Unit operations in food processing - Ear - Earle

INTRODUCTION Method of studying food process engineering Basic principles of food process engineering Conservation of mass and energy Overall view of an engineering process.. BASIC P

UNIT OPERATIONS IN FOOD PROCESSING

R L EARLE with M.D EARLE

An introduction to the principles

of food process engineering

This is the free web edition of a popular textbook known for its simple approach to the diversity and complexity of food processing

First published in 1966 but still relevant today, Unit Operations in Food Processing explains the

principles of operations and illustrates them by individual processes

Each Chapter contains unworked examples to help the student food technologist or process engineer gain a grasp of the subject

Now in electronic form, fully searchable and cross-linked, this online resource will also be a useful quick reference for technical workers in the food industry

The author, Dick Earle (owner of the copyright) gives permission to download and print any part or all of the text for any nonprofit purposes Content can be printed by individual page, or as complete Chapters Funding, publication and hosting for the book is provided by the New Zealand Institute of Food Science

& Technology (NZIFST)

This web edition of Unit Operations in Food Processing is given by Dick and Mary Earle, with the support

of the NZIFST, as a service to education in food technology, and to the wider food industry

Unit Operations in Food Processing - the Web Edition

http://www.nzifst.org.nz/unitoperations

Unit Operations in Food Processing Copyright © 1983, R L Earle :: Published by NZIFST (Inc.)

ABOUT THE BOOK

The Web Edition

Process engineering is a major contributor to food technology, and provides important and useful tools for the food technologist to apply in designing, developing and controlling food processes Process engineering principles are the basis for food processing, but only some of them are important and commonly encountered in the food industry This book aims to select these important principles and show how they can be quantitatively applied in the food industry It explains, develops and illustrates them at a level of understanding which covers most of the needs of the food technologist in industry and of the student working to become one It can also be used as an introduction to food engineering.

When this book was first published in 1966, there were almost no books available in food process engineering This book met an extensive need at its modest standard and cost It was widely distributed and used, all over the world Subsequently other textbooks have emerged and the available literature and data have grown enormously In particular there are excellent books covering advanced food engineering and also specialist areas of food processing

However there still seems to be a need for an introductory, less specialised book at an accessible level With the hard copy book in English having been out of print for some time, it seemed appropriate to make the book widely available through a free Web site.

So what is largely the text of the 2nd Edition with corrections and only minor changes has been converted

to a user-friendly computer-based learning source on the World Wide Web Here it will be freely available for consultation or copying, indeed for any use save commercial reproduction It is contributed as a service to the food industry It can be used not only as an interactive learning text for the student, but also

as a quick reference for people in industry who from time to time have a specific need for a method of calculation The contents are interlinked so that specific information, examples and figures can easily be found.

The book is intended to introduce technological ideas and engineering concepts, and to illustrate their use Data, including properties and charts, are provided, but for definitive design details may need to be independently checked to ensure requisite precision Every effort has been made to provide clear explanations and to avoid errors, but errors may occur including in the translation to the Web Also greater precision and clarity may well be achievable So feedback from users will be most welcome, and should be directed to The Editor

Obviously this book is the product of much more than just the efforts of the author whose name appears

on the title The ideas developed have been built up over the years by a multitude of researchers, inventors, scientists, engineers and technologists, far too numerous to list Some have been identified in the text and references, and some of these have made individual contributions; the material they made available has provided the essence of the book, the facts and figures and diagrams It is hoped that they have been accurately quoted and nowhere misinterpreted

Pergamon Press first published the book giving it clear layout and wide distribution at a reasonable price

A number of colleagues helped with improvements for the second edition More extensive acknowledgement of these contributors has been made in the Prefaces and elsewhere in the earlier editions The thanks and gratitude of the author to all who have provided material remain undiminished Prof Buncha Ooraikul and Prof Paul Jelen encouraged putting it onto the Web, as it was still being used

Editions even for the Web do not come without cost So particular mention for this Web edition must be made of the New Zealand Institute of Food Science and Technology which contributed finance and hosting, and of Chris Newey who converted it to the new form Chris found that translation of printed text carrying many tables, equations, superscripts and subscripts into Web format moved well beyond the capacity of the optical character recognition, and it gave him a great deal of work before final emergence

in the convenient html and swf forms I am very grateful to him for his extensive and very worthwhile contribution

As in the earlier editions, even more so in this, appearance would never have occurred without the cheerful, unstinting, and technically invaluable help of my wife Mary We will all be rewarded by this site being both useful, and well and widely used.

Richard L.Earle

Palmerston North, New Zealand 2003

About the Author

R L Earle, Emeritus Professor, Massey University, Palmerston North, New

Zealand

Dick Earle trained as a chemical engineer, and in research in food

technology, before entering the New Zealand meat industry His interests

were particularly in refrigeration and energy usage, heat transfer and freezing,

and byproduct and waste processing

Dick joined Massey University in 1965, initially in food technology, and later

founding the biotechnology discipline, which had special interests in the

processing of biologically-based materials

He has published several books jointly with his wife (Dr) Mary Earle on

product development and reaction technology, and many technical papers

and reports He is a Distinguished Fellow of the Institution of Professional Engineers New Zealand

(IPENZ) Dick and Mary Earle have recently established a scholarship for the support and

encouragement of postgraduate research into aspects of technology in New Zealand universities.

The Print Editions

This book is now out of print It was originally published by Pergamon Press:

ISBN 0-08-025537-X Hardcover

ISBN 0-08-025536-1 Flexicover

Copyright

Copyright © 1983-2004 R L Earle All Rights Reserved

Copyright remains with the author, however, the author gives permission to The New Zealand Institute of Food Science & Technology (Inc.) (NZIFST) for free use and display of this material on the internet, and permission to all site visitors for the free use and copying of all or part of the text for non-commercial purposes, subject to acknowledgement of the source (which is, unless otherwise indicated):

Unit Operations in Food Processing, Web Edition, 2004.

Publisher: The New Zealand Institute of Food Science & Technology (Inc.)

Authors: R.L Earle with M.D Earle.

