Drying and Dehydration

Drying and Dehydration Abstract. This chapter reviews basic concepts of drying and dehydration, including ass balance analyses. Equilibriu oisture content, water activity, and related paraeters are discussed. Drying ethods and drier types are briefly discussed. Keywords. Dehydration, driers, drying, equilibriu oisture content, evaporation, water activity. 10.1 Introduction Reoving water fro food and agricultural products constitutes a significant portion of the processing activity for persons working in the food and agricultural processing industries. Two ajor oisture reoval ethods are drying (or dehydration) to produce a solid product and evaporation to produce a ore concentrated liquid. The words drying and dehydration are often used interchangeably, especially when referring to food products; however, only the word drying is coonly used when referring to processing of non-food products. Applications range fro on-far drying of grain, fruits, and vegetables to large scale coercial drying of fruits, vegetables, snack food products, ilk products, coffee, and other products. Although certain basic factors are involved in all drying processes, the equipent and techniques vary greatly depending upon the product and other factors. In this unit, we will consider soe basic factors affecting drying and briefly exaine soe drying ethods. Evaporation is the reoval of soe water fro a liquid product to produce a ore concentrated liquid. Applications include concentration of ilk, fruit juices, and syrup products. Most evaporation systes are large-scale coercial operations, although sall-scale far operations still exist for production of aple, sorghu, and sugar cane syrups. The governing principles of evaporator operation will be briefly exained in this unit. 10.2 Moisture Content No agricultural product in its natural state is copletely dry. Soe water is always present. This oisture is usually indicated as a percent oisture content for the product. Two ethods are used to express this oisture content. These ethods are wet basis () and dry basis (M). In addition, the content ay be expressed as a percent or as a decial ratio. We will use all four fors (wet basis, dry basis, percent, and decial ratio) in analyzing oisture or food products. The general governing equations for indicating oisture content are:

260 Food & Process Engineering Technology = w w = (10.01) w + d t w M = (10.02) where: = decial oisture content wet basis (wb) M = decial oisture content dry basis (db) d = ass of dry atter in the product w = ass of water in the product t = total ass of the product, water plus dry atter The percent oisture content is found by ultiplying the decial oisture content by 100. In addition, relationships between wet and dry oisture content on a decial basis can be derived fro Equations 10.01 and 10.02. Those relationships are: M M = or = (10.03) 1 1+ M Use of the wet basis easureent is coon in the grain industry where oisture content is typically expressed as percent wet basis. However, use of the wet basis has one clear disadvantage the total ass changes as oisture is reoved. Since the total ass is the reference base for the oisture content, the reference condition is changing as the oisture content changes. On the other hand, the aount of dry atter does not change. Thus, the reference condition for dry basis easureents does not change as oisture is reoved. For a given product, the oisture content dry basis is always higher than the wet basis oisture content. This is obvious fro a coparison of Equations 10.01 and 10.02. The difference between the two bases is sall at low oisture levels, but it increases rapidly at higher oisture levels. A final note regarding oisture content relates to high oisture aterials such as fruits and vegetables. Many of these products have oisture contents near 0.90 (or 90%) (wb). On a dry basis this would be 900% if expressed as a percentage. For products of this type, oisture is often given as ass of water per unit ass of dry product, the decial basis we discussed earlier. d Exaple 10.1 A bin holds 2000 kg of wet grain containing 500 kg of water. This grain is to be dried to a final oisture content of 14% (wb). a. What are the initial and final oisture contents of the grain (wet basis, dry basis, decial and percent)? b. How uch water is reoved during drying?

