1.7 General Questions

CHAPTER 2: ORGANIZING INFORMATION

2.1 Introduction

2.2 The “Unit Operation” Approach

2.3 Process Flow Diagrams2.3.1 What is a Process Flow Diagram?2.3.2 Preparing a Process Flow Diagram

2.4 Tables and Graphs

2.5 Organization in Problem Solving2.5.1 The Need for Organization2.5.2 Mass Balances

2.6 Practice Problems

CHAPTER 3: DIMENSIONAL ANALYSIS APPROACH TO PROBLEM SOLVING

3.1 Introduction

3.2 Conversion Factors

3.3 Dimensional Analysis

3.4 Sample Calculations

3.5 Practice Problems (with answers)

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An Introduction to Food Dehydration and Drying Outline: Page iii

CHAPTER 4: WET AND DRY BASIS MOISTURE

4.1 Definitions of Wet and Dry Basis Moistures

4.2 Wet and Dry Basis Moisture Conversions4.2.1 Wet Basis Moisture Sample Calculations4.2.2 Dry Basis Moisture Sample Calculations4.2.3 Additional Wet Basis Moisture Calculations4.2.4 Additional Dry Basis Moisture Calculations

4.3 Methods of Moisture Determination4.3.1 Introduction4.3.2 Moisture Balances4.3.3 Vacuum Oven Method4.3.4 Sample Calculations of Vacuum Oven Moisture Tests4.3.5 Miscellaneous Drying Calculations

4.3.5.1 Sample Problem #14.3.5.2 Sample Problem #2

4.4 Case Study: Effects of Feed Moisture on Finished Product Moisture


CHAPTER 5: SOURCES OF INFORMATION

5.1 Introduction

5.2 General References

ACKNOWLEDGEMENTS

During the preparation of the “Introductionto Food Dehydration and Drying” coursemanual, I have had the privilege ofworking with a multi-national team ofdedicated professionals who aremembers of the International Union ofFood Science and Technology (IUFoST),the International Academy of FoodScience and Technology (IAFoST), andthe South African Association of FoodScience and Technology (SAAFoST).The support, encouragement, andenthusiasm of these individuals has beenphenomenal and I would like to publiclyacknowledge their efforts.

Dr. Daryl Lund (USA) is the Chair of theDistance Education Task Force (amongstmany other duties Food Science relatedduties). It is under his watchful eye andinsightful guidance that the concept ofthis Distance Education Initiative isbecoming a reality.

Dr. Walter Spiess (Germany) is a PastPresident of IUFoST and was the drivingforce in establishing the DistanceEducation Initiative for Sub-SaharanAfrica during his term as President. Hehas continued to be a strong supporterand advocate of our activities.

Mrs. Judith Meech (Canada) is theSecretary General and Treasurer ofIUFoST. She has been a tremendousresource in establishing contacts andworking on the administrative aspects.

Dr. Ralph Blanchfield, MBE (UnitedKingdom) is the President of IAFoST. Hehas provided continuing support for thisInitiative, and reviewed the draft copy ofthis course.

The South African Association of FoodScience and Technology (SAAFoST)collectively has been a dedicatedsupporter of this effort. Within SAAFoST,are three gentlemen who havecontributed greatly to making thingshappen. Mr. Owen Frisby, ExecutiveDirector, has worked tirelessly to promotethis program. The time he took to showme the food industry in South Africaduring the week I spent with him in 2006will never be forgotten. Dr. AubreyParsons and Dr. Pieter Van Twiskprovided in-depth comments regardingthe draft course material. Their input hasdone much to ensure the relevance of thematerial for our target audience.

Dr. Samuel Sefa-Dedeh (Ghana) isserving with me as Co-Chair of theModule Development Committee. Hisassistance during my visit to Africa didmuch assist me in understanding the foodindustry in Sub-Saharan Africa.

I would also like to thank TheHonourable Dr. Ruth Oniang’o (Kenya)and faculty members at Jomo KenyattaUniversity of Agriculture andTechnology in Nairobi, Kenya for theirkindness and assistance during my visit inAugust 2006.

Finally, I would like to acknowledge thesupport of my wife, Jane, without whosepatience and understanding I would nothave been able to do this work.

Donald G. MercerApril 2007.

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An Introduction to Food Dehydration and Drying Chapter 1: Page 1.


CHAPTER 1: GETTING STARTED

1.1 Learning Objectives

The purpose of this manual is to providea basic understanding of dehydration anddrying as it is used in the food industryand in various sectors of the agricultureindustry.

In Chapter 1, we will examine the reasonsfor drying food, and then look at thefactors influencing drying. We willconclude Chapter 1 with a discussion ofthe effects of drying on the product.

A difficulty that many people have inunderstanding drying involves thegathering and organization of informationassociated with their specific dryingprocess. Chapter 2 is designed to assistyou in these key activities. Theapproaches outlined in Chapter 2 can beapplied to most areas of food processingand are not limited to drying.Construction of process flow diagramsand the “Unit Operations” concept will beintroduced here. In order to illustrate this,a case study example that goes beyondjust the drying of a product is included forillustrative purposes.

Once the information regarding a processhas been gathered and organized in amanner that makes it clearer and easierto understand, the next step is to developrelationships between the raw materialsentering the process and the finishedproducts leaving the process. In a dryingprocess, this generally involves a seriesof mathematical calculations to determineamounts of moisture removed during the

time that the materials are subjected tovarious drying conditions. They may alsoinvolve heat calculations. It is absolutelyessential that you develop a structuredapproach to the mathematics involved.The mathematics is usually not toocomplex. However, if you do not follow alogical and systematic approach to yourproblem solving, it is easy to becomeconfused and ultimately frustrated.Chapter 3 has been included to introducethe “Dimensional Analysis” approach toproblem solving. This is a relativelystraight-forward, yet highly effectivemethod of attacking mathematicalproblems that involves using numbersalong with their associated units ordimensions such as kg or kg per hour,etc.

Once you feel comfortable with theorganization of information from Chapter2 and the use of Dimensional Analysisfrom Chapter 3, you will be ready to moveon to Chapter 4 which introduces theconcepts of wet basis moistures and drybasis moistures. Sample calculations andcase studies are included to familiarizeyou with these concepts and theirapplications to drying process work.

After completing the “Introduction to FoodDehydration and Drying”, you should:

C be familiar with the reasons why food isdried and the factors that influencefood drying.

C be able to “break down” a food processinto its component “unit operations”.

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C be able to prepare a process flowdiagram for a typical food process andunderstand the basic functions of theunit operations involved in it.

C be able to understand and apply the“Dimensional Analysis” approach insolving food processing relatedproblems (as well as to be able todetect calculation errors on the basis ofthe units or dimensions involved).

C be able to calculate (or estimate) themoisture content of a finished productbased on information regarding itsinitial moisture content and the amountof water removed.

C be able to understand the pitfallsinvolved in using wet basis moisturevalues when working with the watercontent of food materials.

1.2 Background

1.2.1 Definition of “Drying”

For the purpose of this work, we will referto “drying” as the removal of water fromany material, such as a food product.This may be done by means of exposingthe material to heated air in a chamberwhich we will call a “dryer”; or it may beaccomplished through natural meanssuch as solar drying where the sunprovides the energy necessary to removethe water from a food product.

The term “dehydration” should beconsidered as being essentially the sameas drying. Although some food scientistsand food processing engineers mayconsider that there are subtle differencesbetween the two terms, we will not worryabout these differences here.

1.2.2 Historical Development

Drying is one of the oldest forms of foodprocessing. Its origins most certainly pre-date recorded history. It is not hard toimagine early hunters and gatherersfinding berries that had become dried onthe vines. These dried berries wouldhave possessed characteristics that weredifferent from the fresh undried fruit, butmost importantly, they would haveremained edible for longer periods of timethan they did in their undried form. Otherforms of food may have also been driednaturally by laying them out in the sun.

With the discovery of fire, a new means ofdrying was available. It offered numerousadvantages over sun-drying since it wasmore reliable, more controllable, andmuch more convenient. Meat could bedried over a fire and gain the addedadvantages of cooking and smoking.

If we come forward to the 21 Century,st

we can see tremendous advances havebeen made in all areas of foodprocessing. Not the least of these areasis food “dehydration” or “drying”.Although the drying process still basicallyinvolves the removal of water, moderndrying equipment has replaced primitiveopen fires and scientific knowledge hasadded to our understanding of it. Eventhough sophisticated drying equipment isavailable in many parts of the world, sundrying, drying over an open fire, andsmoking are still viable ways of drying andpreserving certain foods.

In spite of all of the modern developmentsin the field of drying, drying still retainsmuch of its “art”. To produce high qualityproducts which meet the demands oftoday’s consumers, it is simply notenough to know the theoretical aspects of

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drying - one also has to know the “art” ofapplying this knowledge to the materialbeing dried.

1.3 Reasons for Drying Foods

1.3.1 Spoilage Reduction

The most important reason for dryingfood is to prevent or reduce spoilage andthereby create a product that remainsedible for a longer period of time than itwould in its undried form. A suitable levelof moisture is required by spoilagemicroorganisms in order to sustain theirgrowth and solublize nutrients that theyrequire. Without this moisture,microorganisms will either lie dormant ordie.

One of the most commonly observedspoilage microorganisms is “mold”.Colonies of mold growth often appears as“fuzzy” grey spots on the surface of afood product.

While mold is generally quite visible to theconsumer, other spoilage microorganismsmay not be so easy to see.

Clostridium botulinum is an earth-bornebacteria that can be found on thesurfaces of vegetables and root crops etc.If allowed to grow in a food product, it canproduce a toxin that has serious healthimplications which we know as foodpoisoning or “botulism”. Similarly,microorganisms like Salmonella, Listeria,and E. coli can produce potentially deadlytoxins in foods if measures are not takento properly process and preserve thefood.

By lowering the moisture content of afood product to a sufficiently low level,

additional advantages can be obtained inaddition to spoilage reduction. These arediscussed in the following sections.

1.3.2 Enhanced Storage-Life

Water removal can increase the durationof time over which a food remains edible.While this is essentially the same asspoilage reduction, it is worthy of mentionas a reason for drying a food product. Asan example, dried berries will last longerthan in their fresh undried form. In manycases, water can be added back intodried products, such as the dried berries,to rehydrate them and regain many of theattributes of the initial product.

1.3.3 Changed Storage Conditions

All foods require certain conditions toprevent their spoilage. It is also desirableto preserve their taste, nutritionalattributes, and other properties, such asaroma. In order to accomplish this,special storage conditions such asfreezing or refrigeration may benecessary. If the food can be dried andstored at ambient or room temperature, itreduces the expense of maintainingspecialized storage conditions. It alsoincreases the potential market for aproduct when such storage andtransportation conditions are not required.

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1.3.4 Weight Reduction

In today’s world, food production areasare often far from the areas where thefood will be consumed. Much of theweight of many food products isattributable to the presence of water. Ifwater can be removed, the weight can bereduced and the cost of shipping theproduct can also be reduced. Often,there is a volume change associated withfood drying. As water is removed, thefood shrinks in size, thereby taking upless room and allowing more product tobe shipped in a given volume.

Shipping water from one location toanother is viewed as an undesirable andunnecessary expense. The removal ofwater from orange juice to create orangejuice concentrate is a good example.With orange juice, the water is removedin a concentration process that is not atypical drying process. However, theexample is an appropriate one to showhow water removal can be used to goodadvantage. Removal of water to create aconcentrate with 65% solids instead ofthe usual 9% to 10% solids in freshsqueezed orange juice reduces both theshipping weight and volume. Once theconcentrate arrives at its intendeddestination, or is purchased by theconsumer, water can be added back intoit to dilute it to the desired final watercontent prior to consumption.

1.3.5 Increased Convenience

Many consumers want foods that areconvenient. This is especially true incases where there may not be enoughtime available to prepare a meal fromfresh unprocessed products. Examplesof this would include dehydrated potatoflakes or instantized rice. The consumermay take the desired portion of driedproduct, mix it with boiling water, allow itto stand until the moisture is taken intothe dried product, and then serve it . Withrecent technological advances, suchproducts are often better in quality thansimilar products in their in fresh form.

Another benefit of dehydrated foods isthat waste may be reduced duringpreparation by the consumer. In the caseof dehydrated potato flakes, there is noneed for the consumer to peel thepotatoes and remove undesirableblemishes or “bad spots”. All of theproduct is usable in its purchased form.

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1.3.6 Changed Properties:

Dried foods may offer different propertiesfrom their undried forms. Thesedifferences may create unique productsthat have applications which are quitenovel compared to the original material.In addition, through drying, value may beadded and new markets may be createdto increase the use and commercial valueof a particular raw material.

Consider the following examples:

Dried products such as raisins (from driedgrapes) and prunes (from dried plums)offer taste and texture differences fromtheir original form. Sun dried raisins areused in a different manner than grapesdue to their changed properties that haveresulted from water removal.

Dried apple slices can be used incombination with dried cereals to addincreased taste and flavour. They alsocreate a textural change from the cerealflakes. While the dried apples do notregain the characteristics of fresh appleswhen they pick up moisture from the milk,their soft chewy texture is quite appealingand pleasant to many individuals.

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1.4 Methods of Water Removal and Dehydration

We will be discussing this topic in muchgreater depth as we move on to theIntermediate Level material. However, inorder to appreciate some of theintroductory material, it would beadvantageous to have someunderstanding of the various methods ofwater removal in food processing.

1.4.1 Traditional Methods

Most traditional drying methods rely onthe application of heat by direct or indirectmeans.

In the case of direct heat application, theheat is usually brought to the product byair which has been previously heated tothe desired temperature. It is the functionof the heated air to evaporate moisture ina product and transport the evaporatedmoisture out of the dryer. Air can beheated by passing it through flames froma gas burner or similar source of heat. Aircan also be heated by passing it overmetal tubes through which steam iscirculated. The hot air is then blownacross or through the product being dried.This is a variation on the simplestexample of direct heat application whichwould be drying something over an openfire. Many commercially available dryersare based on direct heat application.

An example of indirect heat applicationmight be the use of a drying oven whichhas metal walls heated from the outside.The heat radiating from the metal ovenwalls then heats the air and product in theoven which in turn dries the product. Thisindirect heat application is generally notas efficient, nor as effective, as direct

heat application.

1.4.2 Non-Traditional Methods:

Non-traditional drying methods may alsobe referred to as “emerging” dryingtechnologies. As drying technologiesbecome more advanced, we are seeingmethods of water removal which areaimed at minimizing the application ofheat. When drying food products that aresensitive to heat, this is especiallyimportant in order to prevent the qualityof the food from being reduced.

We are also seeing methods that addressenergy efficiency. In these cases, theapplication of heat is highly controlled soas to avoid waste and apply heat in amanner that matches the needs of theproduct rather than simply applyingexcessive amounts of heat to drive off asmuch water as possible in the shortestpossible time.

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1.5 Factors Influencing Drying

1.5.1 Introduction

For purpose of the following discussion,let us consider materials that containwater on the outside surface anddistributed throughout the interiorstructure of the material as well.

We will divide the factors influencingdrying into two basic groups: ProductAttributes and Drying EquipmentAttributes.

