Lewis Physical Properties of Foods and Food Processing Systems

1. Auflage 1990
ISBN: 978-1-84569-842-3
Verlag: Elsevier Science & Techn.
Format: EPUB
Kopierschutz: 6 - ePub Watermark

E-Book, Englisch, 480 Seiten

Reihe: Woodhead Publishing Series in Food Science, Technology and Nutrition

This book is an invaluable introduction to the physical properties of foods and the physics involved in food processing. It provides descriptions and data that are needed for selecting the most appropriate equipment in food technology and for making food processing calculations.

Autoren/Hrsg.

Lewis, M J

Leseproben

1 Units and dimensions
1.1 INTRODUCTION
Confusion often arises from the diverse system of weights and measures at present operating in both the UK and overseas. It is possible to buy petrol in gallons (gal) or litres (1), beer in pints, vegetables in pounds (lb) and butter in grams (g). Temperatures are still measured in both degrees Fahrenheit (°F) and degrees Celsius (°C); imagine our surprise if the weather forecaster announced temperatures in kelvins (K). It has been common usage tor pressure to be measured in pounds-force per square inch (lbf in- 2) (often referred to colloquially as psi) e.g. when checking car tyres, but more recently units such as bars and kilograms-force per square centimetre (kgf cm- 2) are becoming more widespread. Electrical energy is measured in kilowatt hours (kWh) units, our gas in therms and the power of our motor vehicles in brake horsepower (hp). Although most science subjects in schools are now taught using the International System of Units (SI), it often comes as a shock to students embarking on courses of higher education to be confronted with the centimetre gram second (cgs) or the Imperial system of units (ft lb s). Most textbooks and articles in earlier scientific journals published in the English language are written using Imperial units; it was not until the 1970s that contributions in food science textbooks began to use SI units. Although American textbooks and papers still quote mainly Imperial units, this is now also changing. The food industry is also steeped in tradition and it will take a considerable time period for all instruments, instruction manuals and personnel to be converted to SI units. In theory, in the UK, the change from Imperial units to the SI system should have been completed by the early 1980s; in practice, this conversion is nowhere near complete and we now have to contend with a mixed system of units. It is necessary for students of applied science to be familiar with the two metric system of units (i.e. cgs and SI) and the Imperial system, in order to derive the maximum benefit from the available literature. This first chapter is written with the first objective in mind. However, throughout the book the major items of theory will be presented in SI units, and relevant conversion factors will be given, where necessary. 1.2 FUNDAMENTAL UNITS
The fundamental dimensions for the main system of measurements are mass [M], length [L], time [T] and temperature [?]. The fundamental units in the main systems (together with the corresponding abbreviations in parentheses) are summarized in Table 1.1. Table 1.1 Fundamental units for the three main systems of measurement. Mass kilogram (kg) gram (g) pound (lb) Length metre (m) centimetre (cm) foot (ft) Time second (s) second (s) second (s) hour (s) Temperature kelvin (K) or kelvin (K) degree degree Celsius (°C) degree Celsius (°C) Fahrenheit (°F) Electric current, luminous intensity and the amount of substance (in units of moles (mol)) are also included amongst the fundamentals the SI system. The Imperial system also includes force F as one of the fundamentals, whereas in the SI system it is included amongst the derived units. The fundamental units will now be discussed in more detail. 1.3 MASS [M] (kg)
Mass is defined as the amount of matter in a body. The international prototype kilogram is a simple cylinder of platinum–iridium alloy, with height equal to diameter, which is kept by the International Bureau of Weights and Measures at Sèvres, near Paris. The mass of this is taken to represent one kilogram (1 kg). It is important to distinguish between mass and weight. Strictly speaking, weight is defined as the force acting on an object as a result of gravity. Consequently the weight of an object will change as the gravitational force changes, whereas the mass will remain constant. It is often assumed that the acceleration due to gravity is constant at all points on the surface of the Earth and that an object will weigh the same everywhere, for most intents and purposes, this is a reasonable assumption. In practice the mass of an object is determined by comparing the force it exerts with that exerted by a known mass. Consequently we often refer to the weight of an object when we actually mean its mass. Weight, mass or portion control is extremely important in all packaging and filling operations, where the weight is declared on the label. 1.3.1 Mass balances