Chapter 10 Drying and Dehydration 261 Solution: Using Equations 10.01 and 10.02 the initial oisture content is: 500 = w = = 0.25 or 25% wet basis (wb) ANSWER t 2000 w 500 M = = = 0.3333 or 33.33 % dry basis (db) ANSWER d 2000 500 The final oisture content was stated as 14% (wb), or = 0.14. Using Equation 10.03, we can solve for the oisture content on a dry basis: ANSWER 0. 14 0. 14 M = = = 0. 1628 or 16. 28% (db) 1 0. 14 0. 86 The aount of water in the dried grain ay be found using either Equation 10.01 or 10.02 and the appropriate final oisture content noted above. For our exaple, we will use both ethods. We will first use the dry basis oisture of M = 0.1628. Noting that the bin contains 1500 kg of dry atter (2000 500), Equation 10.02 gives: w = d M = 1500 0.1628 = 244 kg Alternatively, using Equation 10.01 and = 0.14, we find the sae solution: ( + ) = 0.14 ( + 1500) = 0.14 + 210 w = w d w w w = 210 w 244 kg 0.86 = Thus, the water reoved is: = 500 244 = 256 kg ANSWER w re Many different techniques are available for easuring the oisture content of a aterial. The technique used in a given instance depends upon the aterial being studied, equipent available, and the tie available for the easureent. The ost straightforward ethod of oisture easureent is to use a drying oven. A saple of the product is heated at a specified teperature and pressure (usually atospheric pressure or a specified vacuu) for a specified tie to reove all oisture (i.e., dry until there is no further weight loss). The loss in ass of the saple represents the oisture reoved fro the product. The teperature, drying tie, and pressure are dependent upon the product being analyzed. Microwave drying ovens and cheical analysis are also used for soe oisture easureent applications. An extensive list of standards for oisture easureent is provided by AOAC (1990). 10.3 Equilibriu Moisture Content A aterial held for a long tie at a fixed teperature and relative huidity will eventually reach a oisture content that is in equilibriu with the surrounding air. This does not ean that the aterial and the air have the sae oisture content. It siply eans that an equilibriu condition exists such that there is no net exchange of oisture between the aterial and the air. This equilibriu oisture content (EMC or M e ) is a function of the teperature, the relative huidity, and the product. These equilibriu oisture relationships are norally expressed atheatically. Nuerous

262 Food & Process Engineering Technology equations have been proposed to represent the EMC curves for various products (Iglesias and Chirfe, 1982; ASAE, 2000). No single equation is suitable for all products; however, ost products can be represented by one of several equations available. Four of these equations are listed below. Halsey s equation (Equation 10.04) and Henderson s equation (Equation 10.05) are two of the less coplicated equations. Teperature is not a paraeter in Halsey s equation. Thus, different constants ust be used for each product and teperature of interest. Note also that the constants are valid only for the equation listed and should not be adapted for other equations. Grains and related products are often represented by the slightly ore coplicated Modified Henderson Equation (Equation 10.06), the Modified Halsey Equation (Equation 10.07), the Modified Oswin Equation, or the Guggenhei-Anderson-DeBoer (GAB) equation (ASAE, 2000). The relative huidity defined by these equations is coonly called the equilibriu relative huidity (ERH). Thus, ERH is the relative huidity for equilibriu between air and a specific product at a given teperature. K ERH = exp Halsey (10.04) N M e N ERH = 1 exp( K t ) Henderson (10.05) M e N ( K( t C ) ERH = 1 exp + ) Modified Henderson (10.06) M e exp( K + C t) ERH = exp Modified Halsey (10.07) N M e where: ERH = relative huidity, decial M e = equilibriu oisture content, percent, dry basis t = teperature, C K, N, C = are constants deterined for each aterial (see Tables 10.01 and 10.02) A typical set of equilibriu oisture curves are presented in Figure 10.01. These curves are coputed using Equation 10.06 and the constants fro Table 10.01 for shelled corn. Note that curves begin at 20% relative huidity and are terinated before reaching 100% relative huidity. Prediction curves are generally written to describe conditions in the iddle ranges of relative huidity. They do not predict well for extree conditions. In addition, reliable experiental data are difficult to obtain in those regions.