As we consider the factors that influencedrying, I will attempt to give examples thatrelate specifically to food, but otherexamples will also be included forillustrative purposes. You should drawupon your experiences in everyday life tosupport your visualization of the conceptsof drying. Even if it is something asseemingly simple as drying wet clothes,there are lessons to be learned that canbe applied to more complex dryingoperations.

1.5.2 Product Attributes

1.5.2.1 Particle Size

Drying is a process that takes place at thesurface of the material for much of thedrying time. Some degree of drying mayalso take place just below the surface, butwe generally should focus on waterremoval as being a surface phenomenon.

When we have large particles, it takeswater at the centre of the particle moretime to travel to the surface and beremoved by the drying air than it would forwater to reach the surface from the centre

of a smaller particle. For this reason, wemust adjust our drying times to accountfor the size of the particle.

Water travels from the interior of theproduct to the surface by a processknown as “diffusion”. The rate ofdiffusion, or how much water reaches thesurface in a given period of time issomething that is dependent on theparticle size and the structure of thematerial (see below). Under normaldrying conditions, there is very little thatcan be done to speed up the rate ofdiffusion and you must match your dryingprocedure with the particle size and typeof material being dried.

If we are drying hydrated cereal grainssuch as rice or wheat, the more plump(i.e., bigger diameter) grains will takelonger to dry than the smaller diametergrains due to the distance that the waterhas to travel from the inner core to reachthe surface.

FIGURE 1-1: Particle Size Differences

As you can visualize from the two circles(or spheres) in Figure 1-1, any water atthe centre of the larger particle will havefurther to travel to reach the surface thanit would in the smaller one. This meansthat it will take longer for the water toreach the surface of the larger productparticle and be removed.

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1.5.2.2 Particle Shape

Shape is another important considerationin drying any material. Shape is alsoimportant when it comes to cooking amaterial and getting heat into it.

There are really only three basic shapesof objects that we have to deal with indrying. They are shown in Figure 1-2.

i. spheres ii. cylinders

iii. flat plates or slabs

FIGURE 1-2: Basic Shapes of Objects

Spherical objects are those that are roundlike a ball. Peas are spherical, as aremost berries. With spherical objects, themaximum possible volume is containedwithin the smallest possible surface areafor a given weight of material. As a result,spherical particles can be rather difficultor slow to dry. As stated above, it takestime for moisture at the interior of thesphere to diffuse to the particle surfacewhere evaporation by the hot air or otherdrying medium will occur. With sphericalobjects, the total surface area isconsidered to be important in the removalof water during drying.

Cylindrical objects are long but have a

circular cross-section. A carrot would beconsidered to be a cylinder, even thoughone end of it might be larger in diameterthan the other end. Wieners andsausages are cylindrical, as are cerealgrains such as rice. When moisture isremoved from a cylindrical material, watermust diffuse from the centre of thecylinder and travel outwards along theradius to reach the surface just as it did inthe case of the spherical object. Themain difference between drying aspherical object and drying a cylindricalone is that the ends of the cylinder aregenerally not considered as being veryimportant in the drying process. Theends represent a relatively small fractionof the surface area for water removalwhen compared to the sides of thecylinder. When drying a cylindrical object,there is usually more surface area perunit weight of material than there is for aspherical object, so drying is somewhatfaster with cylinders than with spheres. Itshould be noted that large diametercylinders will take longer to dry thansmaller diameter cylinders.

Flat plates or slabs may be thought of aslooking like a flat piece of wood that hasflat top and bottom surfaces and a flatsurface around its edge. A steak or apiece of flat cookie dough may beconsidered as being slabs. In drying,water would be lost from the top andbottom surfaces. The surfaces aroundthe edge would play a minor role in waterremoval. Because the surface area of aslab is often quite large in relation to itsthickness, it may be easier to dry slab-shaped objects than it would be to dryspherical or cylindrical objects.

Each of the three basic shapes has acharacteristic dimension that determinesthe relative drying rate. In the case of

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spherical objects, it is the radius (i.e., halfof the diameter) that is important. Forcylinders, it is also the radius that is mostimportant. For flat plates or slabs, thecharacteristic dimension is one-half thethickness since water only has to diffusefrom the centre of the slab to either thetop or bottom surfaces of the slab.

If we consider an everyday example, thinkof drying a wet sock that is rolled up intoa ball and an identical wet sock that isspread out fully. The rolled-up sock willresemble a sphere and the flat sock willlook like a flat plate or slab. You probablyknow from experience that the sock thatis spread out will dry faster than therolled-up sock, which shows theimportance of shape in influencing drying.

1.5.2.3 Composition, Structure, and Porosity

The influence of composition andstructure on drying is quite pronounced.A tight, dense structure will tend to holdmoisture inside the material more than aloose open structure will. Large porespenetrating throughout the interior of aproduct or material will allow diffusion ofmoisture to proceed at a faster rate thanin the case where there are very few orvery narrow pores.

A starchy material will hold more moisturewhen the starch is gelatinized and the“starch matrix” is swollen with absorbedwater. A fibrous material may loosemoisture more readily than a less fibrousmaterial due to the routes available formoisture to escape from the interior to thesurface.

The importance of composition, structure,and poros i t y shou ld not be

underestimated. These are quitecomplex properties that may even changeduring the drying process. As water isremoved, the “starch matrix” of somethinglike hydrated rice may collapse in on itselfand change the diffusional properties ofthe material. Temperature can alsochange starches from a non-gelatinized toa gelatinized form and affect the dryingprocess.

One of the most important things torecognize here is that before purchasinga dryer to dry any product, you should doa series of thorough investigative trialsusing all available resources, includingpilot-scale drying equipment if that ispossible. You should work with dryersuppliers to determine how your productbehaves throughout the drying processand leave nothing to chance.

1.5.2.4 Moisture Content

Perhaps one of the most obvious factorsaffecting drying is the actual moisturecontent of the material itself. In mostdrying applications, we have a final targetmoisture to which we want to reduce theproduct moisture. If the initial moisture ofthe starting material is high, then moremoisture will have to be removedcompared to cases when its moisturecontent is somewhat lower. This will takeadditional drying time which must betaken into account when establishingoperating conditions for the dryer.

In setting up a dryer, the initial moisturecontent and amount of product beingprocessed per unit time (e.g., kilogramsor “kg” of product at a given moisturecontent going into a dryer each hour) arekey variables to consider.

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1.5.2.5 Surface Characteristics

Surface characteristics are important fromseveral perspectives. First, we shouldconsider that in many cases, such asberries, the surface is designed to retainmoisture. If we want to dry berries, wehave to realize that nature has intendedthe berry to protect the seeds to ensurethat the plant reproduces. In order tokeep a supply of moisture around theseed, the surface of the berry is oftenquite tough, or waxy, or otherwise forminga barrier to moisture loss. If we wish todry the berry, then we must recognize thisfact and take measures to overcome thisbarrier to moisture loss. We could slicethe berries to allow the drying medium(i.e., the air) to reach the fleshy moist partof the berry. Alternately, we could rupturethe skin of the berry to allow pointsthrough which the moisture could escapeduring drying.

If the surface of a product is quite roughand pitted, this may allow moisture toescape better than if the surface wasrelatively smooth.

1.5.2.6 Specific Surface Area

Specific surface area is an indication ofthe surface area present in a unit weightof material. It could be measured in unitsof square centimetres per gram or squaremetres per kilogram (m /kg), or some2

other appropriate set of units. Basically,it acknowledges that drying is a surfacephenomenon and the more surface that isavailable for moisture loss, the faster thedrying process can proceed.

We use this principle frequently when weare cooking food to speed the heatingprocess. For example, when we want to

boil potatoes for a meal, we usually cutlarge potatoes into smaller pieces so theywill cook faster. The cutting processcreates more surface area for the heat topenetrate than was available before thepotato was cut. The cutting also reducesthe distance the heat has to travel toreach the centre of the piece of potato.The same thing is true with drying. If wehave smaller pieces, we have moresurface area available from which themoisture can escape. In addition to this,we have the reduced thickness of thepieces which reduces the time it takes formoisture from the centre of the material toreach the surface and be evaporated bythe drying air.

1.5.2.7 Specific Heat Capacity

Although we will not be using specificheat capacities in our early discussionsabout drying technology, it does have aneffect on drying, and it should bementioned here for the sake ofcompleteness.

pSpecific heat capacity (C ) is a measureof how much heat it takes to raise thetemperature of one kilogram of a materialby one Celsius degree. It is expressed inunits of kilojoules per kilogram perCelsius degree (kJ / kg C°). The higherthe water content of a product, the higherits specific heat capacity. Pure water hasa specific heat capacity of 4.187 kJ / kgC°. Potatoes, which are approximately78% water by weight have a specific heatcapacity of 3.450 kJ / kg C°. Okra, havinga moisture content of about 90% water byweight, has a specific heat capacity of3.852 kJ / kg C°. This means that it takesmore heat to raise the temperature of agiven weight of okra by a certaintemperature than it does to raise the

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temperature of the same weight ofpotatoes by the same number of degrees.Or, putting it another way, if we apply thesame amount of heat to equal weights ofpotatoes and okra which are at the samestarting temperature, the potatoes willhave a higher final temperature than theokra.

The effect of specific heat capacity ondrying may not be as obvious as some ofthe other factors discussed here. In mostcases, when material is first brought intoa dryer, its temperature must be raised toa certain level to enhance the dryingprocess. When we introduce a productinto the dryer, it may take longer to getproducts with higher specific heatcapacities up to the desired temperature

pthan it would for products with lower Cvalues. This can lengthen the timerequired for drying. Although the effectsof specific heat capacities may not alwaysbe noticeable, they should be mentionedhere for due consideration.

1.5.2.8 Seasonal Variation:

Seasonal variation from one crop year tothe next is a factor that often confrontsfood processors. Consider cereal grainssuch as rice. In a good crop year, the ricekernels may be large and quite plump. Ina poor growing season, the kernels maybe significantly smaller and not as plump.This size variation can have quite aneffect on how the kernels dry. If you haveyour dryer set up to process rice kernelsfrom a rather poor growing season andthen change to a new shipment of ricefrom a very good growing season, you willprobably find that the kernels have nothad enough moisture removed. This isbecause moisture inside the kernels withthe bigger diameter requires more time to

reach the surface of the kernel than it didin the smaller kernels. If your dryer is setup to give the proper drying time to thesmaller kernels, the larger kernels willtend to be higher in final moisture untilyou make the necessary adjustments tothe dryer.

1.5.2.9 Cultivar or Varietal Differences

Not all varieties of a particular agriculturalcommodity have identical properties. Thekernel sizes of grains of rice will varydepending on whether the rice is a long-grain or short-grain variety. Thecharacteristics of the starches may alsovary between varieties of rice and otherstarchy foodstuffs such as cassava orpotatoes. These differences will affecthow water is held within the product,which will ultimately lead to differencesbetween the drying properties of twovarieties of the same plant.

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1.5.3 Drying Equipment Attributes

Just as the properties of the materials orproducts being dried have an impact onthe overall drying process, so also do theattributes of the equipment that is beingused to do the drying. We will not go intoany great detail at this point, since we willbe discussing various factors about dryeroperation and design at a later time.However, we will introduce several ofthese factors at this time.

1.5.3.1 Type of Dryer:

Keep in mind that all materials havesomewhat different drying characteristicsthat require special consideration. Dryingparsley flakes is vastly different thandrying corn for animal feed. Drying wheyto recover its solid components is muchdifferent than sun-drying tomatoes. Dryingpotatoes to produce potato flakes isdifferent than drying tobacco leaves in akiln. The list goes on and on. The pointhere is that there are many types ofdryers available from manufacturers toaddress the needs of a particular product.If you purchase a dryer to dry parsleyflakes, you should not use it to drypotatoes for potato flakes. If you do usea dryer for a purpose other than for whatit was designed or intended to be used,do not be surprised if it does not workproperly for you and your final product isnot of the quality you had hoped toobtain.

1.5.3.2 Dryer Design Features

Just as the actual type of dryer can affectthe drying of a product, features that areincorporated into the design of a dryercan also have an impact. Some dryersuse a moving wire mesh belt to convey

product through a drying chamber. Withthis arrangement, particles of product arestacked up on top of each other and air ispassed through the bed of material toremove moisture. To overcome theeffects of surface-to-surface contactamong the food particles and to create abetter overall contact of the drying air andthe product surface, some dryers may beequipped with a vibrating mechanism thatagitates the particles of food as theytravel through the dryer. As the particlesof food bounce upwards and fall backdown onto the vibrating surface of thedryer, they experience closer contact withthe drying air than they would experiencein a flat bed on a moving conveyor.

1.5.3.3 Air Temperature

Most drying applications in the foodindustry use hot air as the drying medium.As stated above, the air can be heated bypassing it through the flames from aburner assembly much like that used in ahome gas or oil-fired heating furnace. Insome cases, air may be heated bypassing it across electrical heatingelements. Regardless of the source ofthe heat, the temperature of the air is akey factor in drying any material. Thehigher the temperature of the air, thegreater its water holding capacity will be,which in turn means that its ability toremove water from a product will begreater than if cooler air was used

While it may be tempting to use thehighest possible air temperature in aparticular dryer, extreme caution must beexercised in this regard. First of all, someproducts are very sensitive to heat andtheir quality diminishes substantially ifthey are exposed to excessivetemperatures. If temperatures are toohigh, the product may be scorched or

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toasted. A product like parsley can onlybe dried at temperatures in the area of60°C for this reason. Although it maytake a long time to remove the moisturefrom sprigs of parsley at this temperature,anything above this limit will seriouslydamage the quality of the dried parsleyflakes.

Another potentially negative effect ofexcessive temperatures is somethingcalled “case hardening”. If the wetmaterial is exposed to too much heat, itsouter surface may become very dry in ashort period of time and create a shell orcrusty layer around the material. Thishardened surface will trap the moisturethat was at the centre of the product andprevent it from reaching the outer surfaceof the particle where it normally wouldevaporate. Since the moisture cannotescape, it will remain inside the “casehardened” piece of material. After thecase hardened product is in storage for awhile, the moisture that is trapped in it willbegin to create problems. Over time, themoisture can diffuse through the surfaceshell and condense inside the package orstorage bin where it can promote moldgrowth etc. It can also cause thestructure of the dried food particles orpieces to change over time. Often theouter shell will remain intact but the innerstructure of the particle collapses in onitself due to the effects of the moistureinside it. When the consumer goes touse the product, the first thing that theynotice is that the product does notfunction as they expected and that thetexture of it has changed drastically fromthe normal texture. Care must be takento apply heat at a rate that does not caseharden the product - unless some degreeof case hardening is actually desired.

We have discussed air as the heat

delivery medium for most dryingapplications. However, we should notlose track of the fact that water can alsobe removed from food materials by othermethods as well. Let’s consider whathappens when products are fried in hotoil. When the product containingmoisture is dropped into the hot oil, thewater will heat up rapidly and will beconverted into a vapour or “steam”. Thevapour then escapes into the oil andsubsequently leaves the oil and goes intothe atmosphere above the oil fryer. Theloss of water leaves the product with alower moisture content and the hot oilcreates a crisp or browned surface. It isthe escaping water vapour which isresponsible for the violent actionobserved when food is initially droppedinto an oil fryer.

