drying is generally by conduction, sometimes by radiation. The materials to be dried are contained in a trough-shaped conveyor belt, made from mesh,.

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CHAPTER 7 DRYING Drying is one of the oldest methods of preserving food. Primitive societies practised the drying of meat and fish in the sun long before recorded history. Today the drying of foods is still important as a method of preservation. Dried f oods can be stored for long periods without deterioration occurring. The principal reasons for this are that the microorganisms which cause food spoilage and decay are unable to grow and multiply in the absence of sufficient water and many of the enzymes w hich promote undesired changes in the chemical composition of the food cannot function without water. Preservation is the principal reason for drying, but drying can also occur in conjunction with other processing. For example in the baking of bread, appl ication of heat expands gases, changes the structure of the protein and starch and dries the loaf. Losses of moisture may also occur when they are not desired, for example during curing of cheese and in the fresh or frozen storage of meat, and in innumerab le other moist food products during holding in air. Drying of foods implies the removal of water from the food. In most cases, drying is accomplished by vaporizing the water that is contained in the food, and to do this the latent heat of vaporization mus t be supplied. There are, thus, two important process – controlling factors that enter into the unit operation of drying: (a) transfer of heat to provide the necessary latent heat of vaporization, (b) movement of water or water vapour through the food material and then away from it to effect separation of water from food. Drying processes fall into three categories: Air and contact drying under atmospheric pressure. In air and contact drying, heat is transferred through the foodstuff either from heated air or from heated surfaces. The water vapour is removed with the air. Vacuum drying. In vacuum drying, advantage is taken of the fact that evaporation of water occurs more readily at lower pressures than at higher ones. Heat transfer in vacuum drying is generally by conduction, sometimes by ra diation. Freeze drying. In freeze drying, the water vapour is sublimed off frozen food. The food structure is better maintained under these conditions. Suitable temperatures and pre ssures must be established in the dryer to en sure that sublimation occurs.

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BASIC DRYING THEORY Three States of Water Pure water can exist in three states, solid, liquid and vapour. The state in which it is at any time depends on the temperature and pressure con ditions and it is possible to i llustrate this on a phase diagram, as in Fig. 7.1. Figure 7.1 Phase diagram for water If we choose any condition of temperature and pressure and find the corresponding point on the diagram, this point will lie, in general, in one of the three labelled r egions, solid, liquid, or gas. This will give the state of the water under the chosen conditions. Under certain conditions, two states may exist side by side, and such conditions are found only along the lines of the diagram. Under one condition, all thre e states may exist together; this condition arises at what is called the triple point, indicated by point O on the diagram. For water it occurs at 0.0098 o C and 0.64 kPa (4.8 mm of mercury) pressure. If heat is applied to water in any state at constant pre ssure, the temperature rises and the condition moves horizontally across the diagram, and as it crosses the boundaries a change of state will occur. For example, starting from condition A on the diagram adding heat warms the ice, then melts it, then warms the water and finally evaporates the water to condition A ‘. Starting from condition B , situated below the triple point, when heat is added, the ice warms and then sublimes without passing through any liquid state.

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Liquid and vapour coexist in equilibrium only under the conditions along the line OP . This line is called the vapour pressure/temperature line. The vapour pressure is the measure of the tendency of molecules to escape as a gas from the liquid. The vapour pressure/temperature curve for water is sh own in Fig. 7.2, which is just an enlargement for water of the curve OP of Fig. 7.1. Figure 7.2. Vapour pressure/temperature curve for water Boiling occurs when the vapour pressure of the water is equal to the total pressure on the water surface. The boiling point at atmospheric pressure is of course 100 o C. At pressures above or below atmospheric, water boils at the corres ponding temperatures above or below 100 o C, as shown in Fig. 7.2 for temperatures below 100 o C. Heat Requirements for Vaporization The energy, which must be supplied to vaporize the water at any temperature, depends upon this temperature. The quantity of energy required per kg of water is called the latent heat of vaporization , if it is from a liquid, or latent heat of sublimation if it is from a solid. The heat energy required to vaporize water under any given set of conditions can be calculated from the latent heats given in the steam table in Appendix 8, as steam and water vapour are the same thing.