Chapter 10 Drying and Dehydration 263 Table 10.01. Constants for Modified Henderson and Halsey Equations. (Fro ASAE, 2000.) Grain K N C Equation Beans, pinto 4.4181 1.7571 0.011875 10.07 Beans, white 0.1633 1.567 87.46 10.06 Canola eal 0.000103 1.6129 89.99 10.06 Corn, shelled 6.6612 10-5 1.9677 42.143 10.06 Popcorn 1.5593 10-4 1.5978 60.754 10.06 Peanut, kernel 3.9916 2.2375 0.017856 10.07 Pupkin seed, adsorption* 3.3725 10-5 3.4174 1728.729 10.06 Pupkin seed, desorption* 3.3045 10-5 3.3645 1697.76 10.06 Rice, ed grain 3.5502 10-5 2.31 27.396 10.06 Soybean 2.87 1.38 0.0054 10.07 Wheat, hard red *kernels 4.3295 10-5 2.1119 41.565 10.06 Table 10.02. Selected EMC relationships for food products. Constants are valid only for the equation nuber listed. (Fro Iglesias and Chirfe, 1982.) Product Teperature rh range Equation K N Apple 30 C 0.10-0.75 10.05 0.1091 0.7535 Apple [a] 19.5 C 0.10-0.70 10.04 4.4751 0.7131 Banana 25 C 0.10-0.80 10.05 0.1268 0.7032 Chives 25 C 0.10-0.80 10.04 11.8931 1.1146 Grapefruit 45 C 0.10-0.80 10.05 0.1519 0.6645 Mushroos 20 C 0.07-0.75 10.04 7.5335 1.1639 Mushroos [a] 25 C 0.10-0.80 10.04 11.5342 1.1606 Peach 20-30 C 0.10-0.80 10.05 0.0471 1.0096 Peach 40 C 0.10-0.80 10.05 0.0440 1.1909 Peach 50 C 0.10-0.80 10.05 0.0477 1.3371 Pear 25 C 0.10-0.80 10.05 0.0882 0.7654 Toato 17 C 0.10-0.80 10.04 10.587 0.9704 [a] Desorption isother data. All other data is for sorption (oisture addition) easureents.

264 Food & Process Engineering Technology Equilibriu Moisture (% db) 35 30 25 20 15 10 5 0 0 20 40 60 80 100 Relative Huidity (%) 0 C 10 C 30 C 50 C Figure 10.01. Equilibriu oisture curves for shelled corn. Coputed fro ASAE data (ASAE, 2000). Exaple 10.2 Copute the equilibriu oisture content for popcorn at 20 C and 50% relative huidity. Solution: We first rearrange Equation 10.06 to express equilibriu oisture in ters of the other paraeters: ln (1 ERH) ln(1 ERH ) M e = N = K( t + C) K( t + C) We now substitute values of K, N, and C fro Table 10.01 into the equation and solve for M e at 20 C and 50% relative huidity: M e ln = = 1. 5593 0. 6931 = 0. 01259 ( 1 0. 5) 4 10 ( 20 + 60. 754) 0. 6258 = 1 1. 5978 ( 0. 5) ) 0. 62586 ( 55. 047) = 12. 19% (db) ANSWER 1 N ln = 0. 01259 0. 62586

Chapter 10 Drying and Dehydration 265 10.4 Water Activity (a w ) The aount of water in food and agricultural products affects the quality and perishability of these products. However, perishability is not directly related to oisture content. In fact, perishability varies greatly aong products with the sae oisture content. A uch better indicator of perishability is the availability of water in the product to support degradation activities such as icrobial action. The ter water activity is widely used in the food industry as an indicator of water available in a product. See Labuza (1984) or Troller and Christian (1978) for further inforation. Water activity (a w ) is defined as: P = φ φ w a w or aw = since = Pws where: φ = the relative huidity, decial P w = the partial pressure of water vapor at the specified conditions P ws = the partial pressure of water vapor at saturation and the teperature specified Thus, water activity is the equilibriu relative huidity (ERH) in decial for for a product at a given teperature and oisture content. Figure 10.01 can then be considered as a plot of equilibriu oisture content as a function of a w, if we change the x-axis scale to decial for. A ore coon ethod for representing the effect of a w is shown in Figure 10.02. The shape of this graph is typical of water activity graphs although the oisture content scale ay vary greatly aong products. The key feature of this relationship is that actual oisture content increases rapidly with a w at the higher values of water activity. 4.0 3.5 P P w ws X (kg water/kg ds) 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 Water Activity (a w ) Figure 10.02. Water activity curve for fresh, diced sweet potato cubes.