Water can also be removed usingconcentrated sugar solutions in a processknown as “osmotic dehydration”. Briefly,when food materials such as sliced fruitsor berries are soaked in a concentratedsugar solution (about 60% sucrose) at anelevated temperature (50° to 60°C),moisture is drawn out of the cells of theproduct and into the concentrated sugarsolution through “osmosis”. Over time,the sugar solution becomes diluted fromthe water pulled out of the product beingdried and measures must be taken torestore the sugar solution to its originalconcentration.

1.5.3.4 Retention Times:

Just as the temperature of the drying aircan affect the final product quality, so toocan the length of time during which theproduct is exposed to the hot air. Weknow from our experiences cooking foodin our kitchen ovens that it takes less time

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to cook something at a highertemperature than it does at a lowertemperature. Temperature and time areinter-related in drying in basically thesame way. It should take less time to drya material at a higher temperature than itwould at a lower temperature. However,there are some complicating factors.

We cannot just hit a product with as muchheat as possible and expect it to dryproperly. We must recognize that it takestime for moisture to travel from the centreof a food particle to the outside surfacewhere it can be evaporated by the hot airin the dryer. We must balance the time inthe dryer at a specific temperature withthe ability of the moisture to diffuse out ofthe food material. In the case of parsleyat 60°C, it takes a number of hours to dryit to the desired final moisture level.Other products may only take 20 or 30minutes to dry under their particular set ofdrying conditions.

1.5.3.5 Relative Humidity

The relative humidity of air is a measureof how much moisture the air is holdingcompared to how much moisture it couldpossibly hold if it was saturated withmoisture under those conditions. On ahot humid summer day, the airtemperature around us might be 30°Cand the relative humidity could be 90%.This means that the air is holding 90% ofthe maximum amount that it couldpossibly hold at that temperature. Itsability to pick up water throughevaporation would be much less than ifwe had air at 30°C with a relative humidityof 20% or 30%. If we try to dry a materialwith air that already contains a highmoisture content, we will not be able toremove as much water as we could with

air containing a lower moisture content.

If we take the 90% relative humidity air at30°C and heat it to 60°C, its relativehumidity would be about 20% since thehotter the air, the more moisture it canhold. This is one very important reasonfor heating air in any drying application.

Anyone who is involved in food dryingshould own and use a device formeasuring relative humidity (RH). TheseRH meters are a valuable tool inmonitoring the moisture content of airentering a dryer and exhaust air leaving adryer. In the “Advanced Drying Course”,we will examine the topic of“psychrometrics” which allows for theestimation of the properties of air atvarious temperatures and moisturecontents. The relative humidity andtemperature of the air are essentialpieces of information in this procedure.

1.5.3.6 Volumetric Air Flowrate

Volumetric air flowrates are a measure ofthe amount of air delivered into the dryingchamber of a dryer in any given timeperiod, and are expressed in units ofvolume per unit time (e.g., cubic metresper minute or cubic feet per second, etc.).They can be controlled by adjusting thefan speeds on the dryer to deliver more orless air volume in a given period of time.Volumetric air flowrates are an alternateway of delivering more heat to the dryer.In many dryers, you can either increasethe temperature of the air to introducemore heat; or you can increase thevolume of air passing through the dryer;or you can do a combination of both.

Adjusting the volumetric air flowrate isparticularly useful when dealing with

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materials that are heat sensitive. If weconsider the parsley example once again;we may need more heat to dry a load ofparsley, but we cannot increase thetemperature due to potential heatdamage to the parsley. By increasing thevolume of air we blow through the dryer,we could accomplish much the samething and increase the drying capacity ofthe dryer.

1.5.3.7 Linear Air Velocity

The linear velocity of the air is directlylinked to the volumetric air flowrate. It isthe speed at which the air is movingthrough the dryer, and would beexpressed in metres per second, or feetper second, or some other similar units ofspeed.

Since most dryers have fixed dimensions(i.e., the size of the dryer does notchange), the cross-sectional area of thedryer is constant. If you increase thevolume of air being blown through thedryer, its linear velocity (i.e., speed) willincrease proportionately. If you decreasethe volumetric flowrate of the air, itsspeed will decrease.

The main effect of linear air velocity in adryer is that if it becomes too great, it mayhave sufficient force to lift pieces of thematerial being dried and blow themaround inside the dryer. Not only can thisresult in lost product being blown right outof the dryer, but it can result in unevendrying. If the material is spread in a bedof uniform thickness across the dryingsurface, the air can blow holes in this bedof material. The holes in the bed thenbecome paths of low resistance to the airas it attempts to reach the outlet orexhaust vents of the dryer. Product

around these holes may become overlydried while product at other locations inthe dryer is not sufficiently dried.

1.5.3.8 Air Flow Patterns

We will deal with this topic in more detailin a future module. However, it should bementioned at this point that it is extremelyimportant to have the airflow as uniformas possible in a dryer. Sometimes, wemay want the air to flow upwards throughthe product, or downwards through theproduct. In other applications, we maywant the air to flow across the top surfaceof the product.

Dryer design is the major contributingfactor to how the air is delivered to theproduct being dried.

1.5.3.9 Seasonal and Daily Variations

Earlier, we mentioned how seasonalvariations in an agricultural commoditysuch as cereal grains can influence theirdrying. We can also notice variations inhow well a dryer performs from oneseason to the next. In Canada, we havesome large swings in temperaturebetween the winter and summer months.In January and February, when ourclimate tends to be most severe, it ispossible to have days when thetemperature is as low as - 40°C (or evenlower). During the summer months ofJuly and August, temperatures can rise toabove 30°C.

When air is brought into a dryer, theburners heat it to the desired temperaturebefore the air goes into the dryingchamber. Cold winter air takes more heatto raise it to the proper temperature than

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does warmer summer air. In cases wherethe dryer is always operating with theburners turned to their maximumcapacity, it may not be possible for theburners to heat the cold winter air to thesame temperature as they could duringthe summer when the air is warmer.

Humidity of the air is another factor toconsider. Because the air is colder inwinter, it does not contain as muchmoisture as the humid warm summer air.While this humidity change may notcreate any major differences in how wella dryer functions during the summer orwinter months, it does have an impact ofwhat is called the “equilibrium moisturecontent” of the product.

All food products contain some level ofmoisture. Even dried wheat, or rice, orother cereal grains contain about 10%water by weight. If we were to completelydry these materials so that they containedabsolutely no moisture and expose themto the ambient air (i.e., the air aroundthem), they would start to pick upmoisture. They would continue to pick upmoisture until there was a balance orequilibrium between their water contentand the water content of the air aroundthem. Even if the product was packagedin a container that allowed some degreeof air passage, there would be anequilibrium relationship establishedbetween the product and the air.

In winter, when the moisture content ofthe air is lower than in the summer, theequilibrium moisture content of most driedproducts is lower than it would be in themore humid summer air. Foodprocessors may need to take theequilibrium moisture content into accountwhen setting their drying specificationsthroughout the year.

Just as there can be seasonal variationsin drying, there can also be daily ordiurnal changes. I have seen instanceswhen product coming off the dischargeend of a dryer has picked up measurableamounts of moisture from the air. Toillustrate this point, consider the followingsituation. On a hot summer afternoon,the operators of a drying facility had allavailable windows open. Since therewere screens on the windows, this wasnot a problem in terms of insects orforeign material getting into the product.Everything went well during the day andthe moisture of the product coming out ofthe dryer was exactly on target. However,late at night, the operator on the midnightshift frequently saw that the finishedproduct moisture increased to the pointwhere the product moisture was at theupper end of its allowable range. Thishappened even though nothing in theactual drying process had changed sincethe afternoon. The problem then becameone of identifying the source of thismoisture increase.

The reason for the moisture increase inthis case was attributed to the air comingthrough the screened windows. Duringthe day, the air was warm enough that itdid not pose a problem. When the aircooled at night, its water holding abilitywas reduced. The cool moist air comingin contact with the warm dry productcreated a situation whereby the moisturefrom the air was taken in by the product,which raised the moisture content of theproduct. By closing the windows early inthe evening, the problem was eliminated.

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1.6 Effects of Drying on Products

1.6.1 Introduction

In an ideal world, drying would have nonegative impact on the final productwhatsoever. We all know that this is notan ideal world and as a result, we mustdo whatever we can to minimize thenegative effects of drying on the variousquality aspects of the product.

In cases where we prepare a dehydratedfood product that is rehydrated beforeeating, the quality attributes of therehydrated food are expected to be asclose to that of the original product aspossible. In some cases, the driedproduct may even be expected to haveimproved attribute when compared to theinitial material. An example of this couldbe instant mashed potato flakes. Theconsumer may feel that for the price ofthe product, not only should they be moreconvenient, but the resultant mashedpotatoes should be lighter or fluffier, orcreamier than the mashed potatoesprepared from whole potatoes at home.Not only that, they should be free fromany black marks and the like that may befound in home-cooked potatoes.

The following is a brief summary of anumber of negative factors that may beintroduced into a food product if drying isnot properly conducted:

1.6.2 Nutritional Degradation

Vitamins that are present in food productsmay be unstable and susceptible to heatdamage. Overheating due to excessivetemperatures or prolonged exposure toheat during drying can cause thisproblem.

1.6.3 Loss of Structural Integrity:

On occasion, improper drying can createstresses in the structure of a material thatcan have serious negative results.

Consider the example of a starch-basedcereal grain that is placed in boiling waterto “cook” it. When heated sufficiently, thestarch can gelatinize to form a rather softand pliable material. It also expands inwater and the water can be taken into thestarch network or matrix. Once thestarch-based cereal grain is “cooked”, itmay be necessary to dry it. As the waterleaves the inner regions of the cerealgrain, the starch matrix may shrink backtowards its original size, or it may stay inits expanded state. Once it is dried andcooled, the gelatinized starch changesfrom a soft, pliable material to a hardglassy or brittle material. If the drying andcooling have been done rapidly, theglassy starch may be under considerablestress. Over time, these stresses maycause the dried product to break apart orshatter.

1.6.4 Reduction in Functionality

“Functionality” is a term used to describethe fact that various components in a foodsystem have a role to perform. Forexample, ungelatinized starch may bepresent in a food (e.g., corn or potatostarch). If ungelatinized starch enters adryer, it may become gelatinized if thetemperature of the material exceeds thegelatinization temperature of the starch.Typically, potato starch gelatinizes at atemperature of around 60°C.

Once it has gelatinized, the starchbehaves much differently than it didbefore gelatinization. Ungelatinized

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starch is not soluble in water; butgelatinized starch is able to take up waterand swell into a rather wet pasty mass.When compressed, the starch gel mayrelease much of this moisture. Care mustbe taken to dry ungelatinized starchymaterials at temperatures below theirgelatinization temperatures if functionalityof the starch is a concern.

1.6.5 Flavour and Aroma Changes

Flavour changes imparted by drying maybe desirable or undesirable dependingupon the end use of the dried product.During drying, heating may create atoasted flavour in the product along withthe associated toasted aroma.

The development of caramel flavours canbe attributed to non-enzymatic browningthrough the Maillard Reaction wherereducing sugars in the product react withamino acids.

Other flavours and aromas may alsodevelop, depending upon the nature ofthe material being dried and the dryingconditions used.

1.6.6 Colour Changes

In the previous section, we mentioned thedevelopment of caramel flavours throughthe Maillard Reaction. This non-enzymatic browning also produces acharacteristic brown colour (hence thename). This is particularly evident in theproduction of maple syrup from the sap ofNorth American sugar maple trees. In thetraditional process, sugars in the tree sapare concentrated from an original level ofless than 2% by weight to over 65% byweight. In the original process, water was

removed by boiling the sap in openkettles over a fire. This prolonged heatingcaused the syrup to turn brown anddevelop desirable flavours.

Some colour changes can be quiteundesirable, however. If a white productis being dried, it may become “toasted”and have a light brown colour whichconsumers find objectionable.

1.6.7 Case Hardening

We have already mentioned casehardening under the heading of “AirTemperature” earlier in this module. It isbeing repeated here to re-emphasize thenegative impact that drying can have on aproduct if it is not done properly.

Case hardening is a term that refers tothe development of a hard shell or crustaround the outside of the food particlesbeing dried. It is caused by excessivedrying of the surface of the food particleand creates a shell that prevents moisturefrom being removed from the interiorportions of the food particle.

In normal drying, moisture is removed byevaporation from the surface of the foodparticles after which moisture from theinner portions of the particle travels to theparticle surface where it, in turn, isevaporated. If the surface moisture isremoved too rapidly, the surface maybecome overly dry and form a hard layerthat prevents any more moisture fromreaching the surface. Since the interiormoisture cannot be removed during thedrying process, it will be trapped insidethe particle where it will remain when theproduct is packaged. Once in thepackage, this moisture will slowly andgradually come to the food particle

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surface where it may promote the growthof mold. Occasionally, the moisture maysoften the shell, or create internalstresses which cause the particle tocollapse in on itself or to break apart.

1.6.8 Leaching of SolubleConstituents

During the drying process, water from theinterior portion of the food particles orpieces travels to the surface where it isevaporated by the hot air inside the dryer.This water may contain dissolvednutrients that it leaves on the food surfacewhen it evaporates. These nutrients arestill present on the food surface at thetime it is packaged, but if the food has tobe rehydrated by soaking it in water orboiling it in water, the nutrients may besolubilized by the water and removedfrom the food. The nutrients are then notavailable for the person consuming thecooked food.

1.7 General Questions

Question 1:

Prepare a list of dried or dehydratedfoods food in your local food store ormarket. (Try to find at least ten).

Question 2:

Prepare a list of dehydrated productswhich your industry uses as ingredients infurther processing steps.

Question 3:

Prepare a list of products in your regionwhich are suitable for dehydration ordrying.

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CHAPTER 2: ORGANIZING INFORMATION

2.1 Introduction

Typically, in the food processing industry,we gather information about how aprocess is operating over a period of time.We do moisture tests on the materialentering a dryer and on the finishedproduct leaving the dryer. Values ofvariables such as temperatures, airvelocities, and retention times aregenerally recorded at regular intervalsover the course of a drying run.

In any process, it is not sufficient tosimply have information of this naturestored away - we must be able to use it tounderstand the manner in which theprocess is functioning. Being able toeffectively organize the information that isgathered is a challenge that often facesthe food processor.

Many food processes appear quitecomplex at first glance. When viewed inits entirety, it may be difficult to knowwhat is happening throughout theprocess. Trying to deal with the entireprocess at one time is probably the mostcommon mistake that people make.When we eat a meal, we do not try to putall of the food in our mouth at the sametime and swallow it in one gulp. Rather,we cut our food into smaller bite-sizedpieces that are easier to chew and easierto digest. If we do this with our food, thenwhy not use the same approach indealing with an industrial food process?The “unit operation” approach will allowus to do just that.

2.2 The "Unit Operation" Approach

Most food processes consist of a seriesof individual steps arranged in a particularsequence to convert starting rawmaterials to the desired finished product.