CONTENTS

ABOUT THE BOOK

The history of Unit Operations in Food Processing, and how it came to be published on the web

CHAPTER 1

INTRODUCTION

Method of studying food process engineering

Basic principles of food process engineering

Conservation of mass and energy

Overall view of an engineering process

Dimensions and units

Dimensions symbols

Units

Dimensional consistency

Unit consistency and unit conversion

Dimensionless ratios specific gravity

Basis and units

total mass and composition

Friction loss

Mechanical energy

Other effects

Bernouilli's equation flow from a nozzle

Viscosity shear forces viscous forces

Newtonian and Non-Newtonian Fluids power law equation

Streamline and turbulent flow dimensionless ratios

Reynolds number

Energy losses in flow

Friction in Pipes Fanning equation Hagen Poiseuille equation Blasius equation pipe roughness Moody graph

Energy Losses in Bends and Fittings

Pressure Drop through Equipment

Equivalent Lengths of Pipe

Compressibility Effects for Gases

Calculation of Pressure Drops in Flow Systems

Measurement of pressure in a fluid manometer tube Bourdon tube

Measurement of velocity in a fluid Pitot tube Pitot-static tube Venturi meter orifice meter

Pumps and fans

Positive Displacement Pumps

Jet pumps

Air-lift Pumps

Propeller Pumps and Fan

Centrifugal Pumps and Fans pump characteristics fan laws

Heat Conductances in Series

Heat Conductances in Parallel

Surface-Heat Transfer Newton's Law of Cooling

Unsteady-State Heat Transfer Biot Number Fourier Number charts

Radiation-Heat Transfer Stefan-Boltzmann Law black body

emissivity grey body absorbtivity reflectivity

Radiation between Two Bodies

Radiation to a Small Body from its Surroundings

Natural Convection Equations vertical cylinders and planes

horizontal cylinders horizontal planes

Forced Convection

Forced-convection Equations inside tubes over plane surfaces

outside tubes

Overall Heat-Transfer Coefficients controlling terms

Heat Transfer from Condensing Vapours

vertical tubes or plane surfaces horizontal tubes

Heat Transfer to Boiling Liquids

Continuous-flow Heat Exchangers parallel flow counter flow

cross flow heat exchanger heat transfer log mean temperature difference

Jacketed Pans

Heating Coils Immersed in Liquids

Scraped Surface Heat Exchangers

Plate Heat Exchangers

Thermal Processing

Thermal Death Time

F values Equivalent Killing Power at Other Temperatures

z value sterilization integration time/temperature curves

Pasteurization milk pasteurization

High Temperature Short Time HTST

Refrigeration, Chilling and Freezing

Refrigeration Cycle temperature/enthalphy chart evaporator

condenser adiabatic compression coefficient of performance

Basic Drying Theory

Three States of Water phase diagram for water

vapour pressure/temperature curve for water

Heat Requirements for Vaporization

Heat Transfer in Drying

Dryer Efficiencies

Wet-bulb Temperatures dry bulb temperature Lewis number

Psychrometric Charts

Measurement of Humidity hygrometers

Equilibrium Moisture Content

Air Drying drying rate curves

Calculation of Constant Drying Rates

Roller or Drum Dryers

Fluidized Bed Dryers

Feeding of Multiple-effect Evaporators

Advantages of Multiple-effect Evaporators

Vapour Recompression

Boiling Point Elevation Raoult's Law Duhring's rule

Duhring plot latent heats of vaporization

Evaporation of Heat-Sensitive Materials

equilibrium distribution coefficients

PART 1: THEORY

Concentrations mole fraction partial pressure Avogadro's Law

Gas-Liquid Equilibria partial vapour pressure Henry's Law solubility of gases in liquids

Solid-Liquid Equiibria solubility in liquids

solubility/temperature relationship saturated solution

supersaturated solution

Equilibrium-Concentration Relationships overflow/underflow equilibrium diagram

Operating Conditions contact stages mass balances

Calculation of Separation in Contact-Equilibrium Processes

combining equilibrium and operating conditions

deodorizing/steam stripping McCabe/Thiele diagram

PART 2: APPLICATIONS

Gas Absorption number of contact stages

Rate of Gas Absorption Lewis and Whitman Theory

Stage-equilibrium Gas Absorption

Extraction and Washing Equipment extraction battery

Crystallization mother liquor

Crystallization Equilibrium growth nucleation

metastable region

seed crystals heat of crystallization

Rate of Crystal Growth

Rate of Flow Through Membranes Van't Hoff equation

Diffusion equations Sherwood number Schmidt number

Membrane Equipment

Distillation Equilibrium relationships

boiling temperature/concentration diagram azeotropes

The velocity of particles moving in a fluid terminal velocity drag coefficient terminal velocity magnitude

Sedimentation Stokes' Law

Gravitational Sedimentation of Particles in a Liquid zones velocity of rising fluid sedimentation equipment

Flotation

Sedimentation of Particles in a Gas

Settling Under Combined Forces

Cyclones- optimum shape efficiency

Impingement separators

Classifiers

Centrifugal separations centrifugal force particle velocity

Liquid Separation radial variation of pressure

radius of neutral zone

Centrifuge Equipment

Filtration rates of filtration filter cake resistance

equation for flow through the filter

Sieving rates of throughput standard sieve sizes

cumulative analyses particle size analysis

industrial sieves air classification

Grinding and cutting

Energy Used in Grinding Kick's Law Rittinger's Law Bond's Law Work Index

New Surface Formed by Grinding shape factors

CHAPTER 12

MIXING

Introduction

Characteristics of mixtures

Measurement of mixing sample size sample compositions

Particle mixing random mixture thorough mixture

mixing index

Mixing of Widely Different Quantities mixing in stages

Rates of Mixing mixing times

Energy Input in Mixing

Liquid mixing propeller mixers Power number Froude number

Mixing equipment

Liquid Mixers

Powder and Particle Mixers

Dough and Paste Mixers

Summary

Problems.

APPENDICES

1 Symbols, units and dimensions

2 Units and conversion factors

3 Some properties of gases

4 Some properties of liquids

5 Some properties of solids

6 Some properties of air and water

7 Thermal data for some food products

8 Steam table - saturated steam

9 (a) Psychrometric charts - normal temperatures

9 (b) Psychrometric charts - high temperatures

10 Standard sieves

11 (a) Pressure/enthalpy chart for refrigerant - R134a

11 (b) Pressure/enthalpy chart for refrigerant - Ammonia

INDEX TO FIGURES

INDEX TO EXAMPLES

REFERENCES

BIBLIOGRAPHY

CHAPTERl INTRODUCTION

This book is designed to give food technologists an understanding of the engineering principles involved in the processing of food products They may not have to design process equipment in detail but they should understand how the equipment operates With an understanding of the basic principles of process engineering, they will be able to develop new food processes and modify existing ones Food technologists must also be able to make the food process clearly understood by design engineers and by the suppliers of the equipment used

Only a thorough understanding of the basic sciences applied in the food industry - chemistry, biology and engineering - can prepare the student for working in the complex food industry

of today This book discusses the basic engineering principles and shows how they are important in, and applicable to, every food industry and every food process

For the food process engineering student, this book will serve as a useful introduction to more specialized studies

METHOD OF STUDYING FOOD PROCESS ENGINEERING

As an introduction to food process engineering, this book describes the scientific principles

on which food processing is based and gives some examples of the application of these principles in several food industries After understanding some of the basic theory, students should study more detailed information about the individual industries and apply the basic principles to their processes

For example, after studying heat transfer in this book, the student could seek information on heat transfer in the canning and freezing industries

To supplement the relatively few books on food-process engmeermg, other sources of information are used, for example:

• Specialist descriptions of particular food industries

These in general are written from a descriptive point of view and deal only briefly with engineering