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EXAMPLE 7.1. Heat energy in ai r drying A food containing 80% water is to be dried at 100 o C down to moisture content of 10%. If the initial temperature of the food is 21 o C, calculate the quantity of heat energy required per unit weight of the original material, for drying under atmosph eric pressure. The latent heat of vapori zation of water at 100 o C and at standard atmospheric pressure is 2257kJkg – 1 . The specific heat capacity of the food is 3.8 kJkg – 1o C – 1 and of water is 4.186 kJkg – 1o C – 1 . Find also the energy requirement/kg water remov ed. Calculating for 1 kg food Initial moisture = 80% 800g moisture are associated with 200 g dry matter. Final moisture = 10 %, 100g moisture are associated with 900 g dry matter, Therefore (100 x 200)/900g = 22.2g moisture are associated with 200 g d ry matter. 1kg of original matter must lose (800 – 22) g moisture = 778 g = 0.778 kg moisture. Heat energy required for 1kg original material = heat energy to raise temperature to 100 o C+ latent heat to remove water = (100 – 21) x 3.8 + 0.778 x 2257 = 300.2 + 1755.9 = 2056kJ. Energy/kg water removed, as 2056 kJ are required to remove 0.778 kg of water, = 2056/0.778 = 2643kJ . Steam is often used to supply heat to air or to surfaces used for drying. In condensing, steam gives up its la tent heat of vaporization; in drying, the substance being dried must take up latent heat of vaporization to convert its liquid into vapour, so it might be reasoned that 1kg of steam condensing will produce 1kg vapour. This is not exactly true, as the steam and the food will in general be under different pressures with the food at the lower pressure. Latent heats of vaporization are slightly higher at lower pressures, as shown in Table 7.1. In practice, there are also heat losses and sensible heat changes th at may require to be considered.

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TABLE 7.1 LATENT HEAT AND SATURATION TEMPERATURE OF WATER Absolute Latent heat of Saturation Pressure vaporization temperature (kPa) (kJkg – 1 ) ( o C) 1 2485 7 2 2460 1 8 5 2424 33 10 2393 46 20 2358 60 50 2305 81 100 2258 99.6 101.35 (1 atm) 2257 100 110 2251 102 120 2244 105 200 2202 120 500 2109 152 EXAMPLE 7.2. Heat energy in vacuum drying Using the same material as in Example 7.1, if vacuum drying is to be carried out at 60 o C under the corresponding saturation pressure of 20kPa abs. (or a vacuum of 81.4 kPa), calculate the heat energy required to remove the moisture per unit weight of raw material. Heat energy required per kg raw material = heat energy to raise temperature to 60 o C + latent heat of vaporization at 20 kPa abs. = (60 – 21) x 3.8 + 0.778 x 2358 = 148.2 + 1834.5 = 1983kJ . In freeze drying the latent heat of sublimation must be supplied. Pressure has little effect on the latent heat of sublimation, which can be taken as 2838 kJ kg – 1 . EXAMPLE 7.3. Heat energy in freeze drying If the food in the two previous examples was to be freeze dried at 0 o C, how much energy would be required per kg of raw material, sta rting from frozen food at 0 o C? Heat energy required per kg raw material = wt. of water vaporised x latent heat of sublimation = 0.778 x 2838 = 2208kJkg – 1 . Heat Transfer in Drying We have been discussing the heat energy requirements for the drying process. The rates of drying are generally determined by the rates at which heat energy can be transferred to the water or to the ice in order to provide the latent heats, though under some circumstances the rate of mass transfer (removal of the water) ca n be limiting. All three of the mechanisms by which heat is transferred – con duction, radiation and convection – may enter into drying. The

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relative importance of the mechanisms varies from one drying process to another and very often one mode of heat tra nsfer predominates to such an extent that it governs the overall process. As an example, in air drying the rate of heat transfer is given by: q = h s A ( T a T s ) (7.1) where q is the heat transfer rate in Js 1 , h s is the surface heat – transfer coe fficient Jm – 2 s – 1o C – 1 , A is the area through which heat flow is taking place, m 2 , T a is the air temperature and T s is the temperature of the surface which is drying, o C. To take another example, in a roller dryer where moist material is spread over the sur face of a heated drum, heat transfer occurs by conduction from the drum to the foodstuff, so that the equation is q = UA ( T d T s ) where U is the overall heat – transfer coefficient, T d is the drum temperature (usually very close to that of the steam), T s is the surface temperature of the food (boiling point of water or slightly above) and A is the area of drying surface on the drum. The value of U can be estimated from the conductivity of the drum material and of the layer of foodstuff. Values of U have been quoted as high as 1800 Jm – 2 s – 1o C – 1 under very good conditions and down to about 60 Jm – 2 s – 1 o C – 1 under poor conditions. In cases where substantial quantities of heat are transferred by radiation, it should be remembered that the surface temperature o f the food may be higher than the air temperature. Estimates of surface temperature can be made using the relationships developed for radiant heat transfer although the actual effect of combined radiation and evaporative cooling is complex. Convection coef ficients also can be estimated using the standard equations. For freeze drying, energy must be transferred to the surface at which sublimation occurs. However, it must be supplied at such a rate as not to increase the temperature at the drying surface abo ve the freezing point. In many applications of freeze drying, the heat transfer occurs mainly by conduction. As drying proceeds, the character of the heat transfer situation changes. Dry material begins to occupy the surface layers and conduction must tak e place through these dry surface layers which are poor heat conductors so that heat is transferred to the drying region progressively more slowly. Dryer Efficiencies Energy efficiency in drying is of obvious importance as energy consumption is such a l arge component of drying costs. Basically it is a simple ratio of the minimum energy needed to the energy actually consumed. But because of the complex relationships of the food, the water, and the drying medium which is often air, a number of efficiency m easures can be worked out, each appropriate to circumstances and therefore selectable to bring out special features