A complete food process may cover agreat deal of floor space in amanufacturing plant. Many processesoccupy several floors in the plant andinvolve large pieces of equipment linkedtogether by conveyors, pumps, and otherassociated hardware.

In the example presented here, we will gobeyond considering only drying. We willlook at a much larger process to illustratethe unit operation concept.

If we were to consider a hypotheticalcereal process where raw grain is broughtto a manufacturing site and made into aformed finished product, we could dividethe overall process into a series ofindividual steps. The output from onestep would become the input for the nextstep and so on. We could then view eachone of these steps on an individual basisto simplify our treatment of the totalprocess. Normally, each step would befocussed upon an individual piece ofequipment, or an individual task whichcan be referred to as a "unit operation".

In the hypothetical cereal process, thegrain might be conveyed into the plantand screened to remove foreign materiallike stones, etc. The bran layer could beremoved from the grain by a milling

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operation and separated from the kernelin the same operation. From here, thebran may go to a secondary process, butour concern is for the kernels of milledgrain.

The milled grain kernels could then beground into flour which could be blendedwith other ingredients, including water, inthe barrel of a large extruder. Aftergrinding, the flour should be screened toremove excessively large particles.These large particles can be sent back tothe grinding operation to be broken downto a smaller size. While in the extruder,the ingredients are thoroughly mixed andheated either by the addition of steam, orby the mechanical energy within theextruder. Heating in the extruder “cooks”the mixture of ingredients and creates acooked dough that is passed through a“die” at the discharge end of the extruderbarrel. The “die” creates a longcontinuous rope-like piece of finishedproduct in the appropriate cross-sectionalshape that can then be cut to the desiredlength by rotating knife blades positionedat the discharge end of the extruderbarrel. As it leaves the high pressureinside the extruder barrel, the productmay expand to give a puffed product.

The shaped cereal in our process thentravels on a conveyor belt into a “coatingreel” where a flavoured syrup is sprayedonto it. After it leaves the coating reel,the cereal enters a small dryer to reducethe moisture of the surface coating. Itthen is screened to remove any smallfragments or large clusters of cerealpieces. Following packaging, the finalcereal product is warehoused prior todistribution to various retail outlets whereit is sold to consumers.

A rambling account of the process as

described in the previous paragraphs issomewhat confusing and hides a greatdeal of information concerning theprocess and the equipment involved. Afar more satisfactory approach todescribing a process is through the use ofa flow chart or “process flow diagram”.

Flow charting begins by listing all thesteps involved, such as:

1. Raw material receiving 2. Screening of grain 3. Milling of grain 4. Grinding 5. Screening 6. Blending of ingredients 7. Heating / Cooking 8. Shaping and Cutting 9. Coating 10. Drying 11. Cooling 12. Screening 13. Packaging 14. Warehousing and Distribution

The majority of the steps listed aboveinvolve specific pieces of equipment foreach specialized task. These are usuallyconsidered as large components madeup of a number of smaller components.A mill would be a good example of a "unitoperation" in our hypothetical process.Essentially, it has one function in theprocess - to remove bran from the grain.The extruder, by comparison, is one pieceof physical equipment that may beconsidered to embody several unitoperations such as blending ofingredients, heating and cooking, andshaping and cutting. Various people willhave their own opinions on how to treatthe extruder in a process flow diagram.However, the main thing is to understandhow the overall process is functioning.

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Many food process engineers like thenames of each unit operation to end in“er” or “or” whenever possible. Examplesof this in our hypothetical cereal processwould be “blender”, “cooker”, “dryer”,“cooler”, etc. A personal preference ofmine is to have the labels for each unitoperation end in “ing” to indicate itsspecific function. This is why you seesuch labels as “screening”, “cooking”, and“drying” in the list of processing steps.

In our example of the "unit operations"approach, we might be tempted to viewthe dryer as a single unit operation. Itsprimary purpose is to remove water fromthe product, as was done after thecoating operation in our process. It mayalso toast the product as well. Somedryer manufacturers include a coolingzone in their dryers; and you may want toconsider cooling as a separate unitoperation, as was done in our processhere. However, you can use your owndiscretion on how you want to addressthis in many cases. In the initial stages,you might not be overly concerned aboutthe dryer's individual parts, unless youwere to be doing an in-depth dryer study.On these occasions, you will have tobreak the dryer down into a series ofsmaller steps within that unit operation toexplain what is happening in the dryingprocess itself.

You also need to realize that coupled withphysical pieces of equipment there arefrequently "reactions" that are occurringwithin them. These may be physical orchemical in nature (or a combination ofboth). There are generally two basicfactors controlling the "reactions" within aprocess, namely, temperature and time.Coupled with temperature and time canbe other factors such as moisturecontent, etc. We will not worry about

these factors here.

2.3 Process Flow Diagrams

2.3.1 What is a Process FlowDiagram?

Having defined the steps involved in theprocess, we can now prepare a processflow diagram or schematic diagram tocommunicate the information moreeffectively than we could by a few pagesof written text. Figure 2-1 is a processflow diagram of our hypothetical cerealprocess which includes the fourteen stepslisted above in our description of theprocess. Several of these steps havebeen combined into one block or “box”which appears as the “extruder” in Figure2-1. If you wish, you can also includesome of the actual processing conditionson the process flow diagram, but that hasnot been done here, in the interest ofclarity.

In dealings that you might have with anyprocess (even non-food processes), thefirst thing you should always try to do is tobreak down the overall process into itsindividual unit operations or componentsteps. This is the first step to being ableto resolve difficulties and addressproblems.

Notice in Figure 2-1 that cooling has beentreated as a separate step from drying. Inmany industrial processes, the targetmoisture is specified after cooling ratherthan after the drying step. This is due tothe fact that in cooling, there is often achange in moisture from that at the end ofthe drying step. Very warm product canpick up moisture from humid ambient airwhich is prevalent in the summer months;or it can lose moisture when cooled with

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cold dry air which we experience in thewinter months. The moisture must bedetermined after all these effects havebeen taken into account. Mostprocessing facilities express the moisturecontent on a “wet basis” which we willcover in Chapter 4.

In a time of conservation, many of thestreams leaving the process (e.g., thedryer exhaust air) would be recycledwithin the process to reduce energyconsumption and waste. Heat recoverymethods allow heat from process wastestreams to be used to warm incomingwater or air. The bran, “fines”, and “overs”(i.e., pieces of product too small or toolarge to be acceptable) would also beused in secondary processes, or productssuch as petfood. A process flow diagramallows you to identify potentially useful“waste” streams.

Hopefully, you can now see how theprocess flow diagram in Figure 2-1 hasbeen a great aid in reducing theconfusion of having to deal with the entireprocess as one single item.

2.3.2 Preparing a Process FlowDiagram

Now that you have seen a process flowdiagram and are familiar with itsappearance and basic construction, wewill step back and review the proceduresfor preparing them.

Setting up a process flow diagram, or“PFD” is usually not excessively difficult,but it is normally something that cannotbe done in just one attempt. It oftenrequires several revisions andmodifications before an acceptablediagram is obtained.

The best way to start a process flowdiagram is to go out into the processingfacility with a pad of paper and a pencil orpen to sketch out the process and writedown the various processing steps. Atthis stage, you should not be concernedabout how “pretty” you diagram looks -you simply want to gather as muchrelevant information about the processsequence as possible.

Having gathered the preliminaryinformation about the process, thesecond step is to construct an initial draftof the PFD. You can draw a series ofrough boxes representing the unitoperations on a sheet of paper and linkthem with arrows to show the productflow. Additional inputs and outputs canalso be included by sketching in arrowsentering and leaving the unit operations.At this stage, you are free to add ordelete items from your draft diagram.

Once you are satisfied that the draft PFDrepresents the process as well aspossible, you should take the diagramback into the processing facility and verifythat all flows and pieces of equipment are

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correct. If temperatures and flowratesetc. appear on your diagram, you shouldconfirm that all values are valid. You mayalso wish to review your process flowdiagram with someone who is familiarwith the process, such as the processoperator, to check that you haverepresented the process in an appropriatemanner. Any corrections or modificationsshould be made directly on the draft copyof the diagram.

When all necessary modifications havebeen made to the draft copy of theprocess flow diagram, it can then bedrawn in its final form using a computerprogram or drawn carefully by hand. Aproper title should be included with thediagram along with the date. The datewill serve as a useful future reference andreduce confusion as to when thearrangement of the process was as it hasbeen drawn. It is also a good idea tolegibly sign or print your name on theprocess flow diagram.

Some people like to have severalversions of process flow diagrams for thesame process. One diagram may showonly the unit operations and productflows, along with major input and outputflows. A second PFD may show values ofkey operating parameters such astemperatures, retention times, andflowrates in addition to the previouslymentioned items. Other PFD’s may beused to show process waste streams, orutility inputs such as steam, heat,electricity, and so on. Each PFD can beused to address the needs of a particulargroup of individuals with whom you maybe discussing the process.

It should be recognized that twoindividuals may not have identicalprocess flow diagrams for the same

process. You and a colleague may havesomewhat differing views on how torepresent the process. If the differencesare small, they may be considered to beinsignificant. As long as the essentialelements of the process are captured andclearly and accurately conveyed to theviewer, the process flow diagram will haveserved its intended purpose. However, itis a good idea to discuss the differencesbetween two differing PFD’s of the sameprocess with your colleague. In this way,an agreement can generally be reachedon how the process should berepresented before producing twodiffering final versions of the diagram.

2.4 Tables and Graphs

Planning is important to the successfulcompletion of any activity. Carefulplanning should take place before anyactual physical work is begun. The morecomplex the project is, the more thoroughand detailed the planning needs to be.

Data gathering is often one of the firststeps in assessing how a process isoperating or in evaluating the effects of achange in operating conditions on theoverall process and finished product.Once you have decided upon theobservations to be taken and thefrequency of them (e.g., one every 15 or30 minutes or once every hour, etc.), youthen need to organize the actual datagathering process. Often, the best way todo this is by means of a “table” that youprepare on paper.

For illustrative purposes, let’s considerthe example of drying a small sample ofsliced carrots in a laboratory dryer. Theweight of the carrots can be monitored bymeans of a scale or balance mounted on

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top of the dryer and linked by a number ofsuspension rods to the rack supportingthe carrots inside the dryer. It is decidedto record the weight of the carrots everyfifteen minutes for seven hours to followthe drying process. Table 2-1 shows thetype of table that would be set up prior tothe drying run to record the information.

Table 2-1: Table for Data Gatheringfrom Drying Test

Material:Date:Test Number:Starting Time:Air Temperature:Additional Information:

Time (minutes) Weight (grams)

0

15

30

45

60

75

90

105

120

etc.

Only the first 120 minutes (2 hours) of thetable have been shown here. Therewould be additional rows for times infifteen minute increments to 420 minutes(seven hours) in the full table. Other

columns could be added if observationsregarding such things as temperatureswere desired.

The initial weight of the carrots would beinserted as the weight at time t = 0minutes. When the test is started, timingwould begin using a stop-watch or otherappropriate means of following the timethat has passed. Every 15 minutes, theweight of the carrots would be recorded.

Table 2-2 shows the weights gathered forthe seven hour drying run in our example.Information regarding the date of thedrying run and some of the particularsassociated with it have been included forfuture reference.

The data presented in Table 2-2 providea starting point from which we can makeobservations in an attempt to developtrends in the drying process, as will bedone in the subsequent drying coursematerial. However, a series of numberson a page is not very informative withoutsome further manipulation.

Most people tend to be “visual”. They canappreciate information in a visual formatmuch better than they can when it ispresented in a purely numerical format.This is one reason why it is often a goodidea to prepare a graph from the datagathered during an experiment. Graphsalso aid in developing mathematicalrelationships as we will see in the“Intermediate” drying course.

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Table 2-2: Table for Data Gatheringfrom Drying Test

Material:Date:Test Number:Starting Time:Air Temperature:Air Flowrate:

Additional Info:

CarrotsMarch 1, 20__Carrots - 1A8:47 am50° C0.2 m / second

0.4 cm thickcircular slices

Time (minutes) Weight (grams)

0 201

15 192

30 182

45 172

60 163

75 153

90 144

105 135

120 126

135 117

150 19

165 102

180 93

195 87

210 80

225 73

240 68

255 62

270 57

285 53

300 49

215 46

330 42

345 40

360 37

375 35

390 33

405 31

420 30

Figure 2-2 shows a plot or graph of“Weight versus Time” for the carrot slicesdried in the test run.

When preparing a graph, “time” isconsidered to be the “independentvariable”. In our case, the weight of thecarrot slices would be considered to bethe “dependent variable”, since the weightof the material depends on the time forwhich it is exposed to the dryingconditions. The independent variable, inthis case “time”, is always placed on thehorizontal axis of a graph. The time axisadvances from a lower value on the left toa higher value on the right. Thedependent variable, in this case theweight of the carrot slices, is placed onthe vertical axis of the graph, as shown inFigure 2-2.

The scale of each axis should be chosento provide a clear view of the data being

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plotted. Choosing the incorrect scale maysquash your data into one portion of thegraph or spread it out so widely that itdoesn’t all fit on the same page. In eithercase, you will not be able to make anymeaningful observations from the graphyou have drawn. Not only do you have toselect the minimum and values for eachaxis, but you must also decide upon thesize of the divisions or increments foreach axis.

In Figure 2-2, the range of values for the“time axis” was selected to cover thelength of time of the drying run and leavea small amount of extra time at the endso that the curve did not touch the rightside of the graph. An appropriate intervalmight be 15 minutes, or thirty minutes, orone hour. To comfortably fit the labels onthe time axis, one hour (i.e., sixtyminutes) was chosen as the major

increment with a minor increment of thirtyminutes. Only the major intervals havebeen labelled. In Figure 2-2, it is easy totell the shorter fifteen minute timeintervals by “interpolation”, or readingbetween the thirty minute divisions.Using a time range of 0 to 480 minuteswith major time intervals of 60 minutesand minor time intervals of 30 minutesgives a reasonable scale for time inFigure 2-2.

The “weight axis” scale spans the initialweight of the carrots (i.e., 201 grams) andthe final weight of the carrots (i.e., 30grams). Extending the weight axis to 250grams gives some space above wherethe actual data curve begins. Usingmajor weight intervals of 50 grams eachwith minor intervals of 10 grams providesa suitable weight axis.

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Whenever possible, or wheneverreasonable, the origin (i.e., zero values)should be shown on a graph. However,there are occasions when this is notappropriate due to the large values ofnumbers that may not approach the zerovalues. For example, we may have 3,500grams of starting material and dry it for ashort time. The final weight might be3,000 grams. In preparing a graph of thisinformation, it would be more suitable tohave a weight range starting at 2,800grams going to a weight of 3,600 gramswhile using weight increments of 50grams or 100 grams. Including 0 gramson this graph would lead to a graph withthe observed data squeezed up near thetop in a very narrow band where notrends could be observed. Whenpreparing graphs, you need to use yourdiscretion as to the range and scale ofeach axis.

Now that Figure 2-2 has been prepared,it can be seen that the rate at which wateris removed (i.e., the change in weight ofthe carrots over time) is not uniform. Theslope of the curve is greater from the startof the run until about 180 minutes into therun than it is after about 180 minutes.