• Textbooks in chemical and biological process engineering

These are studies of processing operations but they seldom have any direct reference

to food processing However, the basic unit operations apply equally to all process industries, including the food industry

BASIC PRINCIPLES OF FOOD PROCESS ENGINEERING

The study of process engineering is an attempt to combine all forms of physical processing into a small number of basic operations, which are called unit operations Food processes may seem bewildering in their diversity, but careful analysis will show that these complicated and differing processes can be broken down into a small number of unit operations For example, consider heating of which innumerable instances occur in every food industry There are many reasons for heating and cooling - for example, the baking of bread, the freezing of meat, the tempering of oils

But in process engineering, the prime considerations are firstly, the extent of the heating or cooling that is required and secondly, the conditions under which this must be accomplished Thus, this physical process qualifies to be called a unit operation It is called 'heat transfer' The essential concept is therefore to divide physical food processes into basic unit operations, each of which stands alone and depends on coherent physical principles For example, heat transfer is a unit operation and the fundamental physical principle underlying it is that heat energy will be transferred spontaneously from hotter to colder bodies

Because of the dependence of the unit operation on a physical principle, or a small group of associated principles, quantitative relationships in the form of mathematical equations can be built to describe them The equations can be used to follow what is happening in the process, and to control and modify the process if required

Important unit operations in the food industry are fluid flow, heat transfer, drying, evaporation, contact equilibrium processes (which include distillation, extraction, gas absorption, crystallization, and membrane processes), mechanical separations (which include filtration, centrifugation, sedimentation and sieving), size reduction and mixing These unit operations, and in particular the basic principles on which they depend, are the subject of this book, rather than the equipment used or the materials being processed

Two very important laws, which all unit operations obey, are the laws of conservation of mass and energy

Conservation of Mass and Energy

then the simple rule holds that "what goes in must come out" Similarly all material entering a unit operation must in due course leave

For example, when milk is being fed into a centrifuge to separate it into skim milk and cream, under the law of conservation of mass the total number of kilograms of material (milk) entering the centrifuge per minute must equal the total number of kilograms of material (skim milk and cream) that leave the centrifuge per minute

Similarly, the law of conservation of mass applies to each component in the entering materials For example, considering the butter fat in the milk entering the centrifuge, the weight of butter fat entering the centrifuge per minute must be equal to the weight of butter fat leaving the centrifuge per minute A similar relationship will hold for the other components, proteins, milk sugars and so on

The law of conservation of energy states that energy can neither be created nor destroyed The total energy in the materials entering the processing plant plus the energy added in the plant must equal the total energy leaving the plant

This is a more complex concept than the conservation of mass, as energy can take various forms such as kinetic energy, potential energy, heat energy, chemical energy, electrical energy and so on

During processing, some of these forms of energy can be converted from one to another Mechanical energy in a fluid can be converted through friction into heat energy Chemical energy in food is converted by the human body into mechanical energy

Note that it is the sum total of all these forms of energy that is conserved

For example, consider the pasteurizing process for milk, in which milk is pumped through a heat exchanger and is frrst heated and then cooled The energy can be considered either over the whole plant or only as it affects the milk For total plant energy, the balance must include: the conversion in the pump of electrical energy to kinetic and heat energy, the kinetic and potential energies of the milk entering and leaving the plant and the various kinds of energy

in the heating and cooling sections, as well as the exiting heat, kinetic and potential energies

To the food technologist, the energies affecting the product are the most important In the case of the pasteurizer, the energy affecting the product is the heat energy in the milk Heat energy is added to the milk by the pump and by the hot water passing through the heat exchanger Cooling water then removes part of the heat energy and some of the heat energy is also lost to the surroundings

The heat energy leaving in the milk must equal the heat energy in the milk entering the pasteurizer plus or minus any heat added or taken away in the plant

Heat energy leaving in milk = initial heat energy

+ heat energy added by pump

The law of conservation of energy can also apply to part of a process For example, considering the heating section of the heat exchanger in the pasteurizer, the heat lost by the hot water must be equal to the sum of the heat gained by the milk and the heat lost from the heat exchanger to its surroundings

From these laws of conservation of mass and energy, a balance sheet for materials and for energy can be drawn up at all times for a unit operation These are called material balances and energy balances

Overall View of an Engineering Process Using a material balance and an energy balance, a food engineering process can be viewed overall or as a series of units Each unit is a unit operation The unit operation can be represented by a box as shown in Fig 1.1

Unit operation

: Further : By· products

• untt Products : operations : Wastes

!! _ • • _ � Energy

Capital Energy Labour Control

Figure 1.1 Unit operation

Into the box go the raw materials and energy, out of the box come the desired products, by­ products, wastes and energy The equipment within the box will enable the required changes

to be made with as little waste of materials and energy as possible In other words, the desired products are required to be maximized and the undesired by-products and wastes minimized Control over the process is exercised by regulating the flow of energy, or of materials, or of both

DIMENSIONS AND UNITS

All engineering deals with definite and measured quantities, and so depends on the making of measurements We must be clear and precise in making these measurements

To make a measurement is to compare the unknown with the known, for example, weighing a material compares it with a standard weight of one kilogram The result of the comparison is expressed in terms of multiples of the known quantity, that is, as so many kilograms

For example, if a rod is 1.18 m long, this measurement can be analysed into a dimension, length; a standard unit, the metre; and a number 1.18 that is the ratio of the length of the rod

to the standard length, 1 m

To say that our rod is 1.18 m long is a commonplace statement and yet because measurement

is the basis of all engineering, the statement deserves some closer attention There are three aspects of our statement to consider: dimensions, units of measurement and the number itself

Dimensions

It has been found from experience that everyday engineering quantities can all be expressed

in terms of a relatively small number of dimensions These dimensions are length, mass, time and temperature For convenience in engineering calculations, force is added as another dimension

Force can be expressed in terms of the other dimensions, but it simplifies many engineering calculations to use force as a dimension (remember that weight is a force, being mass times the acceleration due to gravity)

Dimensions are represented as symbols by: length [L], mass [M], time [t], temperature [T] and force [F]

Note that these are enclosed in square brackets: this is the conventional way of expressing dimensions

All engineering quantities used in this book can be expressed in terms of these fundamental dimensions All symbols for units and dimensions are gathered in Appendix 1

For example:

Length = [L] area = [L] 2 vo lume =

Velocity = length travelled per unit time

Acceleration = rate of change of velocity

Pressure

Density

Energy

Power

= force per unit area

= mass per unit volume

= force times length

= energy per unit time

[L] 3 ill [t]

ill x 1 [t] [t]

[fl [Lf [M]

[L]3 [F] x [L]

[F] x [L]

[t]

=lIJ [tf

As more complex quantities are needed, these can be analysed in terms of the fundamental dimensions For example in heat transfer, the heat-transfer coefficient, h, is defined as the