In batch processing operations, such as mixing, blending, evaporation and dehydration, the law of conservation of mass applies. This can be used in the form of mass balances for evaluating these processes. In a batch mixing process it is possible to perform a total mass balance and a mass balance on each of the components, e.g. fat or protein. Sometimes, as in evaporation, all components are grouped together as total solids. The mass balance states that =output Total and component balances will be illustrated by the following example. 1.3.1.1 Example of mass balance It is required to produce 100 kg of low fat cream (18% fat) from double cream containing 48% fat and milk containing 3.5% fat; all concentrations are given in weight per weight (w/w). (strictly speaking, mass per mass). How much double cream and milk are required? With all such problems it is helpful to represent the process diagrammatically (see Fig. 1.1). Fig. 1.1 Mass balance in a cream standardization unit. Let the mass of the milk and the mass of the double cream required be X kg and Y kg, respectively. Then the total balance is +Yinput=100output   (1.1) The component balance on the fat is X+0.48Y=100×0.18   (1.2) Substituting X from equation (1.1) into equation (1.2) gives 100-Y+0.48Y=183.5-0.035Y+0.48Y=18 Therefore =32.58kgX=67.42kg The blending operation requires 67.4 kg of milk and 32.6 kg of double cream. Standardization is the name given to the process where the fat content of milk products is adjusted by the addition of cream or skim-milk. In a continuous process, the same principles apply but an additional term is introduced to account for the fact that some material may accumulate. The equation becomes –output=accumulation However, many continuous food-processing operations take place under steady-state conditions, once they have settled down and can be analysed as such. A steady state is achieved when there is no accumulation of material, i.e. =output Such operations are analysed in the same way as batch operations, normally on a time basis (hourly), as in the following example. 1.3.1.2 Mass balances (hourly basis) If 100 kg h- 1 of liquid containing 12% total solids is to be concentrated to produce a liquid containing 32% total solids, how much water is removed each hour? Again the process can be represented diagramatically (Fig. 1.2). Fig. 1.2 Mass balance in an evaporation plant. Let mass of water removed be m and mass of concentrate produced be C. Therefore the total balance is =m+C and the solids balance is: ×0.12=C×0.32 It is assumed that the water leaving the evaporator contains no solid. Thus, =37.5kg,m=62.5kg Water needs to be removed at the rate of 62.5 kg h- 1. This fixes the evaporative capacity of the equipment. Such calculations involving total solids are extremely useful in evaporation and dehydration processes. In some cases, chemical or biological reactions take place and an extra term for the production of new components will need to be considered. +production-output=accumulation Mass balances can also be used for evaluating losses occurring during food processing. For example, a large creamery may process 1,000,0001 of milk a day producing butter and skim-milk powder. If the amount and composition of milk processed are known together with the amounts and compositions of butter and skim-milk powder, it is possible to determine the processing losses; most of this will end up in the effluent stream and require extra expensive treatment. For example, let us evaluate the losses occurring during the conversion of 106 l of full cream milk to 40 000 kg of butter and 92 000 kg of skim-milk powder. The input is as follows (note that kg m- 3 is equivalent to gl- 1): milk, 106 l; fat, 35 kg m- 3; milk solids not fat (MSNF), 90 kg m- 3. The output is as follows: butter, 40 000 kg (fat, 84%; MSNF, 1%; water, 15% (w/w)); skim-milk powder, 92 000 kg (fat, 1%; MSNF, 95%, water, 4% (w/w)). The losses occurring are shown in Table 1.2. This table shows that 480 kg of fat and...