We can also see that there is a levellingoff of the weight of the carrots around 420minutes into the run. These trends werenot as evident from the raw data collectedand displayed in Table 2-2. Detailsregarding the handling and interpretationof drying data will be discussed in theIntermediate drying course.

2.5 Organization in Problem Solving

2.5.1 The Need for Organization

There are times when you may be calledupon to solve mathematical problemsassociated with a certain process.Regardless of how complicated orstraight-forward the problem appears tobe, it should be approached in astructured and disciplined manner. Usinga logical sequence of problem-solvingsteps will enhance the probability ofsuccess and reduce the likelihood oferrors in the problem solution.

Use of an orderly approach withexplanatory statements as to whatassumptions were made, and why aspecific mathematical relationship wasestablished, are highly beneficial whenreviewing material that has been put intoa file for several years and is then re-examined. A good way to view yourmathematical solutions to problems is toconsider if you would be able tounderstand what you have done inreviewing your notes five years later. Ifyou are confused with your mathematicalsolutions now, then you will definitely bemore confused if you have to refer backto them at a later date. You should alsoconsider that others may have to readand understand the work that you havedone without the benefit of yourassistance.

Many food processing workers literallyscribble out solutions to problems onscraps of paper which are then droppedinto a file folder. Such notes areessentially useless at a later date sincethey lack any form of logic and often don’teven have a date on them to tell when thework was done. This is the type of thingthat we want to avoid, and it is for this

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reason that Chapter 2 is included in thisstudy manual.

In order to demonstrate how to approachproblems in a disciplined and organizedmanner, several examples using massbalances will be used.

2.5.2 Mass Balances

Mass balances are used by processors ofmost products on a routine basis. Massbalances rely on the fact that the totalmass of materials coming out of aprocess must equal the total mass ofmaterials entering a process. This is thecase for most food processes once theyare set up and running continuously (i.e.,operating under steady-state conditions)and generally holds true unless there isan accumulation of material or someother similar non-steady state eventinside the process. When a process isbeing started up or being shut down,mass balances will not apply. Therefore,when a process is operating understeady-state conditions, we can say:

Total Mass Total Mass Entering = Leaving the Process the Process

Example Problem #1:

If 100 kg of wet material enters a dryerand 70 kg of water are removed duringthe drying process, how much productcan be expected to be leaving the dryer?

Since there is no indication that anymaterial is lost in the drying process andnothing is being retained in the dryer,there will be 30 kg of product leaving thedryer.

This answer was obtained by recognizingthat the total mass leaving the dryer mustbe equal to the total mass of materialentering the dryer. Therefore:

100 kg 70 kg water X kgentering = leaving + leavingdryer dryer dryer

Where X kg is the weight of the productleaving the dryer. In this case, X = 30 kg.

Most people tend to be “visual” in theirapproach to problem solving. This meansthat it is often useful to draw a diagram toshow what is happening in a process andto organize the available information. Byusing a diagram, it is often apparent whatinformation is lacking, or whatrelationships exist among the variousprocessing streams. While it is certainlynot difficult to visualize what is happeningin the example mass balance we areconsidering here, not all problems are aseasy nor as clear to follow. It is a goodidea to use a systematic approach toproblem solving while working onexamples such as these so that whenmore difficult problems are encountered,you have developed the necessary skillsto solve them.

In looking back at the mass balanceexample, we can begin by drawing a“diagram”. This does not have to beanything fancy or detailed. A simple “box”with arrows going into it to show theprocess inputs and arrows coming out ofit to represent the process outputs is allthat is really necessary. As shown inFigure 2-3, 100 kg of raw materials areentering the dryer while 70 kg of water isleaving the dryer. The arrow for theproduct stream has yet to be quantified,but it is easy to visualize that the total of

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the 70 kg water arrow and the productstream leaving the dryer must total the100 kg of material entering the dryer.

Solving for X gives a product weight of 30kg, as stated above.

While it may be tempting to say that aproblem of this nature is too simple tobother drawing a diagram to solve, pleaseuse this as an opportunity to develop yourproblem solving skills so that you will notbe frustrated when it comes to workingwith more difficult problems.

Example Problem #2:

73 kg per hour of ingredient “A”, 26 kg perhour of ingredient “B”, and 37 kg per hourof ingredient “C” are mixed in a devicethat feeds into a drying process. Duringthe drying process, 32 kg per hour ofwater are removed and 102.5 kg per hourof finished product are obtained. It issuspected that some small pieces ofpowdery material are being carried out ofthe dryer along with the air used in drying.How much material is lost each hour inthe drying process?

The first step in solving this problem is toorganize the information using a simplediagram. As shown in Figure 2-4, two“boxes” can be used to represent themixing and drying operations. The threearrows entering the mixing operationindicate the three ingredients that arebeing blended together to give the mixeroutput of 136 kg per hour.

The mixer output of 136 kg per hour thenbecomes the input to the dryer. Heatedair is the only other input into the dryerand it is shown without any values sinceit is not an essential part of the massbalance being done on the material.There are four streams leaving the dryer.The first stream is the exhaust air, whichagain has not been quantified. Thesecond stream is the water loss of 32 kgper hour, and the third stream is theproduct at 102.5 kg per hour. Theamount of material lost in the air leavingthe dryer is shown as the fourth streamleaving the dryer. It has been separatedfrom the other streams to assist insimplifying the solution to the problem.

Solving for X gives a loss of 1.5 kg/hour.

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2.6 Practice Problems

Question 1:

Prepare a process flow diagram of asuitable process within your foodprocessing facility. Include the basic unitoperations, major inputs and outputs, andindicate the values of a number of keyvariables in the process. Retain all youwork, including your rough work, so thatyou can see how your process flowdiagram has evolved with eachsuccessive step in your design.

If you do not have access to an operatingfood process, or if your processinformation is considered confidential in

nature, visit a website such as that of the“Institute of Food Technologists” (IFT)and use an example of a process thatappears there. You may use any processfor any material you wish, as long asthere are multiple processing stepsinvolving a variety of unit operations.

Discuss your process flow diagram withsomeone knowledgeable with the processyou have chosen, and gather his or herfeedback. Incorporate this feedback intoa revised PFD if necessary.

Question 2:

Prepare a process flow diagram todescribe the process for preparing a

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favourite food in your home. It may besomething like preparing rice where youhave a number of steps involved from thetime the rice is harvested until it is cookedand consumed. Discuss your processf l o w d i a g r am w i th s om e o n eknowledgeable in food processing andgather his or her feedback. Incorporatethis feedback into a revised PFD ifnecessary.

Question 3:

From a newspaper, or other appropriatesource, collect the high and lowtemperature readings for a fourteen dayperiod in your area. Prepare a graph oftemperature versus time (i.e., the date)with curves for both the daily maximumand daily minimum temperatures on thesame graph. Use different symbols tomark the maximum and minimumtemperatures. Join the data points for themaximum temperatures with a solid lineand use a dashed line to join theminimum temperature points.

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CHAPTER 3: DIMENSIONAL ANALYSIS APPROACH TO PROBLEM SOLVING

3.1 Introduction

Nearly all food processing applicationsinvolve some degree of mathematics tofully understand the actions taking placewithin a particular unit operation. Thecomplexity of the mathematics can rangefrom almost trivial to extremely difficult.As stated in Chapter 2, an organizeddisciplined approach should be taken inall problem solving activities. This givesyou a better understanding of thecomponents of the problem and reducesthe chances of error.

The following steps may be helpful insolving problems:

i. draw a sketch or "picture" of whatis taking place.

ii. label all known values and identifythe “unknowns”.

iii. establish initial mathematicalrelationships.

iv. refine the relationships to definethe “unknowns”.

v. u s e t h e s e re l a t i o n s h i p s t ocharacterize or “model” theprocess.

The primary focus of this chapter is onhelping you address steps iii and iv, andto help you avoid making unnecessaryerrors in calculations as you proceedthrough your problem solving sequences.

3.2 Conversion Factors

While most countries of the world haveofficially embraced the metric system ofmeasurement (or SI, “SystemeInternationale” units of measurement), theUnited States has maintained its links towhat used to be called the “British” or“Stirling” system of measurement. Theremay be times that you are called upon toconvert from one system of measurementto the other. In this case, a set ofconversion factors is required to assistyou in doing the conversion calculations.In addition, doing conversions such asthese is an effective way of demonstratinga problem solving technique referred toas “Dimensional Analysis”.

A list of basic conversion factors forlength, mass, volume, and temperatureappears below. Conversion factors fortime have also been included forcompleteness. To avoid confusion,conversions for various pressure units,viscosity units, and other morecomplicated quantities have not beenincluded. They will be introduced in theappropriate courses when needed.

While it is not necessary to commit thesevalues to memory, learning certainconversion factors such as “2.54 cmequal one inch” may prove beneficial.

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Length:

1 inch = 2.54 cm

1 foot = 12 inches = 30.48 cm

1 yard = 3 feet = 36 inches = 91.44 cm = 0.9144 metres

1 mile = 5,280 feet = 1.609 kilometres

1 metre = 100 cm

1 metre = 1,000 mm

1 kilometre = 1,000 metres

Mass / Weight:

1 kilogram (kg) = 1,000 grams = 2.20462 pounds (lb)

1 pound (lb) = 16 ounces (oz) = 453.6 grams

1 ounce (oz) = 28.35 grams

1 metric tonne = 1,000 kg

1 ton = 2,000 pounds

Volume:

1 mL = 1 cm3

1 litre = 1,000 millilitres (mL)= 1,000 cubic centimetres

1 cubic metre = 1,000 litres

1 U.S. gallon = 3.785 litres

1 Imperial gallon = 4.546 litres

Temperature Conversions:

Celsius to Fahrenheit:F = (C * 1.8) + 32

Fahrenheit to Celsius:C = (F - 32) / 1.8

Celsius to Kelvin:K = C + 273.15

A temperature difference of one degreeon the Celsius scale is equal to atemperature difference of one degree onthe Kelvin scale.

Time:

1 minute = 60 seconds

60 minutes = 1 hour

24 hours = 1 day

365 days = 1 year (if not a “leap year”)

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3.3 Dimensional Analysis:

A common sources of error in problemsolving is confusion in how to handle thedata that you have been given. You maynot know whether to multiply two numbersor divide them. If you have to divide twonumbers, you may not know whichnumber goes on top (i.e., in thenumerator) and which goes into thedenominator (i.e., the bottom) of afraction. Dimensional analysis virtuallyeliminates this problem, yet very fewpeople seem to use it.

Very simply put, dimensional analysisinvolves putting units or “dimensions” withall numbers used in a mathematicalequation.

Example 1:

In a simple example, we may have toconvert 4.5 feet to the equivalent numberof centimetres. Knowing that there are 12inches in a foot, and 2.54 centimetres inan inch is essential to solving thisproblem.

There are several ways to go aboutsolving this example, but we will convertthe feet to inches first, and then convertthe inches to centimetres.

Number of inches = 4.5 ft x 12 in ft

= 54 inches (Eq'n 1)

Number of cm = 54 inches x 2.54 cminch

= 137.16 cm (Eq'n 2)

In equation 1, the units of feet in thedenominator cancel the units of feet in thenumerator to leave you with inches asyour resultant dimension. This is whatyou want, so you're halfway to solving theproblem. You then multiply 4.5 by 12 toget the numerical part of your answer. Inequation 2, the inches in the denominatorcancel the inches in the numerator andyou are left with centimetres as your finaldimension, so you've done theappropriate mathematical operations.Multiplying 54 by 2.54 will give the finalnumerical portion of the answer.

The example can be handled in one step:

Number of cm =

4.5 ft x 12 in x 2.54 cm ft in

= 137.16 cm (Eq'n 3)

For more complex problems (there aren'tmany simpler than this example), a bettermethod of writing the mathematicalequations is to use a horizontal line and aseries of vertical lines. Terms above thehorizontal line get multiplied by eachother. Terms below the line get dividedinto those above the line. For example,equation 3 can be rewritten as:

Number of centimetres =

4.5 ft | 12 in | 2.54 cm | ft | in

= 137.16 cm (Eq'n 4)

The starting value should be positionedas the first entry on the calculation lineand then you can work from there toobtain the desired conversion.

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Example 2:

Knowing that there are 16 ounces (oz) ina pound (lb); 453.59 grams in a pound;and 1,000 g in a kilogram, convert 250ounces to kilograms.

Number of kilograms =

250 oz | lb | 453.59 g | kg | 16 oz | lb | 1000 g

= 7.09 kg (Eq'n 5)

You can see how the appropriate unitscancel each other to leave only kilograms.“Ounces” (i.e.”oz”) above the line cancelwith “ounces” below the line. “Pounds”(i.e., “lb”) above the live cancel with“pounds” below the line. “Grams” (i.e.,“g”) above the line cancel with “grams”below the line. This leaves “kilograms”(i.e., “kg”) as the final dimension for thecalculation, which is what we are requiredto find. The numbers above the line arethen multiplied together and the numbersbelow the line can also be multipliedtogether. The numbers below the line arethen divided into the numbers above theline to get the final numerical value.

Example 3:

Let’s suppose we get "mixed up" in howto do the conversion from Example 2 andgo about things in the wrong way.Instead of dividing by 16 ounces perpound, we erroneously multiply by 16ounces per pound. Here is whathappens.

Number of kilograms =

250 oz | 16 oz | 453.59 g | kg | lb | lb | 1000 g

= 1,814.36 oz kg/lb (Eq'n 6)2 2

If the size of the number doesn't tip youoff that your answer is wrong, the unitscertainly should. There is no such thingas a "pound squared", or an “ouncesquared”; and even if there were, youwere setting out to determine the numberof kilograms, so you know you've madean error in your conversion.

You should try to develop an appreciationof size and orders of magnitude indifferent units of measurement. We useSI Units for teaching and scientificpurposes because of their inherentadvantages. However, the "industrialworld" uses a variety of measurementunits depending on the type of monitoringdevice they have and also on the basis of"that's how we've always done it". Youmay see temperatures measured in bothCelsius and Fahrenheit; or weightsmeasured in both pounds and kilograms,in the same processing plant. Thesituation is even worse when it comes topressure measurements, but that is aplace where we will not go here.

So, even though we are trying to convertto SI Units, the "real world" is not as quickto respond as we would like.

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3.4 Sample Calculations

In this section, we will switch from the format of two columns per page to a single column.In this way, the calculations will be easier to show and follow.

Example 1:

Problem Statement:

Convert a speed of “500 furlongs per fortnight” to “cm per second”.

Note: These are legitimate units of measurement. However, they do not fit withinthe SI guidelines.

A furlong is a unit of measurement for distance used in horse-racing that isequivalent to one-eighth of a mile. Therefore, 1 furlong = 1/8 mile = 0.125miles

A fortnight is an old English term referring to a two-week period of time.

Solution:

The approach will be to convert "furlongs" to "centimetres" first and then convert"fortnights" to "seconds".

500 furlongs = 500 furlongs | 0.125 miles | 5,280 ft | 12 inches | 2.54 cm per fortnight fortnight | 1 furlong | mile | foot | inch

x 1 fortnight | 1 day | 1 hour | 1 minute 14 days | 24 hours | 60 minutes | 60 seconds

= 8.3154 cm / sec

Therefore: 500 furlongs / fortnight . 8.32 cm / sec.