Units Dimensions are measured in terms of units For example, the dimension of length is measured

in terms of length units: the micro metre, millimetre, metre, kilometre, etc

So that the measurements can always be compared, the units have been defined in terms of physical quantities For example:

• the metre (m) is defmed in terms of the wavelength of light;

• the standard kilogram (kg) is the mass of a standard lump of platinum-iridium;

• the second (s) is the time taken for light of a given wavelength to vibrate a given number of times;

• the degree Celsius eC) is a one-hundredth part of the temperature interval between the freezing point and the boiling point of water at standard pressure;

• the unit of force, the newton (N), is that force which will give an acceleration of

1 m sec-2 to a mass of lkg;

• the energy unit, the newton metre is called the joule (1), and

• the power unit, 1 J S-l, is called the watt (W)

More complex units arise from equations in which several of these fundamental units are combined to defme some new relationship For example, volume has the dimensions [L]3 and

so the units are m3 Density, mass per unit volume, similarly has the dimensions [M]/[L]3, and the units kg/m3 A table of such relationships is given in Appendix 1 When dealing with quantities which cannot conveniently be measured in m, kg, s, multiples of these units are used For example, kilometres, tonnes and hours are useful for large quantities of metres, kilograms and seconds respectively In general, multiples of 103 are preferred such as millimetres (m x 10-3) rather than centimetres (m x 10-2) Time is an exception: its multiples are not decimalized and so although we have micro (10-6) and milli (10-3) seconds, at the other end of the scale we still have minutes (min), hours (h), days (d), etc

Care must be taken to use appropriate multiplying factors when working with these units The common secondary units then use the prefIxes micro (J.!, 10-6), milli (m,lO-3), kilo (k, 103) and mega (M, 106)

Dimensional Consistency All physical equations must be dimensionally consistent This means that both sides of the equation must reduce to the same dimensions For example, if on one side of the equation, the dimensions are [M] [L ]/[T]2, the other side of the equation must also be [M] [L ]/[Tf with the same dimensions to the same powers Dimensions can be handled algebraically and therefore they can be divided, multiplied, or cancelled By remembering that an equation must be dimensionally consistent, the dimensions of otherwise unknown quantities can sometimes be calculated

Knowing that length has dimensions [L] and time has dimensions [t] we have the dimensional equation:

[v] = [L]/[t]

the dimensions of velocity must be [L][tr1

The test of dimensional homogeneity is sometimes useful as an aid to memory If an equation

is written down and on checking is not dimensionally homogeneous, then something has been forgotten

Unit Consistency and Unit Conversion Unit consistency implies that the units employed for the dimensions should be chosen from a consistent group, for example in this book we are using the SI (Systeme Intemationale de Unites) system of units This has been internationally accepted as being desirable and necessary for the standardization of physical measurements and although many countries have adopted it, in the USA feet and pounds are very widely used The other commonly used system is the fps (foot pound second) system and a table of conversion factors is given in Appendix 2

Very often, quantities are specified or measured in mixed units For example, if a liquid has been flowing at 1.3 I /min for 18.5 h, all the times have to be put into one only of minutes, hours or seconds before we can calculate the total quantity that has passed Similarly where tabulated data are only available in non-standard units, conversion tables such as those in Appendix 2 have to be used to convert the units

EXAMPLE 1.2 Conversion of grams to pounds

Convert 10 grams into pounds

From Appendix 2, lIb = 0.4536kg and 1000g = 1kg

be used for the unwanted units to cancel in the conversion

EXAMPLE 1.3 Velocity of flow of milk in a pipe

Milk is flowing through a full pipe whose diameter is known to be 1.8 cm The only measure available is a tank calibrated in cubic feet, and it is found that it takes 1 h to fill 12.4 ft3 What

is the velocity of flow of the liquid in the pipe in SI units?

Therefore v = VIAt

Checking this dimensionally

[L][trl = [L]3[L]-2[trl = [L][trl which is correct

Since the required velocity is in m s-t, volume must be in m3, time in s and area in m2

From the volume measurement

The viscosity of water at 60°F is given as 7.8 x 10-4 lb ft-l S-l

Calculate this viscosity in N s m-2

1 N = 1 kg m s-2 therefore 1 N s m-2 = 1 kg m-l S-l

Required viscosity = 1.16 x 10-3 N s m-2

EXAMPLE 1.5 Thermal conductivity of aluminium: conversion from fps to SI units

The thermal conductivity of aluminium is given as 120 Btu ft-l h-l °Fl Calculate this thermal conductivity in J m-l S-l °Cl

From Appendix 2,

120 Btu frl h-l °Fl X 1055 J x 1 ft x �x 1°F

1 Btu 0.3048m 3600s (5/9)OC Alternatively a conversion factor IBtu ft-l h-l °Flcan be calculated:

Because engineering measurements are often made in convenient or conventional units, this question of consistency in equations is very important Before making calculations always check that the units are the right ones and if not use the necessary conversion factors The method given above, which can be applied even in very complicated cases, is a safe one if applied systematically

A loose mode of expression that has arisen, which is sometimes confusing, follows from the use of the word per, or its equivalent the solidus, / A common example is to give acceleration due to gravity as 9.81 metres per second per second From this the units of g would seem to

be mls/s, that is m s S-l which is incorrect A better way to write these units would be

g = 9.81 mls2 which is clearly the same as 9.81 m S-2

Precision in writing down the units of measurement is a great help in solving problems

Dimensionless Ratios

It is often easier to visualize quantities if they are expressed in ratio form and ratios have the great advantage of being dimensionless If a car is said to be going at twice the speed limit, this is a dimensionless ratio, which quickly draws attention to the speed of the car These dimensionless ratios are often used in process engineering, comparing the unknown with some well-known material or factor

For example, specific gravity is a simple way to express the relative masses or weights of equal volumes of various materials The specific gravity is defmed as the ratio of the weight

of a volume of the substance to the weight of an equal volume of water

SG = weight of a volume of the substance/ weight of an equal volume of water

Dimensionally, SG IEL -;- ill = 1

[�]3 [�]3

where p (rho) is the density of the substance, SG is the specific gravity of the substance and

pw is the density of water

Perhaps the most important attribute of a dimensionless ratio, such as specific gravity, is that

it gives an immediate sense of proportion This sense of proportion is very important to food technologists as they are constantly making approximate mental calculations for which they must be able to maintain correct proportions For example, if the specific gravity of a solid is known to be greater than 1 then that solid will sink in water The fact that the specific gravity

of iron is 7.88 makes the quantity more easily visualized than the equivalent statement that the density of iron is 7880 kg m-3

Another advantage of a dimensionless ratio is that it does not depend upon the units of measurement used, provided the units are consistent for each dimension

Dimensionless ratios are employed frequently in the study of fluid flow and heat flow They may sometimes appear to be more complicated than specific gravity, but they are in the same way expressing ratios of the unknown to the known material or fact These dimensionless ratios are then called dimensionless numbers and are often called after a prominent person who was associated with them, for example Reynolds number, Prandtl number, and Nusselt number; these will be explained in the appropriate section