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Heat given up by air/100 kg potato = 1.0 x (80 – 71) x 49,800 x 1.06 = 4.75 x l0 5 kJ. The latent heat of steam at 70 kPa gauge(170Pa abs) is 2216 kJkg – 1 Heat in steam = 250 x 2216 = 5.54 x l0 5 kJ. Therefore (a) e fficiency based on latent heat of vaporisation only: = (1.81 x 10 5 )/ (5.54 x 10 5 ) = 33% (b) efficiency assuming sensible heat remaining in food after drying is unavailable = (1.9 7 x 10 5 )/ (5.54 x 10 5 ) = 36% (c) efficiency based heat input and output, in drying air (Eqn.7.2) = ( T 1 – T 2 )/( T 1 – T a ) = (80 71)/ (80 24) = 16% Whichever of these is chosen depends on the objective for cons idering efficiency. For example in a spray dryer, the efficiency calculated on the air temperatures shows clearly and emphatically the advantages gained by operating at the highest feasible air inlet temperature and the lowest air outlet temperatures that can be employed in the dryer. Examples of overall thermal efficiencies are: drum dryers 35 – 80% spray dryers 20 – 50% radiant dryers 30 – 40% After sufficient energy has been provided to vaporize or to sublime moisture from the food, some way must be found to remove this moisture. In freeze – drying and vacuum systems it is normally convenient to condense the water to a liquid or a solid and then the vacuum pumps have to handle only the non – condensible gases. In atmospheric drying a current of air is nor mally used. MASS TRANSFER IN DRYING In heat transfer, heat energy is transferred under the driving force provided by a temperature difference, and the rate of heat transfer is proportional to the potential (temperature) difference and to the properties of the transfer system characterized by the heat – transfer coefficient. In the same way, mass is transferred under the driving force provided by a partial pressure or concentration difference. The rate of mass transfer is proportional to the potential (pre ssure or concentration) difference and to the properties of the transfer system characterized by a mass – transfer coefficient. Writing these symbolically, analogous to q = UA T , we have dw = k g A Y (7.3) dt

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where d w is the mass (mois ture) being transferred kgs – 1 in time dt, A is the area through which the transfer is taking place , k g ‘ is the mass transfer coefficient in this case in units kgm – 2 s – 1 , and Y is the humidity difference in kgkg – 1 . Unfortunately the application of mass tran sfer is not as straightforward as heat transfer. One reason is that the movement pattern of moisture changes as drying proceeds. Initially, the mass (moisture) is transferred from the surface of the material and later, to an increasing extent, from deeper within the food to the surface and thence to the air. So the first stage is to determine the relationships between the moist surface and the ambient air and then to consider the diffusion through the food. In studying the surface/air relationships, it is n ecessary to consider mass and heat transfer simultaneously. Air for drying is usually heated and it is also a major heat transfer medium. Therefore it is necessary to look carefully into the relationships between air and the moisture it contains. PSYCHR OMETRY The capacity of air for moisture removal depends on its humidity and its temperature. The study of relationships between air and its associated water is called psychrometry. Humidity (Y) is the measure of the water content of the air. The absolut e humidity is the mass of water vapour per unit mass of dry air and the units are therefore kg kg – 1 . Absolute humidity in charts. Air is said to be saturat ed with water vapour at a given temperature and pressure if its humidity is a maximum under these conditions. If further water is added to saturated air, it must appear as liquid water in the form of a mist or droplets. Under conditions of saturation, the partial pressure of the water vapour in the air is equal to the saturation vapour pressure of water at that temperature. The total pressure of a gaseous mixture, such as air and water vapour, is made up from the sum of the pressures of its constituents, w hich are called the partial pressures. Each partial pressure arises from the molecular concentration of the constituent and the pressure exerted by each gas is that which corresponds to the number of moles present and the total volume of the system. The pa rtial pressures are added to obtain the total pressure. EXAMPLE 7.5. Partial pressure of water vapour If the total pressure of moist air is 100kPa (approximately atmospheric) and the humidity is measured as 0.03kg kg – 1 , calculate the partial pressure of the water vapour. The mole fraction of the water is the number of moles of water to the total number of moles (water + dry air) The molecular weight of air is 29, and of water 18 So the mole fraction of water = 0.03 /18 (1.00/29 + 0.03/18) = 0.0017 /(0.034 + 0.0017) = 0.048 Therefore the water vapour pressure = 0.048 x l00kPa = 4.8kPa .