(Note: The mathematical treatment above had to be broken into two parts sinceit would not fit across the page on a single line.)

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Example #2:

Problem Statement:

Calculate the volume of a tank that is 2.0 m tall and has an interior diameter of 80 cm.Express the volume to the nearest tenth of a litre.

Solution:

Begin by drawing a diagram of the tank and labelling it with the appropriate dimensions asshown in Figure 3-1. Even though this may be a relatively straight-forward problem, it isimportant to develop good problem solving techniques, which is the purpose of thisexercise.

Figure 3-1:Diagram of Tank

Height = 2.0 m

d = 80 cm

The volume of a cylinder is given by the formula:

Volume = B r h = B d h 2 2

4

where: B = 3.14159 (constant used in equations involving circles) r = radius of the cylinder (radius = one-half the diameter) d = diameter of the cylinder h = height of cylinder (or length of cylinder)

The first thing to do is to plan the approach to this problem. Since the final answeris required to be in “litres”, it would be most efficient to calculate the volume of thecylinder based on its dimensions in centimetres. The resulting volume would thenbe in cubic centimetres which can easily be converted to litres on the basis thatthere are 1,000 cm per litre.3

The height of the tank is 2 metres which is equal to 200 cm.

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Volume = B d h = 3.14159 | 80 cm | 80 cm | 200 cm 2

4 4 | | |

= 1,005,309 cm 3

Volume in litres = 1,005,309 cm | litre 3

| 1,000 cm3

= 1,005.3 litres

Therefore, the volume of the tank is 1,005.3 litres (to the nearest tenth of a litre as required).

In the calculation of the tank volume, the “80 cm” diameter appears twice to showthat the diameter must be squared as part of the calculation.

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Example #3:

Problem Statement:

A food processing establishment is renovating its production facility. It intends to put newsanitary tiles on the floors and new washable panels on the walls of the “wet” processingarea. The room is 48 feet long by 36 feet wide and the ceiling is twelve feet high. Thefloor tiles are 9 inches square (i.e., nine inches by nine inches in size) and the wall panelsthat are available are four feet by twelve feet in size. If we do not worry about openingsfor doors and windows in the walls and drains etc. in the floor, how many floor tiles and wallpanels would be required for this job?

Solution:

This is probably one of the few times when it is easier to work in units of feet and inchesthan it is to work with their metric equivalents. The room is twelve feet high which is theheight of a wall panel, and the panels are four feet wide which is a convenient dimensionconsidering the length and width of the room.

Calculation of wall panels:

Draw a diagram of the room and determine what steps you need to take to solve theproblem. A simple rectangle as shown in Figure 3-2 is sufficient for this problem.

Figure 3-2:Top View of “Wet”Processing Area

Height = 12 feet

36 feet

48 feet

As stated above, this problem is suited to using feet as the dimension of choice.

The perimeter of the room (i.e., the distance around the outside of the room) isequal to 168 feet ( i.e., 48 feet + 36 feet + 48 feet + 36 feet = 168 feet).

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Number of wall panels = perimeter of room width of each wall panel

= 168 feet | wall panel | 4 feet

= 42 wall panels

There are also other approaches to solving this problem that work equally as well.For example, you could say that it takes 12 panels to cover each 48 foot wall and9 panels to cover each 36 foot wall. This would give you a total of 42 panels.

Calculation of floor tiles:

Each floor tile is 9 inches by 9 inches. Convert this size to feet.

9 inches = 9 inches | foot = 0.75 feet | 12 inches

Area of the room = length x width = 48 feet x 36 feet = 1,728 ft2

Area per floor tile = length x width = 0.75 feet x 0.75 feet = 0.5625 ft2

Tiles required = area of room = 1728 ft | tile = 3,072 tiles2

area per floor tile | 0.5625 ft 2

Therefore, we would need 42 pieces of wall panelling and 3,072 floor tiles for thisjob.

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Use the “dimensional analysis” approachto answer the following questions. Besure to show all of the steps involved inyour solutions in order to develop goodproblem solving skills. Draw diagrams toassist you whenever they seemappropriate.

Question 1:

Calculate the capacity of a rectangulartrough (with straight vertical sides) that is3 feet - 9 inches long by 1 foot - 6 incheswide by 15 inches deep. Express youranswer to the nearest tenth of a litre.(Convert ing the dimensions tocentimetres as a first step will assistgreatly in solving this problem)

Answer: 199.1 litres

Question 2:

An industrial dryer is set up to remove6,500 pounds of water every hour. Howmany kg of water and how many metrictonnes of water will it remove in an eighthour production shift?

Answer: 23,587 kg or 23.59 tonnesper eight-hour shift.

Question 3:

Many chemicals are sold in 55 U.S. gallondrums. What is the capacity in litres?

Answer: 208.2 litres

Question 4:

A dryer has two drying chambers eachwith its own temperature setting. Zone 1is set to run with air at a temperature of250°F and Zone 2 is set to run at 180°F.What are these temperatures in degreesCelsius?

Answer: Zone 1 = 121°CZone 2 = 82°C

Question 5:

Air is entering a drying chamber at aspeed of 4.2 miles per hour. Convert thisspeed to cm per second and metres persecond.

Answer: 187.8 cm / second1.88 m / second

Question 6:

A recipe calls for 4 ounces of sugar in 1.3Imperial gallons of water. Convert thismixing ratio to grams of sugar per litre ofwater.

Answer: 19.22 grams / litre

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Question 7:

How many seconds are in one year?(Not a “leap year”).

Answer: 31,536,000 seconds

Question 8:

The speedometer in a car reads 65 milesper hour. How fast is this in kilometresper hour and metres per second?

Answer: 104.59 km / hour29.05 m / second

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CHAPTER 4: WET AND DRY BASIS MOISTURE

4.1 Definitions of Wet and Dry BasisMoistures

There are two ways in which the moisturecontent of a food product may beexpressed.

The more frequently used method is tostate the water content on a “wet” basis,which may be defined by the followingequation:

Wet Basis Moisture =

Weight of Water x 100% Total Weight of Wet Material

= % Moisture (wet basis)

A less commonly used method ofdescribing the moisture content of aproduct is one that is based on the weightof the dry material contained in theproduct. This “dry” basis moisture maybe defined as being:

Dry Basis Moisture =

Weight of Water Weight of Dry Material

= g water / g dry solids

Although wet basis moistures are themost frequently used methods ofexpressing the amount of water present ina material, they can also be somewhatmisleading and confusing if they are notused in the proper manner.

It should be noted that wet basismoistures are expressed with water as afraction of the total weight of the materialpresent. As more water is added, thetotal weight of the material increases, asdoes the weight of the water. This meansthat the numerator (i.e., top) anddenominator (i.e., bottom) of the fractiondefining wet basis moisture are bothchanging as water is added or removedfrom a mixture. There is no constantbasis in this definition, so you cannotcompare two wet basis moistures directly.

Several practice problems (with answersprovided) will be used to illustrate anumber of points regarding water removalfrom products, plus water addition toproducts. As we will also see in the“Case Study” at the end of this chapter,the difference of 5% moisture betweentwo products with moisture levels of 80%and 85% wet basis moistures issignificantly different from a 5% moisturedifference between two products havingmoisture levels of 10% and 15% wetbasis moistures, which is a mistakecommonly made by workers involved indrying products.

Dry basis moistures are not ascommonly used in defining targetmoisture values. However, they do havea constant basis which allows you tocompare two dry basis moistures directly.For example, a product containing 0.70 gwater per g dry product contains twice theamount of water present in a productcontaining 0.35 g water per g dry product.

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The graph of "Wet Basis vs Dry BasisMoisture" shown as Figure 4-1 illustrateshow wet basis and dry basis moisturevalues are related.

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To demonstrate how misleading it can beto express moistures on a wet basis,consider the modified version of the "WetBasis vs Dry Basis Moisture" graphshown in Figure 4-2.

In Figure 4-2, vertical lines have beendrawn at intervals of 1 gram of water pergram of dry solids (i.e., at 1.0, 2.0, 3.0,and 4.0 grams of water per gram of drysolids). At the point where these verticallines contact the curve of “wet basis vsdry basis moistures”, lines have beendrawn horizontally across to the wet basismoisture axis.

Let us consider a four step process ofadding moisture to some dry solids.

Step 1: If we start with a totally dry or“bone dry” solid containing absolutely nowater (i.e., 0.0 grams of water per gram ofdry solids) and add 1.0 gram of water, wewill have a dry basis moisture of 1.0 gwater per g dry solids. On the wet basismoisture axis, this corresponds to goingfrom 0% to 50% wet basis moisture. So,an initial addition of 1 gram of water to 1gram of dry solids gives a 50.0% changein the wet basis moisture.

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Step 2: We will now take the productfrom Step 1 with 1.0 gram of water pergram of dry solids (or 50% wet basismoisture) and add another gram of waterto it, we now have 2.0 g water per g drysolids as a dry basis moisture. Looking atFigure 4-2, a dry basis moisture of 2.0 gwater per g dry solids corresponds to awet basis moisture of 66.7%. Eventhough we added 1.0 gram of water to thematerial in Step 2, just as we did in Step1, the wet basis moisture only increasedby 16.7% (i.e., from 50% to 66.7%)compared to an increase of 50% wetbasis moisture for the same amount ofwater being added in Step 1.

Step 3: Next, we will take the productfrom Step 2 with 2.0 grams of water pergram of dry solids (or 66.7% wet basismoisture) and add another gram of waterto it. We now have 3.0 g water per g drysolids as a dry basis moisture. Again,looking at Figure 4-2, we can see that adry basis moisture of 3.0 g water per gdry solids corresponds to a wet basismoisture of 75.0%. This addition of 1.0gram of water to the material in Step 3gave us an increase in the wet basismoisture of only 8.3% (i.e., from 66.7% to75.0%). Compare this to an increase of50% wet basis moisture for the sameamount of water being added in Step 1,and a 16.7% increase in Step 2.

Step 4: In our final step, we will take theproduct from Step 3 with 3.0 grams ofwater per gram of dry solids (or 75.0%wet basis moisture) and add anothergram of water to it. Now, we have 4.0 gwater per g dry solids as a dry basismoisture. Once again, by looking atFigure 4-2, we can see that a dry basismoisture of 4.0 g water per g dry solidscorresponds to a wet basis moisture of80.0%. This addition of 1.0 gram of water

to the material in Step 3 gave us anincrease in the wet basis moisture of only5.0% (i.e., from 75.0% to 80.0%).Compare this to an increase of 50% wetbasis moisture for the same amount ofwater being added in Step 1, and a16.7% increase in Step 2, and an 8.3%increase in Step 3.

Hopefully, you can now see that when weadd uniform amounts of water to amaterial, the wet basis moisture does notchange by uniform amounts. This is dueto the fact that wet basis moistures arebased on the total weight of the materialincluding the water that is added. For thisreason, we do not have a constant basison which to base the comparison of themoisture content of two materials. Thismeans that when we do any calculationsof water content using wet basismoistures, we must convert them to a drybasis before we can do any meaningfulcomparisons.

It is now an appropriate time to show howto convert from wet basis to dry basismoistures and from dry basis to wet basismoistures.

For the detailed calculations presentedbelow, we will change the format of thetext. By not having two columns perpage, the calculations may be shownsomewhat more clearly than when twocolumns are used.

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4.2 Wet and Dry Basis Moisture Conversions

4.2.1 Wet Basis Moisture Sample Calculations

Convert the following wet basis moistures to dry basis moistures:

Example 1: 74% wet basis moisture (typically the moisture content of yams)Example 2: 93% wet basis moisture (typically the moisture content of tomatoes)Example 3: 11% wet basis moisture (typically the moisture content of dried beans)Example 4: 95% wet basis moisture (typically the moisture content of lettuce)

Recall the definition of dry basis moisture:

Dry Basis Moisture = Weight of Water Weight of Dry Solids

Example 1: For 74% wet basis moisture: (Convert to dry basis moisture)

Since no initial weight of product is given, start by considering 100 kg of startingmaterial.

Convert the wet basis moisture from a percent to a decimal fraction. 74% willbecome 0.74 for our calculations.

Weight of water = 100 kg material | 0.74 kg water | kg starting material

= 74 kg water

Weight of solids = 100 kg starting material - 74 kg water

= 26 kg solids

Dry Basis Moisture = 74 kg water = 2.85 kg water / kg dry solids 26 kg solids

Or: 2.85 g water / g dry solids etc.

Therefore, a product like yams with a wet basis moisture of 74% would have a dry basismoisture of 2.85 kg water / kg dry solids, or 2.85 g water / g dry solids. (Note that theweight units can vary except whatever units are used for the water should be used forthe solids to maintain a consistent comparison).

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As in Example 1, consider 100 kg of starting material


= 93 kg water


= 7.0 kg solids



Therefore, a product like tomatoes with a wet basis moisture of 93% would have a drybasis moisture of 13.3 kg water / kg dry solids, or 13.3 g water / g dry solids.




= 11 kg water


= 89 kg solids



Therefore, a product like dried beans with a wet basis moisture of 11% would have adry basis moisture of 0.124 kg water / kg dry solids, or 0.124 g water / g dry solids.

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= 95 kg water


= 5.0 kg solids



Therefore, a product like lettuce with a wet basis moisture of 95% would have a drybasis moisture of 19.0 kg water / kg dry solids, or 19.0 g water / g dry solids. This istruly an incredible amount of water per unit weight of dry solids!

Compare this to tomatoes with a wet basis moisture of 93% which contain 13.3 gramsof water per gram of dry solids. At these moisture levels, a difference of 5.7 grams ofwater per gram of dry solids causes a difference of only 2% in the wet basis moisture.

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4.2.2 Dry Basis Moisture Sample Calculations

Convert the following dry basis moistures to wet basis moistures:

Example 1: 3.6 kg water / kg dry solidsExample 2: 4.7 grams of water per gram of dry solidsExample 3: 0.075 grams of water per gram of dry solids

Recall the definition of wet basis moisture:

Wet Basis Moisture = Weight of Water X 100% Total Weight of Material

Example 1: 3.6 kg water / kg dry solids: (Convert to wet basis moisture)

Note that this is actually a dry basis moisture.

As a starting position, consider 1 kg of solids, and work from there.

Water contained in 1 kg dry solids = 3.6 kg water

Total weight with added water = 1.0 kg solids + 3.6 kg water

= 4.6 kg total weight

Percent wet basis moisture = 3.6 kg water X 100% 4.6 kg total weight

= 78.26 %

Therefore, the wet basis moisture is approximately 78.3%

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Example 2: 4.7 grams of water per gram of dry solids: (Convert to dry basis moisture)

To begin this conversion process, consider 1 gram of solids.

Water contained in 1 g dry solids = 4.7 g water

Total weight with added water = 1.0 g solids + 4.7 g water

= 5.7 g total weight

Percent wet basis moisture = 4.7 g water X 100% 5.7 g total weight

= 82.46 %


Example 3: 0.075 grams of water per gram of dry solids: (Convert to dry basismoisture)

As we did in Example 2, to begin this conversion process, consider 1gram of solids.