When evaluating dimensionless ratios, all units must be kept consistent For this purpose, conversion factors must be used where necessary

Precision of Measurement Every measurement necessarily carries a degree of precision, and it is a great advantage if the statement of the result of the measurement shows this precision The statement of quantity should either itself imply the tolerance, or else the tolerances should be explicitly specified For example, a quoted weight of 10.1 kg should mean that the weight lies between 10.05 and 10.149 kg

Where there is doubt it is better to express the limits explicitly as 10.1 ± 0.05 kg

The temptation to refme measurements by the use of arithmetic must be resisted For example, if the surface of a rectangular tank is measured as 4.18 m x 2.22 m and its depth estimated at 3 m, it is obviously unjustified to calculate its volume as 27.8388 m3 which is what arithmetic or an electronic calculator will give A more reasonable answer would be 28 m3• Multiplication of quantities in fact multiplies errors also

In process engineering, the degree of precision of statements and calculations should always

be borne in mind Every set of data has its least precise member and no amount of mathematics can improve on it Only better measurement can do this

may be physically meaningless For much of process engineering three significant figures are all that are justifiable

SUMMARY

1 Food processes can be analysed in terms of unit operations

2 In all processes, mass and energy are conserved

3 Material and energy balances can be written for every process

4 All physical quantities used in this book can be expressed in terms of five fundamental dimensions [M] [L] [t] [F] [T]

5 Equations must be dimensionally homogeneous

6 Equations should be consistent in their units

7 Dimensions and units can be treated algebraically in equations

8 Dimensionless ratios are often a very graphic way of expressing physical relationships

9 Calculations are based on measurement, and the precision of the calculation is no better than the precision of the measurements

PROBLEMS

1 Show that the following heat transfer equation is consistent in its units:

q= UAIlT where q is the heat flow rate (J S-l), U is the overall heat transfer coefficient (J m-2s-loC-I),

A is the area (m2) and IlT is the temperature difference (OC)

2 The specific heat of apples is given as 0.86 Btu lb-l °FI Calculate this in J kg-lac-I

(3.4%)

6 In determining the rate of heating of a tank of 20% sugar syrup, the temperature at the beginning was 20°C and it took 30min to heat to 80°C The volume of the sugar syrup was 50 ft3 and its density 66.9 lbft-3 The s�ecific heat of sugar syrup is 0.9 Btu lb-l of -1 (a) Convert the specific heat to kJ kg-l °e

(b) Determine the average rate of heating, that is the heat energy transferred in unit time,

in SI units (kJs-I)

((a) 3.7 kJ kg-l °el (b) 187 kJs-I)

7 The gas equation is PV = n RT

If P the pressure is 2.0 atm, V the volume of the gas is 6 m3, R the gas constant is 0.08206 m3atm mole-l K-I and T is 300 degrees Kelvin, what are the units of n and what is its numerical value?

(0.49 moles)

8 The gas law constant R is §iven as 0.08206 m3 atm mole-l K-I Find its value in:

(a) ft3mm Hg lb-mole-l K- ,

(b) m3 Pa mole-l K-I,

(c) Joules g-mole-l K-I

Assume 1 atm = 760mm Hg = 1.013x105 Nm-2• Remember 1 joule = INm and in this book, mole is kg mole

((a) 999 ft3mm Hg lb-mole-l K-I (b) 8313 m3 Pa mole-l K-I(c) 8.313 J g-mole-l K-I)

9 The equation determining the liquid pressure in a tank is z = Ppg where z is the depth, P

is the pressure, p is the density and g is the acceleration due to gravity Show that the two sides of the equation are dimensionally the same

10 The Grashof number (Gr) arises in the study of natural convection heat flow If the number is given as:

D3p2Bg�T J.l2 verify the dimensions of � the coefficient of expansion of the fluid The symbols are all defmed in Appendix 1

( [Trl)

CHAPTER 2 MATERIAL AND ENERGY BALANCES

Material quantities, as they pass through food processing operations, can be described by material balances Such balances are statements on the conservation of mass Similarly, energy quantities can be described by energy balances, which are statements on the conservation of energy If there is no accumulation, what goes into a process must come out This is true for batch operation It is equally true for continuous operation over any chosen time interval

Material and energy balances are very important in the food industry Material balances are fundamental to the control of processing, particularly in the control of yields of the products The frrst material balances are determined in the exploratory stages of a new process, improved during pilot plant experiments when the process is being planned and tested, checked out when the plant is commissioned and then refined and maintained as a control instrument as production continues When any changes occur in the process, the material balances need to be determined agam

The increasing cost of energy has caused the food industry to examine means of reducing energy consumption in processing Energy balances are used in the examination of the various stages of

a process, over the whole process and even extending over the total food production system from the farm to the consumer's plate

Material and energy balances can be simple, at times they can be very complicated, but the basic approach is general Experience in working with the simpler systems such as individual unit operations will develop the facility to extend the methods to the more complicated situations, which do arise The increasing availability of computers has meant that very complex mass and energy balances can be set up and manipulated quite readily and therefore used in everyday process management to maximise product yields and minimise costs

BASIC PRINCIPLES

If the unit operation, whatever its nature is seen as a whole it may be represented diagrammatically as a box, as shown in Fig 2.1 The mass and energy going into the box must balance with the mass and energy coming out

Raw materials

Energy in

Heot, Work, Chemica l , Electrical ERI ERZER3

Unit operatio n

Stored materia ls

Stored energy ESI EszEs3

Products out mp1 mP2mp3

Waste products

rTW1rTWfT'W2

Energy in products Epi EpzE P3

Energy in wastes Ew,EwzEw3

Energy losses to surroundings ELI EL E L3

Figure 2.1 Mass and energy balancejoeperreau@aol.com The law of conservation of mass leads to what is called a mass or a material balance

Mass In Raw Materials

= Mass Out + Mass Stored

= Products + Wastes + Stored Materials

:EmR :E mp + :Emw + :Ems (where:E (sigma) denotes the sum of all terms)

:EmR mR! + IIlR2 + IIlR3 Total Raw Materials

:Emp mpl + mn + mp3 Total Products

:Emw mWl + mW2 + mW3 Total Waste Products

:Ems = msl + ms2 + ms3 Total Stored Products

If there are no chemical changes occurring in the plant, the law of conservation of mass will apply also to each component, so that for component A:

rnA in entering materials = rnA in the exit materials + rnA stored in plant For example, in a plant that is producing sugar, if the total quantity of sugar going into the plant

is not equalled by the total of the purified sugar and the sugar in the waste liquors, then there is something wrong Sugar is either being burned (chemically changed) or accumulating in the plant or else it is going unnoticed down the drain somewhere In this case:

where mAU is the unknown loss and needs to be identified So the material balance is now:

Raw Materials = Products + Waste Products + Stored Products + Losses

where Losses are the unidentified materials

Just as mass is conserved, so is energy conserved in food-processing operations The energy coming into a unit operation can be balanced with the energy coming out and the energy stored