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The relative humidity (RH) is defined as the ratio of the partial pressure of the water vapour in the air ( p ) to the partial pressure of saturated water vap our at the same temperature ( p s ). Therefore: RH = p / p s and is often expressed as a percentage = 100 p / p s EXAMPLE 7.6. Relative humidity If the air in Example 7.5 is at 60 o C, calculate the relative humidity. From steam tables, the saturation pres sure of water vapour at 60 o C is 19.9 kPa. Therefore the relative humidity = p / p s = 4.8/19.9 = 0.24 or 24% . If such air were cooled, then when the percentage relative humidity reached 100% the air would be saturated and this would occur a t that temperature at which p = p s = 4.8 kPa. Interpolating from the steam tables, or reading from the water vapour pressure/temperature graph, this occurs at a temperature of 32 o C and this temperature is called the dew – point of the air at this particula r moisture content. If cooled below the dew – point, the air can no longer retain this quantity of water as vapour and so water must condense out as droplets or a fog, and the water remaining as vapour in the air will be that corresponding to saturation at t he temperature reached. The humidity Y can therefore be related to the partial pressure p w of the water in air vapour by the equation: Y = l8 p w /[29 (P p w )] (7.4) where P is the total pressure. In circumstances where p w is small compared with P, and this is approximately the case in air/water systems at room temperatures, Y 18 p w / 29 P. Corresponding to the specific heat capacity, c p , of gases, is the humid heat, c s of moist air. It is used in the same way as a specific heat capacity, the entha lpy change being the mass of dry air multiplied by the temperature difference and by the humid heat. The units are Jkg – 1o C and the numerical values can be read off a psychrometric chart. It differs from specific heat capacity at constant pressure in that i t is based only on the mass of the dry air. The specific heat of the water it contains is effectively incorporated into the humid heat which therefore is numerically a little larger than the specific heat capacity to allow for this. Wet – bulb Temperatures A useful concept in psychrometry is the wet – bulb temperature, as compared with the ordinary temperature, which is called the dry – bulb tempera ture. The wet – bulb temperature is the temperature reached by a water surface, such as that registered by a therm ometer bulb surrounded by a wet wick, when exposed to air passing over it. The wick and therefore the

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thermometer bulb decreases in temperature below the dry – bulb temperature, until the rate of heat transfer from the warmer air to the wick is just equal to the rate of heat transfer needed to provide for the evaporation of water from the wick into the air stream. Equating these two rates of heat transfer gives h c A ( T a T s ) = k g ‘A ( Y s Y a ) where a and s denote actual and saturation temperatures and hum idities; h c is the heat transfer coefficient and k g ‘ the mass transfer coefficient from the air to the wick surface; is the latent heat of evaporation of water. Therefore h c /k g ‘ = ( Y s Y a )/ ( T a T s ) As the relative humidity of the air decreases, so the difference between the wet – bulb and dry – bulb temperatures, called the wet – bulb depression, increases and a line connecting wet – bulb tempera ture and relative humidity can be plotted on a suitable chart. When the air is saturated, the wet – bulb temper ature and the dry – bulb temperature are identical. Therefore if ( T a T s ) is plotted against ( Y s Y a ) remembering that the point ( T s , Y s ) must correspond to a dew – point condition, we then have a wet – bulb straight line on a temperature/humidity chart slopi ng down from the point ( T s , Y s ) with a slope of: – ( k g ‘/h c ) A further important concept is that of the adiabatic saturation condition. This is the situation reached by a stream of water, in contact with the humid air. Both ultimately reach a tempera ture at which the heat lost by the humid air on cooling is equal to the heat of evaporation of the water leaving the stream of water by evaporation. Under this condition with no heat exchange to the surroundings, the total enthalpy change (kJkg – 1 dry air) H = c s ( T a T s ) + ( Y s Y a ) = 0 c s = – ( Y s Y a )/ ( T a T s ) = – h c /k g ‘ where c s is the humid heat of the air. Now it just so happens, for the water/air system at normal working temperatures and pressures that for practical purposes t he numerical magnitude of the ratio: h c (known as the Lewis number) = 1 (7.5) c s k g This has a useful practical consequence. The wet bulb line and the adiabatic saturation line coincide when the Lewis number = 1. It is now time to examin e the chart we have spoken about. It is called a psychrometric chart.

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