Water contained in 1 g dry solids = 0.075 g water

Total weight with added water = 1.0 g solids + 0.075 g water

= 1.075 g total weight

Percent wet basis moisture = 0.075 g water X 100% 1.075 g total weight

= 6.98 %


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4.2.3 Additional Wet Basis Moisture Calculations

Often we are required to determine the wet basis moisture of materials from informationother than the dry basis moisture.

When working with liquid products such as sugar solutions, we are frequently told thesolids content of the solution as a percentage of the total weight (i.e., % solids), or inunits of “degrees Brix”. “Degrees Brix” (or ° Brix) have nothing at all to do withtemperature. “Degrees Brix” is just another way of expressing the percent solids,usually sugar, in a solution. A 65° Brix sucrose solution would contain 65% by weightsucrose and the remaining 35% would be water.

In other cases, we may be required to calculate the moisture content on a wet basis ofa mixture prepared by blending a solid material of known moisture with some additionalwater. Although we may not often think of liquid solutions in terms of drying, theprinciples and calculations involved are quite similar.

For illustrative purposes, let us consider the following examples for which we are tocalculate the wet basis moistures:

Example 1: 38° Brix sucrose solution (Convert to wet basis moisture)

Example 2: 250 g water + 630 g powdered material with no moisture(Convert to wet basis moisture)

Example 3: 55 kg water + 30 kg powdered material with 15% moisture (wet basis)(Convert to wet basis moisture)

Example 1: 38° Brix solution: (Convert to wet basis moisture)

Note As stated above, “Brix” is an indication of the percent solids byweight present in a solution. The solids are usually sugars.

A 38° Brix solution will contain 38% solids by weight. We will express thispercentage as a decimal fraction (i.e., 0.38) in our calculations.

Since no initial weight of the solution is given, we will consider 100 kg as aweight for the starting solution

Weight of solids = 100 kg solution | 0.38 kg solids | kg solution

= 38 kg

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Weight of water = 100 kg solution - 38 kg solids = 62 kg water

% wet basis moisture = 62 kg water X 100% = 62% 100 kg solution

Or: Since 38° Brix means that the solids content is 38% by weight, thewater content by weight will be:

100% - 38 % solids = 62% water

Therefore, a 38° Brix solution is 62% water on a wet basis.


To calculate the wet basis moisture, we need to know the total weight ofthe mixture and the weight of water in it. Since there is no water presentin the powder, our only source of water is the 250 grams that we add as aliquid.

Total water present = 250 g water (no water in the powdered material)

Total material in mixture = weight of water + weight of powder

= 250 g + 630 g = 880 g


= 250 g water x 100% 880 g total weight

= 28.41%

Therefore, the wet basis moisture of this mixture is approximately 28.4%

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In this case, we must take into account the water that is present in thepowered material as well as the water that we are adding in its liquid form.

To calculate the wet basis moisture, we need to know the total weight ofthe mixture and the weight of water in it. We will begin by determininghow much water is present in the powdered material.

Water in 30 kg powder = weight of powder x water fraction = 30 kg x 0.15 = 4.5 kg

Total water present = water in powdered material + water as liquid= 4.5 kg + 55 kg= 59.5 kg water

Total material in mixture = weight of water + weight of powder

= 55 kg + 30 kg = 85 kg


= 59.5 kg water x 100% 85 kg total weight

= 70.0%

Therefore, the mixture of the powder and the water has a wet basismoisture of 70.0%.

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4.2.4 Additional Dry Basis Moisture Calculations

Just as we may have to calculate wet basis moistures based on a variety of information,we also may have to do similar calculations of dry basis moistures.

For comparative purposes, we will use the same three examples as we used in the wetbasis moisture calculations and we will calculate their dry basis moistures. The threeexamples were:




Here are the dry basis moisture calculations.


A 38° Brix solution will contain 38% solids by weight

Consider 100 kg of starting solution

Weight of solids = 100 kg solution | 0.38 kg solids | kg solution

= 38 kg

Weight of water = 100 kg solution - 38 kg solids = 62 kg water

Dry Basis Moisture = Weight of Water Weight of Dry Material

= 62 kg water 38 kg solids

= 1.63 kg water / kg dry solids

Therefore, a 38° Brix sugar solution has a dry basis moisture of 1.63 kg water /kg dry solids, or it could also be expressed as 1.63 g water / g dry solids.

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To calculate the dry basis moisture, we need to know the total weight ofwater and the weight of the dry solids in the mixture. We will begin bydetermining how much water and dry solids are present in the powderedmaterial.

Since we are told that the powder is dry and contains no moisture,

Water in powder = 0.0 kg

Solids in powder = 630 g (i.e., the total weight of the dry powder)

Next, we need to find the total weight of the water present in the mixture.

Total weight of water = weight of + weight of water liquid water in powder

= 250 g + 0 g = 250 g

Therefore, we have 630 g solids and 250 water in the mixture

Dry basis moisture = weight of water weight of dry solids

= 250 g water 630 g dry solids

= 0.397 g water / g dry solids

Therefore, we have a dry basis moisture of 0.397 g water / g dry solids in ourmixture.

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To calculate the dry basis moisture, we need to know the total weight ofwater and the weight of the dry solids in the mixture. We will begin bydetermining how much water and dry solids are present in the powderedmaterial. (Please note that many of these calculations are repeated fromthe same example done to determine the wet basis moisture of thismixture)

Water in powder = 30 kg powder | 0.15 kg water = 4.5 kg water | kg powder

Solids in powder = 30 kg powder - 4.5 kg water = 25.5 kg solids

Or: We can calculate the weight of solids using the percentage of waterpresent and subtracting it from 100%. In the equation below, wehave converted the percents to decimal fractions.

Solids in powder = 30 kg powder | (1.0 - 0.15) kg solids | kg powder

= 25.5 kg solids

Therefore, we have 25.5 kg solids and 4.5 kg water present in 30 kg of thepowder.

Total water present = water in powdered material + water as liquid= 4.5 kg + 55 kg= 59.5 kg

Total dry solids present = solids in powder = 25.5 kg

Dry basis moisture = weight of water weight of dry solids

= 59.5 kg water 25.5 kg solids

= 2.33 kg water / kg dry solids

Therefore, the dry basis moisture of this mixture is 2.33 kg water / kg dry solids.

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4.3 Methods of MoistureDetermination

4.3.1 Introduction

In the previous sections, we have lookedat methods of converting from onemoisture basis to another and we havedone calculations of the moisturecontents for several mixtures. However,we have not yet discussed how themoisture contents of various products canbe determined.

If we have an apple, how do wedetermine its moisture content ? Or, whatis the moisture content of hay used tofeed cattle before and after it is dried ?These are questions that we need toaddress before we proceed too muchfurther in our discussions of drying.

There are several ways of determininghow much moisture (i.e., water) is presentin a material. Water determinationmethods such as laboratory titrations willnot be examined here. We will focus ontwo basic methods that a commonly usedin industry. These are vacuum oven andmoisture balance methods.

At the outset, it should be explained thatboth the vacuum oven and moisturebalance methods actually measure thetotal amount of water and other volatilematerial present in the sample. Volatilecompounds are those that evaporatewhen the sample is heated. For example,if there was water and some alcoholpresent in a sample, both of these wouldevaporate and give a weight loss to thesample. In our testing, we assume thatall weight loss is due to the evaporation ofwater. If there was a high level of alcoholpresent in the sample, we would have anerror in our moisture determination.

However, most food materials have sucha low weight of volatile compounds otherthan water that we can generally assumethat the total weight loss in our testing isdue to the evaporation of water.

4.3.2 Moisture Balances

Moisture balances are extremely usefulfor determining the moisture content ofvarious products. Many of them arehighly automated and are very easy touse. However, they tend to be rathercostly.

When using a moisture balance, a smallsample (perhaps 5 grams) is placed on aweighing pan inside the drying chamberof the moisture balance. The moisturebalance registers the weight of the initialsample at the start of the test. Once thedrying chamber is closed, heat is appliedto the sample by means of an infraredheating lamp. As water is evaporated,the moisture balance compares theweight of the sample being dried to itsinitial weight and the wet basis moisture isindicated on its display panel. When theweight of the sample no longer changes,it is assumed that all of the water hasbeen evaporated. The moisture balancethen turns off the heating lamp anddisplays the final wet basis moisture ofthe sample. Depending on the nature ofthe sample and its initial weight, themoisture determination test can takeanywhere from a few minutes to an houror so.

Modern moisture balances are highlyreliable and are quite accurate in theirmoisture determinations.

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4.3.3 Vacuum Oven Method:

The vacuum oven method is consideredto be the “official” method of moisturedetermination. Exacting standards andprocedures have been developed by theAssociation of Official Analytical Chemists(AOAC) for moisture determinationtesting. Their methodology should befollowed. The explanation given here ofhow vacuum oven moisture tests areconducted is a simplified overview of theprocedure.

When conducting a vacuum ovenmoisture test, samples are accuratelyweighed in previously dried and weighedaluminum weighing pans. Weighing isgenerally done on sophisticated balancesthat are capable of weighing to one ten-thousandth of a gram. Since the weightsare so accurate, special handling of thesamples is explained in great detail in thetesting procedures. Forceps or tweezersare always used to move and lift theweighing pans.

The product samples are then placed in aspecially designed drying oven which hasbeen previously heated to the specifieddrying temperature (ca. 105° C). In orderto assist in the evaporation of water fromthe sample, a vacuum is applied to thedrying oven. The samples remain in theoven for a considerable time period (16 to24 hours). After they have been in theoven for the appropriate amount of time,the vacuum that has been drawn on theoven is released and the sample pans areremoved and placed in a desiccator tocool. If the samples are left out in theroom air while they are still hot, they canpick up moisture from the air and this willaffect the test results.

Once cooled, each weighing pan and its

dried contents are accurately weighed. Atthis point, the moisture content of thesamples can then be calculated. Samplecalculations of several vacuum oven testsare shown below.

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4.3.4 Sample Calculations of Vacuum Oven Moisture Tests

Test 1: A sample of an apple is tested for its moisture content.

Weight of empty weighing pan = 1.5052 gWeight of wet sample and pan = 6.8191 gWeight of sample and pan after drying = 2.3536 g

Weight of initial apple sample = weight of wet sample - weight of empty pan and pan

= 6.8191 g - 1.5052 g

= 5.3139 g

Weight of dried apple sample = weight of dry sample - weight of empty pan and pan

= 2.3536 g - 1.5052 g

= 0.8484 g

Weight of water in sample = weight of wet sample - weight of dry sample

= 5.3139 g - 0.8484 g

= 4.4655 g

Wet Basis Moisture = Weight of Water x 100% Total Weight of Wet Material

= 4.4655 g X 100%5.3139 g

= 84.034%

Therefore, the moisture content of the apple would be reported as 84.03%, on a wetbasis (to two decimal places).

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Test 2: A sample of hay going into a hay dryer is tested for its moisture content.

Weight of empty weighing pan = 1.5813 gWeight of wet sample and pan = 4.3208 gWeight of sample and pan after drying = 3.5072 g

Weight of initial hay sample = weight of wet sample - weight of empty pan and pan

= 4.3208 g - 1.5813 g

= 2.7395 g

Weight of dried hay sample = weight of dry sample - weight of empty pan and pan

= 3.5072 g - 1.5813 g

= 1.9259 g

Weight of water in sample = weight of wet sample - weight of dry sample

= 2.7395 g - 1.9259 g

= 0.8136 g

Wet Basis Moisture = Weight of Water x 100% Total Weight of Wet Material

= 0.8136 g X 100%2.7395 g

= 29.699%

Therefore, the moisture content of the hay would be reported as 29.70%, on a wetbasis (to two decimal places).

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4.3.5 Miscellaneous Drying Calculations

It is almost impossible to show all of the types of calculations that would be encounteredin drying applications. There are some examples, however, that may be useful for you tosee. You may wish to consider the following sample drying problems which involve the useof wet basis moisture values.

Please note that these problem solutions may look rather long and involved due to the factthat every step in the calculations has been shown and explanatory notes have beenadded for clarity.

4.3.5.1 Sample Problem #1

Problem Statement:

A processor has 525 kg of fresh green peas containing 74% moisture (wet basis).How much moisture must be removed from these peas to produce a final dried peaproduct with 12% moisture? What would the final weight of the dried peas be?

Solution:

Before proceeding with any problem of this nature, it is a good idea to organize yourinformation in a simple diagram. While this problem may not appear to be too difficult,developing the habit of drawing a diagram will be a definite benefit to you when it comestime to solving more difficult problems which involve much more information.

Figure 4-3 summarizes the known information and allows us to determine which values weneed to calculate.

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As a general rule, it is always a good idea to determine the amount of solids and theamount of water entering the drying process. We will do this first.

Weight of water in peas = weight of peas x water fraction in peas = 525 kg x 0.74 = 388.5 kg

Weight of solids in peas = weight of peas x solids fraction in peas = 525 kg x 0.26 = 136.5 kg

As we can see in Figure 4-3, the only thing that we know about the peas leaving thedryer is that their moisture content is 12% by weight. At this point, we know nothingabout the weight of the dried peas as they leave the dryer, nor do we know howmuch water is removed in the process. Basically, we have two unknown quantities,but, as we shall see, they are inter-related.

In order to proceed, we need to make a simplifying assumption. Since we are nottold that there are any losses of solids during the drying process, we will assumethat the weight of dry solids leaving the dryer is equal to the weight of the dry solidsentering the dryer. This is most frequently the case in drying. Sometimes, thedrying air may carry solids out of the dryer; but if this is the case, you will have todetermine the solids loss before you can proceed with your calculations.

Weight of solids leaving dryer = weight of solids entering dryer = 136.5 kg

The moisture content of the peas leaving the dryer is 12%. This means that of thetotal weight of dried peas leaving the dryer, 88% is solids (i.e., 100% - % water).

Since we do not know the weight of the peas leaving the dryer, we will let this weightbe X kg.

Weight of peas leaving dryer = X kg

Since 88% of this weight is solids and the solids weigh 136.5 kg, we can say:

0.88 X = 136.5 kg

Solving for X: X = 136.5 kg = 155.1 kg 0.88

This means that we have 155.1 kg of dried peas leaving the dryer.

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Weight of water = weight of dried peas - weight of solids in peas in dried peas

= 155.1 kg - 136.5 kg = 18.6 kg

Weight of water removed = weight of water in peas - weight of water in peasentering dryer leaving dryer

= 388.5 kg - 18.6 kg

= 369.9 kg

Therefore, 369.9 kg of water must be removed from the peas in the dryer and youwould get 155.1 kg of dried peas from the process.

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4.3.5.2 Sample Problem #2

Problem Statement:

A processor has 25 kg of dried onion flakes containing 11% moisture (wet basis).How much water must be added to these onions to rehydrate them and raise theirmoisture content to 85% moisture? What would the final weight of the onions be?

Although this is not a drying process, the problem does illustrate how to work withwet basis moistures when you are adding water to a material. For that reason, it isincluded here.

Solution:

Once again, we will start by drawing a diagram of the process (see Figure 4-4).