Energy In Energy Out + Energy Stored

where:

�ER ERl + ER2 + ER3 + = Total Energy Entering

�Ep Epl + EP2 + Ep3 + Total Energy Leaving with Products

�Ew EWl +EW2 + EW3 + Total Energy Leaving with Waste Materials

�EL ELl + EL2 + EL3 + Total Energy Lost to Surroundings

�Es ESl + ES2 + ES3 + Total Energy Stored

Energy balances are often complicated because forms of energy can be interconverted, for example mechanical energy to heat energy, but overall the quantities must balance

Basis and Units Having decided which constituents need consideration, the basis for the calculations has to be decided This might be some mass of raw material entering the process in a batch system, or some mass per hour in a continuous process It could be: some mass of a particular predominant constituent, for example mass balances in a bakery might be all related to 100 kg of flour entering; or some unchanging constituent, such as in combustion calculations with air where it is helpful to relate everything to the inert nitrogen component; or carbon added in the nutrients in a

fermentation system because the essential energy relationships of the growing micro-organisms are related to the combined carbon in the feed; or the essentially inert non-oil constituents of the oilseeds in an oil-extraction process Sometimes it is unimportant what basis is chosen and in such cases a convenient quantity such as the total raw materials into one batch or passed in per hour to a continuous process are often selected Having selected the basis, then the units may be chosen such as mass, or concentrations which can be by weight or can be molar if reactions are important

Total mass and composition

Material balances can be based on total mass, mass of dry solids, or mass of particular com­ponents, for example protein

EXAMPLE 2.1 Constituent balance of milk

Skim milk is prepared by the removal of some of the fat from whole milk This skim milk is found to contain 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1 % fat and 0.8% ash If the original milk contained 4.5% fat, calculate its composition, assuming that fat only was removed

to make the skim milk and that there are no losses in processing

Basis: 100 kg of skim milk This contains, therefore, 0.1 kg of fat Let the fat which was removed from it to make skim milk be x kg

Total original fat = (x + 0.1) kg

Total original mass = (100 + x) kg

and as it is known that the original fat content was 4.5% so

So the composition of the whole milk is then:

fat = 4.5% , water 90.5 = 86.5 %, protein = � = 3.3 %, carbohydrate = � = 4.9%

the solute in a volume of the solution, in this book expressed as kg mole in I m3 of the solution The mole fraction is the ratio of the number of moles of the solute to the total number of moles

of all species present in the solution Notice that in process engineering, it is usual to consider kg moles and in this book the term mole means a mass of the material equal to its molecular weight

in kilograms In this book, percentage signifies percentage by weight (w/w) unless otherwise specified

EXAMPLE 2.2 Concentrations

A solution of common salt in water is prepared by adding 20 kg of salt to 100 kg of water, to make a liquid of density 1323 kg m-3• Calculate the concentration of salt in this solution as a (a) weight fraction, (b) weight/volume fraction, (c) mole fraction, (d) molal concentration

Mole fraction of salt

and so mole fraction of salt

= 0.058

(d) The molar concentration (M) is 220.5/58.5 = 3.77 moles in 1 m3

Note that the mole fraction can be approximated by the (moles of salt/moles of water) as the number of moles of water are dominant, that is the mole fraction is close to 0.34/5.56 = 0.06 1

As the solution becomes more dilute, this approximation improves and generally for dilute

solutions the mole fraction of solute is a close approximation to the moles of solute/moles of solvent

In solid/liquid mixtures, all these methods can be used but in solid mixtures the concentrations are normally expressed as simple weight fractions

With gases, concentrations are primarily measured in weight concentrations per unit volume, or

as partial pressures These can be related through the gas laws Using the gas law in the form:

The SI unit of pressure is N m-2 called the Pascal (Pa) As this is of inconvenient size for many purposes, standard atmospheres (atm) are often used as pressure units, the conversion being 1 atm = 1.013 x 105 Pa, or very nearly 1 atm = 100kPa

EXAMPLE 2.3 Air composition

If air consists of 77% by weight of nitrogen and 23% by weight of oxygen calculate:

(a) the mean molecular weight of air,

(b) the mole fraction of oxygen,

(c) the concentration of oxygen in mole m-3 and kg m-3 if the total pressure is1.5 atmospheres and the temperature is 25°C

(a) Taking the basis of 100 kg of air: it contains 77 moles of N2 and � moles of 02

Total number of moles = 2.75 + 0.72 = 3.47 moles

So mean molecular weight of air 100 = 28.8

3.47 Mean molecular weight of air = 28.8 (b) The mole fraction of oxygen 0.72

2.75 + 0.72

0.72 3.47 Mole fraction of oxygen = 0.21 (Note this is also the volume fraction)

= 0.21

( c) In the gas equation, where n is the number of moles present: the value of R is 0.08206 m3 atm mole-l KI and at a temperature of 25°C = 25 + 273 = 298 K., and where V = 1 m3

and so

1.5 x 1 = n x 0.08206 x 298

n = 0.061 moles weight of air in 1 m3 = n x mean molecular weight

= 0.061 x 28.8

= 1.76 kg and of this 23% is oxygen, so weight of oxygen in 1 m3

by frrst calculating the number of moles of gas using the gas laws, treating the volume as the volume of the liquid, and then calculating the number of moles of liquid directly

EXAMPLE 204 Carbonation of a soft drink

In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent

of 3 volumes of gas to one volume of water at O°C and atmospheric pressure Calculate (a) the mass fraction and (b) the mole fraction of the C02 in the drink, ignoring all components other than C02 and water

Basis 1 m30f water

Volume of carbon dioxide added = 1000 kg =3 m3

From the gas equation

And so

pV =nR T

1 x 3 = n x 0.08206 x 273

n = 0.134 moles Molecular weight of carbon dioxide = 44

And so weight of carbon dioxide added = 0.134 x 44 = 5.9 kg

(a) Mass fraction of carbon dioxide in drink = 5.9/(1000 + 5.9) = 5.9 x 10-3

(b) Mole fraction of carbon dioxide in drink = 0.134/(1000/18 + 0.134) = 2.41 x 10-3

Types of Process Situations Continuous processes

In continuous processes, time also enters into consideration and the balances are related to unit

time Consider a continuous centrifuge separating whole milk into skim milk and cream If the

material hold-up in the centrifuge is constant both in mass and in composition, then the quantities

of the components entering and leaving in the different streams in unit time are constant and a

materials balance can be written on this basis Such an analysis assumes that the process is in a

steady state, that is flows and quantities held up in vessels do not change with time

EXAMPLE 2.5 Materials balance in continuous centrifuging of milk

If 35,000kg of whole milk containing 4% fat is to be separated in a 6 hour period into skim milk

with 0.45% fat and cream with 45% fat, what are the flow rates of the two output streams from a

continuous centrifuge which accomplishes this separation?

Basis I hour's flow of whole milk

Mass out

Let the mass of cream be x kg then its total fat content is 0.45x The mass of skim milk is

(5833 - x) and its total fat content is 0.0045(5833 - x)

Materials balance on fat:

5833 x 0.04 0.0045(5833 - x) + 0.45x

The time unit has to be considered carefully in continuous processes as normally such processes

operate continuously for only part of the total factory time Usually there are three periods, start

up, continuous processing (so-called steady state) and close down, and it is important to decide

what material balance is being studied Also the time interval over which any measurements are taken must be long enough to allow for any slight periodic or chance variation

In some instances a reaction takes place and the material balances have to be adjusted accordingly Chemical changes can take place during a process, for example bacteria may be destroyed during heat processing, sugars may combine with amino acids, fats may be hydrolysed and these affect details of the material balance The total mass of the system will remain the same but the constituent parts may change, for example in browning the sugars may reduce but browning compounds will increase An example of the growth of microbial cells is given Details

of chemical and biological changes form a whole area for study in themselves, coming under the heading of unit processes or reaction technology

EXAMPLE 2.6 Materials balance of yeast fermentation

Baker's yeast is to be grown in a continuous fermentation system using a fermenter volume of 20m3 in which the flow residence time is 16 h A 2% inoculum containing 1.2 % of yeast cells is included in the growth medium This is then passed to the fermenter, in which the yeast grows with a steady doubling time of 2.9h The broth leaving the fermenter then passes to a continuous centrifuge, which produces a yeast cream containing 7% of yeast, 97% of the total yeast in the broth Calculate the rate of flow of the yeast cream and of the residual broth from the centrifuge The volume of the fermenter is 20m3 and the residence time in this is 16 h so the flow rate through the fermenter must be:

20/16 = 1.25 m3h-1 Assuming the broth to have a density substantially equal to that of water, i.e 1000 kgm-3,

Mass flow rate = 1250kg h-l Yeast concentration in the liquid flowing to the fermenter

= (concentration in inocu1um)/( dilution of inoculum)

(1.2/100)/(10012 ) 2.4 x 104kgkgl Now the yeast mass doubles every 2.9 h, so in 2.9h, 1kg becomes 1 x 2lkg (1 generation)

In 16h there are 16/2.9 5.52 doubling times

1kg yeast grows to 1 x 25.5kg 45.9 kg

Yeast leaving fermenter 2.4 x 104 x 45.9 kgkg-l

Yeast leaving fermenter initial concentration x growth x flow rate

2.4 x 104 X 45.9 x 1250 13.8 kgh-l

From the centrifuge flows a (yeast rich) cream with 7% yeast, this being 97% of the total yeast: The yeast rich cream = (13.8 x 0.97) x 100/7 191 kgh-l

and the broth (yeast lean) stream is (1250 - 191) 1059kgh-1

which contains (13.8 x 0.03 ) 0.4 1 kgh-l yeast and the yeast concentration in the residual broth 0.4111059

0.039%

Materials balance over the centrifulfe

(Yeast in cream 13.4) 188

A materials balance, such as in Example 2.6 for the manufacture of yeast, could be prepared in much greater detail if this were necessary and if the appropriate information were available Not only broad constituents, such as the yeast, can be balanced as indicated but all the other constituents must also balance

One constituent is the element carbon: this comes with the yeast inoculum in the medium, which must have a suitable fermentable carbon source, for example it might be sucrose in molasses The input carbon must then balance the output carbon, which will include the carbon in the outgoing yeast, carbon in the unused medium and also that which was converted to carbon dioxide and which came off as a gas or remained dissolved in the liquid Similarly all of the other elements such as nitrogen and phosphorus can be balanced out and calculation of the balance can be used to determine what inputs are necessary knowing the final yeast production that is required and the expected yields While a formal solution can be set out in terms of a number of simultaneous equations, it can often be easier both to visualize and to calculate if the data are tabulated and calculation proceeds step by step gradually filling out the whole detail Blending

Another class of situations that arises are blending problems in which various ingredients are combined in such proportions as to give a product of some desired composition Complicated examples, in which an optimum or best achievable composition must be sought, need quite elaborate calculation methods, such as linear programming, but simple examples can be solved

by straightforward mass balances

EXAMPLE 2.7 Blending of minced meat

A processing plant is producing minced meat, which must contain 15% of fat If this is to be made up from boneless cow beef with 23% of fat and from boneless bull beef with 5% of fat, what are the proportions in which these should be mixed?

Let the proportions be A of cow beef to B of bull beef

Then by a mass balance on the fat,

A(0.23 - 0.15) A(0.08) AlB

It is possible to solve such a problem formally using algebraic equations and indeed all material balance problems are amenable to algebraic treatment They reduce to sets of simultaneous equations and if the number of independent equations equals the number of unknowns the equations can be solved For example, the blending problem above can be solved in this way

If the weights of the constituents are A and B and proportions of fat are a, b, blended to give C of composition e:

then for fat and overall Aa +Bb A + B Ce C

of which A and B are unknown, and say we require these to make up 100 kg of C then

100 (e-b) (a-b)

100 (0.15 - 0.05) (0.23 -0.05)

100 0.10 0.18 55.6kg 44.4

as before, but the algebraic solution has really added nothing beyond a formula which could be useful if a number of blending operations were under consideration

Layout

In setting up a material balance for a process a series of equations can be written for the various individual components and for the process as a whole In some cases where groups of materials maintain constant ratios, then the equations can include such groups rather than their individual constituents For example in drying vegetables, the carbohydrates, minerals, proteins etc., can be grouped together as 'dry solids', and then only dry solids and water need be taken through the material balance

EXAMPLE 2.8 Drying yield of potatoes

Potatoes are dried from 14% total solids to 93% total solids What is the product yield from each

1000 kg of raw potatoes assuming that 8% by weight of the original potatoes is lost in peeling

Basis 1000kg potato entering

As 8% of potatoes are lost in peeling, potatoes to drying are 920 kg, solids 129 kg

Peelings

solids 11 water 69 Water evaporated Total losses Total

EXAMPLE 2.9 Extraction

1000 kg of soya beans, of composition 18% oil, 35% protein, 27.1 % carbohydrate, 9.4% fibre and ash, 10.5% moisture, are:

(a) crushed and pressed, which reduces oil content in beans to 6%;

(b) then extracted with hexane to produce a meal containing 0.5% oil;

(c) fmally dried to 8% moisture

Assuming that there is no loss of protein and water with the oil, set out a materials balance for the soya bean constituents

Basis 1000kg

Mass in:

Oil = 1000 x 18/100 = 180 kg

Protein = 1000 x 35/100 = 350 kg

Other non-oil constituents = 470 kg

Carbohydrate, ash, fibre and water are calculated in a similar manner to fat and protein

Mass out:

(a) Expressed oil

In original beans, 820kg of protein, water, etc., are associated with 180 kg of oil