We will begin by determining the weight of water and solids in the onion flakes at the startof the process. Since the onion flakes contain 11% moisture, they will contain 89% solids.

Weight of water in flakes = weight of flakes x water fraction in flakes = 25 kg x 0.11 = 2.75 kg

Weight of solids in flakes = weight of flakes x solids fraction in flakes = 25 kg x 0.89 = 22.25 kg

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We know nothing about the rehydrated onions except that there final moisturecontent is 85% by weight. However, since we are just adding water to the onionflakes, it would be safe to assume that no dry material is lost in this process. Bymaking this assumption, we can say:

Weight of onion solids = Weight of onion solids = 22.25 kg after water is added before water is added

If we let the total weight of the onion flakes after the water is added be X kg, we cansay that 85% of the total weight is water and 15% of the total weight is dry solids.

Expressing this as a mathematical equation gives us the following:

0.15 X = 22.25 kg

Solving for X: X = 22.25 kg = 148.33 kg 0.15

Water present in = weight of rehydrated - weight of dry solidsrehydrated onions onions in rehydrated onion

= 148.33 kg - 22.25 kg

= 126.08 kg

Water added to dried onion flakes = water at end - water at start

= 126.08 kg - 2.75 kg

= 123.33 kg

Therefore, the processor must add 123.33 kg of water to the dried onion flakes.The final weight of the rehydrated onion flakes will be 148.33 kg.

You could also take the final weight of the hydrated onion flakes (148.33 kg) minusthe initial weight of the dried onion flakes (25.0 kg) to get the weight of water added(123.33 kg).

Please note that in these two sample problems, the key step is the assumption that nosolids are lost in the process. Although this is generally true, there may be exceptions. Asalready stated, if there is a loss of solids, you must identify the size of this loss before youcan proceed with any calculations.

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4.4 Case Study: Effects of Feed Moisture on Finished Product Moisture

Background:

This case study is based on actual experiences that I have had in dealing with operatorsof drying processes. I have changed all of the details in this case study to create afictitious process. However, this example is highly relevant to drying processes and shouldnot be dismissed too lightly.

Case Study Scenario:

You are the food process engineer for a fictitious company which has a drying operationthat produces dried carrot cubes for use in dehydrated soup mixes. The small pieces ofdiced carrot entering the dryer at the start of a production run have a moisture content of86% as expressed on a wet basis.

At the start of a drying run, the process operator carefully adjusts the dryer conditions soas to obtain a finished product moisture of 11%. After a few hours, the supply of dicedcarrots starts to run low and the operator requests an additional load of carrot cubes forher dryer.

When the new load of carrots arrives, it has a moisture content of 88%. The new batchof carrot cubes is sent through the dryer with identical operating conditions to the initialbatch of carrots.

When you hear about the change to the new batch of carrot cubes, you go to the dryingarea and discuss the production with the dryer operator. The operator tells you that sheexpects the moisture of the second batch of dried carrots to be about 13% moisture. Youdisagree, but the operator says you’re wrong and will not change the operating conditionsof the dryer. Her reasoning is that the first batch went in at 86% moisture and came outat 11% moisture; so if the moisture of the incoming carrots is increased by two percent, themoisture of the dried carrots should also increase by two percent. A 13% moisture wouldstill be acceptable, so no adjustments need to be made to the dryer.

When you check back later to see how well the second batch of carrots has beenprocessed, the operator is standing there shaking her head in disbelief. The moisturecontent is well above the expected 13%. The operator cannot figure it out and looks to youfor an explanation.

The feed-rate of the carrots is 175 kg of raw carrot cubes per hour in both cases.

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Questions:

Is the operator correct in keeping the dryer conditions the same, even though addedmoisture is coming through? (Or is there a big surprise in store?)

What is the weight of finished product produced on an hourly basis?

What would you predict the final product moisture to be if conditions were to be leftunchanged?

Solution:

Step 1: Always draw a picture before proceeding (see Figure 4-5). This will helpidentify the variables of interest and will organize the data for solving theproblem.

Step 2: Determine the amount of water and solids present in the carrots whereverpossible in the drying operation. You can then do a “mass balance” to followwhere the water and solids go in the process.

Initial Feed: 175 kg / hour of material at 86% moisture (wet basis)

Water in carrots = 175 kg / hour x 0.86 = 150.5 kg / hour

Solids in carrots = 175 kg / hour x (1.0 - 0.86) = 24.5 kg / hour

Or: Solids = 175 kg / hour - 150.5 kg / hour = 24.5 kg / hour

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Initial Product: The dried product moisture is 11% (wet basis), which meansthat the dried product will have a solids content of 89% byweight

Solids leaving dryer = 24.5 kg / hour (assuming no losses)

Let weight of dried product = X kg / hour

0.89 X = 24.5 kg / hour

X = 24.5 kg / hour = 27.53 kg / hour 0.89

Water in dried product = weight of product - solids in product = 27.53 kg / hour - 24.50 kg / hour = 3.03 kg / hour

Therefore, 3.03 kg of water and 24.50 kg of solids would be in the finishedproduct produced leaving the dryer each hour.

27.53 kg of finished product would be produced.

Water removed in = Water in carrots - Water in carrots dryer going into dryer coming out of dryer

= 150.5 kg / hour - 3.03 kg / hour

= 147.47 kg / hour

This means that the dryer is set up to remove 147.47 kg of water from thecarrot cubes every hour.

Step 3: Calculate the water and solids in the new carrot cubes entering the dryerafter the moisture content has changed.

New Feed: 175 kg / hour of material at 88% moisture (wet basis)

Water in carrots = 175 kg / hour x 0.88 = 154.0 kg / hour

Solids in carrots = 175 kg / hour x (1.0 - 0.88) = 21.0 kg / hour

Or: Solids = 175 kg / hour - 154.0 kg / hour = 21.0 kg / hour

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Step 4: Calculate the amount of water leaving the dryer after the moisture content ofthe feed has increased.

From Step 2, we know that the dryer is capable of removing a certain amount ofwater under its current operating set-up. This amount is 147.47 kg / hour.

Weight of water in carrots = Weight of water in carrots - Weight of water leaving the dryer entering dryer removed in dryer

= 154.0 kg / hour - 147.47 kg / hour

= 6.53 kg / hour

Step 5: Calculate how the weight of the dried carrot cubes leaving the dryer

Weight of carrots = Weight of dry solids in carrots + Weight of water in carrots leaving dryer leaving dryer leaving dryer

= 21.0 kg / hour + 6.53 kg / hour (from steps 3 and 4)

= 27.53 kg / hour

Step 6: Calculate the wet basis moisture of the finished product.

From Step 4, there are 6.53 kg/hr of water leaving the dryer in the product.

From Step 5, we know that 27.53 kg/hr of dried carrot cubes leave the dryer.

Wet basis moisture = Water in Product x 100%Total Product Weight

= 6.53 kg/hr x 100% 27.53 kg/hr

= 23.72%

Therefore, the finished product moisture will be approximately 23.7% after themoisture content of the dryer feed material is increased from 86% to 88%.

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Summary:

Under “normal” operating conditions,27.53 kg of product would be producedon an hourly basis with an 11% moisturecontent.

When the moisture content of the feedmaterial jumps from 86% to 88%, thefinished product moisture will increasefrom 11% to 23.7%, if no changes aremade to the dryer’s setpoints.

The dryer operator was in for a bigsurprise since 13% moisture wasexpected and 23.7% moisture wasactually obtained.

The Moral of the Story:

Don’t be lulled into a false sense ofsecurity by using “wet basis moistures”when you should be looking at dry basismoistures.

Dry basis moistures have a consistentbasis of comparison. You are alwaysexpressing the moisture level on a “unitweight basis of dry material”.

Wet basis moistures do not have aconsistent basis of comparison. As themoisture increases, the total weight of thewet material increases. This is whatcauses the problems for many dryeroperators.

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Some of these questions do not involvedrying as a specific topic. However, theydo involve calculating wet and dry basismoistures as well as determining weightsof solids and water in various products.These are important factors in doingfuture drying problems.

Question 1:

Calculate the amount of solids in 425 kgof frozen orange juice concentrate with awater content of 59.8%.

Answer: 170.85 kg (solids content is40.2% by weight).

Question 2:

What is the volume of 150 kg of a sugarsyrup that has a specific gravity of 1.358?

Answer: approx. 110.5 litres

Question 3:

What are the weights of solids and waterin 57.2 kg of 10.5° Brix apple juice?

Answer: 6.0 kg solids and 51.2 kg water. (Remember to convert the volume to itsequivalent weight as a first step).l

Question 4:

Calculate the weights of water and solidsin 13.8 kg litres of 75° Brix corn syrup.

Answer: 10.35 kg solids and 3.45 kgwater.

Question 5:

Calculate the percent solids and Brix ofthe solution that you would get by mixingthe 57.2 kg of apple juice in Question 3and the 13.8 kg of corn syrup in Question4.

Answer: 23.0% solids which isequivalent to 23.0° Brix. (Hint: You needthe total weight of solids and the totalweight of the final solution for this answer.You have already calculated thenecessary values in questions 3 and 4).

Question 6:

What are the solids and water content of150 kg of grape juice having 13.5%solids?

Answer: 20.25 kg of solids and 129.75kg of water.

Question 7:

If 17.5 kg of dry granular sugar wasadded to the grape juice in Question 6,what would the percent solids become?

Answer: 22.5% solids. (Remember thatthe percent solids would be the totalweight of the solids divided by the totalweight of the solution times 100%).

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Question 8:

Calculate the wet basis moisture and drybasis moisture of mangoes having 19%solids.

Answer: 81% wet basis moisture and4.26 grams of water per gram of drysolids.

Question 9:

What is the wet basis moisture (i.e., %water) of a product if 25 kg of it contains3.75 kg of solids?

Answer: 85% moisture

Question 10:

What is the dry basis moisture of papayascontaining 91% moisture on a wet basis?

Answer: 10.1 grams of water per gramof dry solids (or 10.1 kg of water per kg ofdry solids).

Question 11:

We start with 18 kg of ripe tomatoescontaining 93% water by weight. If weremove 15 kg of water by sun-drying,what is the final wet basis moisture?

Answer: 58% (You will have to find thewater and solids at the start. Thensubtract the water removed. This will giveyou the water left after drying. The finalweight of the dried tomatoes will be theweight of the water after drying plus theweight of the solids. Normally, no solidswill be lost during drying.)

Question 12:

How much water must be removed from50 kg of lima beans with a moisturecontent of 67% to get a final moisture of11%? What would the weight of the driedlima beans be?

Answer: 31.46 kg of water must beremoved and 18.54 kg of dried limabeans would be obtained. (The trickhere is to find the final weight of theproduct. Let it be X kg. Remember thatthe weight of the solids at the start of thedrying equals the weight of solids afterdrying. You should find that you have2.04 kg of water in the dried lima beanswhich you can subtract from the initialwater content to get the amount of waterremoved).

Question 13:

a. Calculate the amount of water thatmust be removed from 250 kg of limabeans with an initial moisture content of68% (wet basis) to reach a final moisturecontent of 48% moisture (wet basis).

Answer: 96.2 kg water removed

b. Calculate the amount of water thatmust be removed from 250 kg of limabeans with an initial moisture content of68% (wet basis) to reach a final moisturecontent of 28% moisture (wet basis).

Answer: 138.9 kg water removed

c. Compare the answer for “Part a”where the moisture content was reducedby 20% (wet basis) to the answer for “Partb” where the moisture content of thesame weight of identical product wasreduced by 40%. Even though the

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reduction in wet basis moisture in “Part b”is double that in “Part a”, is the amount ofwater removed also doubled.

Answer: No. Therefore, you shouldnever base comparisons of water removalon the changes in wet basis moistures.

Question 14:

a. Calculate the amount of water thatmust be added to 100 kg of totally drysolids (i.e., bone dry solids) to reach afinal moisture of 30% on a wet basis.

Answer: 42.9 kg water must be added.

b. Calculate the amount of water thatmust be added to 100 kg of totally drysolids (i.e., bone dry solids) to reach afinal moisture of 60% on a wet basis.

Answer: 150.0 kg water must be added.

c. Compare the answer for “Part a”where the moisture content wasincreased by 30% (wet basis) to theanswer for “Part b” where the moisturecontent of the same weight of identicalproduct was increased by 60%. Eventhough the increase in wet basis moisturein “Part b” is double that in “Part a”, is theamount of water added also doubled.

Answer: No. As stated in question 13c,you should never base comparisons ofwater removal on the changes in wetbasis moistures.


CHAPTER 5: SOURCES OF INFORMATION

5.1 Introduction

The following is a list of informationsources relating to food drying and foodprocessing in general. It is not intendedto be an exhaustive listing of availabletextbooks, etc.

The Internet / World-Wide-Web is avaluable source of information andshould definitely be considered as aprimary resource. Dryer manufacturersand equipment suppliers often haveuseful websites with details on how tocontact them for further information. Due to the fact that website addressesare constantly changing while new onesappear and others may disappear, nospecific website addresses are givenhere.

Scientific journals also offer in-depthstudies of food drying. These papersare often highly specific and complex intheir mathematical treatment of aparticular drying phenomenon. Forthese reasons, scientific journal articlesare not listed here.

5.2 General References:

“Dehydration of Foods”; Gustavo V.Barbosa-Canovas and Humberto Vega-Mercado; Chapman and Hall; New York,1996. (ISBN 0-412-06421-9)

“Food Preservation and Safety -Principles and Practice”; Shirley J.VanGarde and Margy Woodburn; IowaState University Press, Ames, Iowa,1994. (ISBN 0-8138-2133-9)

“Food Science - Fifth Edition”; Norman N. Potter and Joseph H.Hotchkiss; Chapman and Hall; NewYork, 1995. (ISBN 0-412-06451-0)

“Food Science and FoodBiotechnology”; Gustavo F. Gutierrez-Lopez and Gustavo V. Barbosa-Canovas, editors; CRC Press; NewYork, 2003. (ISBN 1-56676-892-6)

“Food Process Engineering - Theoryand Laboratory Experiments”; Shri K.Sharma, Steven J. Mulvaney, and SyedS. H. Rizvi; Wiley-Interscience; NewYork, 2000. (ISBN 0-471-32241-5)

“Food Processing Technology:Principles and Practice”; P.J. Fellows;Taylor and Francis, New York, 2000. (ISBN 0-8493-0887-9)

“Fruit and Vegetable Processing”; WJongen; Taylor and Francis; New York,2002. (ISBN 0-8493-1541-7)

”How to Dry Foods”; Deanna DeLong;HP Books, Division of BerkleyPublishing Group; New York, 1992. (ISBN 1-55788-050-6)

“Introduction to Food Engineering -Second Edition”; R. Paul Singh andDennis R. Heldman; Academic Press;New York, 1993. (ISBN 0-12-646381-6)

“Perry’s Chemical Engineers’Handbook - Seventh Edition”; RobertH. Perry and Don W. Green editors;McGraw-Hill, New York, 1997.

(ISBN 0-07-049841-5, InternationalEdition ISBN 0-07-115448-5)

“Principles of Food Processing”; Dennis R. Heldman and Richard W.Hartel; Chapman and Hall, New York,1997. (ISBN 0-412-99451-8)

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