In pressed material, 94 parts of protein, water, etc., are associated with 6 parts of oil

Total oil in expressed material

Loss of oil in press

(b) Extracted oil

= 820 x 6/94 = 52.3 kg

= 180 - 52.3 = 127.7 kg

In extracted meal 99.5 parts of protein, water, etc., are associated with 0.5 parts of oil

Total oil in extracted meal = 820 x 0.5/99.5 = 4.1 kg

Loss of oil to hexane = 52.3 - 4.1 = 48.2 kg

(c) Water

In the extracted meal, 8 parts of water are associated with 92 parts of oil, protein, etc

Weights of dry materials in fmal meal = 350 + 271 + 94 + 4.1 = 719.1 kg

MATERIALS BALANCE BASIS 1000 kg SOYA BEANS ENTERING

Oil

Mass in (kg)

180

350 Protein

Carbohydrate Ash and fibre 271 94

105 Water

Mass out (kg) Expressed oil

Oil in hexane Total oil Total meal Consisting of:

Protein Carbohydrate Ash and fibre Water

Oil

350

271

94 62.5

4 1 Water lost in drying Total

ENERGY BALANCES

127.7 48.2 175.9

78 1.6

42.5 1000.0

Energy takes many forms such as heat, kinetic energy, chemical energy, potential energy but because of interconversions it is not always easy to isolate separate constituents of energy balances However, under some circumstances certain aspects predominate In many heat balances, other forms of energy are insignificant; in some chemical situations, mechanical energy

is insignificant and in some mechanical energy situations, as in the flow of fluids in pipes, the frictional losses appear as heat but the details of the heating need not be considered We are seldom concerned with internal energies

Therefore practical applications of energy balances tend to focus on particular dominant aspects and so a heat balance, for example, can be a useful description of important cost and quality aspects of a food process When unfamiliar with the relative magnitudes of the various forms of energy entering into a particular processing situation, it is wise to put them all down Then after some preliminary calculations, the important ones emerge and other minor ones can be lumped

together or even ignored without introducing substantial errors With experience, the obviously minor ones can perhaps be left out completely though this always raises the possibility of error Energy balances can be calculated on the basis of external energy used per kilogram of product,

or raw material processed, or on dry solids or some key component The energy consumed in food production includes:

direct energy which is fuel and electricity used on the :fu.rm, and in transport and in factories, and

in storage, selling, etc.; and

indirect energy which is used to actually build the machines, to make the packaging, to produce the electricity and the oil and so on

Food itself is a major energy source, and energy balances can be determined for animal or human feeding; food energy input can be balanced against outputs in heat and mechanical energy and chemical synthesis

In the SI system there is only one energy unit, the joule However, kilocalories are still used by some nutritionists, and British thermal units (Btu) in some heat-balance work

The two applications used in this book are heat balances, which are the basis for heat transfer, and the energy balances used in analysing fluid flow

Heat Balances

The most common important energy form is heat energy and the conservation of this can be illustrated by considering operations such as heating and drying In these, enthalpy (total heat) is conserved As with the material balances, so enthalpy balances can be written round the various items of equipment, or process stages, or round the whole plant, and it is assumed that no appreciable heat is converted to other forms of energy such as work

Heat to surroundings

Heat from electricity

Heat from fuel combustion

Heat from mechanical sources Heat stored

Heat in raw materials

Heat out in products

various quantities of materials involved, their specific heats, and their changes in temperature or state (as quite frequently, latent heats arising from phase changes are encountered) Fig 2.2 illustrates the heat balance

Heat is absorbed or evolved by some reactions in food processing but usually the quantities are small when compared with the other forms of energy entering into food processing such as sensible heat and latent heat Latent heat is the heat required to change, at constant temperature, the physical state of materials from solid to liquid, liquid to gas, or solid to gas Sensible heat is the heat which when added or subtracted from food materials changes their temperature and thus can be sensed The units of specific heat (c) are J kg-l °Cl and sensible heat change is calculated

by multiplying the mass by the specific heat and the change in temperature, m c !1T and the unit

is J The unit of latent heat is J kg-l and total latent heat change is calculated by multiplying the mass of the material, which changes its phase, by the latent heat Having determined those factors that are significant in the overall energy balance, the simplified heat balance can then be used with confidence in industrial energy studies Such calculations can be quite simple and straightforward but they give a quantitative feeling for the situation and can be of great use in design of equipment and process

EXAMPLE 2.10 Heat demand in freezing bread

It is desired to freeze 10,000 loaves of bread, each weighing 0.75 kg, from an initial room temperature of 18°C to a fmal store temperature of -18°C If this is to be carried out in such a way that the maximum heat demand for the freezing is twice the average demand, estimate this maximum demand, if the total freezing time is to be 6 h

If data on the actual bread is unavailable, in the literature are data on bread constituents, calculation methods and enthalpy/temperature tables

(a) Tabulated data �Appendix 7) indicates specific heat above freezing 2.93 kJ kg-l °Cl, below freezing 1.42 kJ kg- °Cl, latent heat of freezing 115 kJ kg-l and freezing temperature is -2°C

Total enthalpy change (M!) = [18 - (-2)] 2.93 + 115 + [-2 - (-18)] 1.42

(b) Formula (Appendix 7) assuming the bread is 36% water gives:

specific heat above freezing

4.2 x 0.36 + 0.84 x 0.64 = 2.05kJkg-l °Cl specific heat below freezing

2.1x 0.36 + 0.84 x 0.64 latent heat

0.36 x 335

= 1.29kJkg-l °Cl

= 121kJkg-l Total enthalpy change (M!) = [18 - (-2)]2.04 + 121 +[-2-(-18)]1.29 = 183kJkg-l (c) Enthalpy/temperature data for bread of 36% moisture (Mannheim et al., 1957) suggest:

= 210.36 kJkg-l

= 65.35 kJkg-l

So from + 180e to - 180e total enthalpy change (Mf) = 145kJkg -1 (d) The enthalpy/temperature data in Mannheim et al 1957 can also be used to estimate

"apparent" specific heats as M /llt = c and so using the data:

Toe -20.6 -17.8 15.6 18.3 HkJkg-1 55.88 65.35 203.4 210.4 Giving C-18 = M 65.35 -55.88 = 3.4 kJkg-10C-1,

Taking Mas 150kJ kg-1 then:

Total heat change = 150 x 10,000 x 0.75 = 1.125 x 106kJ

In some cases, it is adequate to make approximations to heat balances by isolating dominant terms and ignoring less important ones To make approximations with any confidence, it is necessary to be reasonably sure about the relative magnitudes of the quantities involved Having once determined the factors that dominate the heat balance, simplified balances can then be set

up if appropriate to the circumstances and used with confidence in industrial energy studies This simplification reduces the calculation effort, focuses attention on the most important terms, and helps to inculcate in the engineer a quantitative feeling for the situation

EXAMPLE 2.11 Dryer heat balance for casein drying