266 Food & Process Engineering Technology The physical significance of a w ay not be iediately clear fro the above equation. Thus, we will exaine a conceptually siple ethod of deterining a w. If we take a saple of a food product and place it in an enclosed container at a fixed teperature, the product will exchange oisture with the air surrounding it. After a period of tie, as with the equilibriu oisture exaple noted in the previous section, an equilibriu condition will occur. The product no longer has any net change in oisture. The water activity is equal to the decial relative huidity at that condition. 10.5 Controlling Factors for Drying Two separate phenoena are involved in drying. First, oisture ust ove fro the interior of a aterial to the surface of that aterial. Second, the surface water ust be evaporated into the air. These two steps involve two very different phenoena. Moveent of water fro the interior to the surface ust occur in one of two anners capillary action or diffusion. Moveent by capillary action would only occur during early stages of drying. As the drying process continues, internal oisture oveent would occur by olecular diffusion of water vapor within the aterial. Reoval of water fro the surface involves evaporation of water fro the surface into the surrounding air. The evaporation rate depends upon the condition of drying air and the concentration of water at the surface. Air drying involves the passing of air over the object(s) to be dried. Typically, the air is heated prior to entering the drying region. Consider the drying process for a high oisture product such as an apple slice. The surface of the slice will be visibly covered with water iediately after slicing. As this water evaporates, the surface becoes slightly dry. Moisture cannot ove fro the interior of the slice as rapidly as it can evaporate at the surface. Thus the governing factor in later stages of drying is the diffusion rate of oisture within the slice. Factors affecting the drying rate will vary slightly depending upon the type of drying syste used. However, in general, the following factors ust be considered: 1. nature of the aterial: physical and cheical coposition, oisture content, etc.; 2. size, shape, and arrangeent of the pieces to be dried; 3. wet-bulb depression (t t wb ), or relative huidity, or partial pressure of water vapor in the air (all are related and indicate the aount of oisture already in the air); 4. air teperature; and 5. air velocity (drying rate is approxiately proportional to u 0.8 ). Another factor that ust be considered in drying solid aterials is case hardening. This proble can occur if the initial stage of drying occurs at low relative huidity and high teperature. Under these conditions, oisture is reoved fro the surface of the aterial uch faster than it can diffuse fro within the aterial. The result is foration of a hardened relatively ipervious layer on the surface of the aterial. Foration of such a layer causes subsequent drying to be uch slower than it would otherwise be.

Chapter 10 Drying and Dehydration 267 10.6 Air Drying Methods Many different drying systes have been developed to eet the needs for drying different aterials. A few of the ore coon systes are described in the following sections. 10.6.1 Bin Drying Bin drying systes are coon in on-far grain drying operations. The bin is filled with grain and drying air is forced up through the grain fro a plenu chaber beneath the perforated floor of the bin. The grain on the botto is dried first. As drying progresses a layer of drying grain separates the dried grain fro the undried grain. This region of drying grain is called a drying front. This drying front progresses upward through the bin of grain until all grain is dried. Figure 10.03 shows a representation of this process. Other on-far drying ethods include stirring of grain in the bin during drying and use of continuous flow dryers to dry grain before storage. The arrangeent shown in Figure 10.03 is very useful for studying drying systes since it can be readily analyzed for ass and energy balances. Undried Product Drying Front Dried Product Perforated Floor Fan Figure 10.03. Bin drying diagra showing the drying front with about half the product dried. 10.6.2 Cabinet Drying Cabinet dryers are usually sall, insulated units with a heater, circulating fan, and shelves to hold the product to be dried. The sall dehydration units sold for hoe use are sall-scale exaples of this type of dryer. Different designs are used, but the general procedure is to force heated air over ultiple trays. Sall-scale cabinet dryers are typically single pass units. However, greater energy efficiencies can be obtained if soe of the heated air is recirculated. This is especially true in later stages of drying when the oisture reoval rate is low and the exit air retains considerable oistureholding capacity. Figure 10.04 shows the basic operation of a cabinet dryer with recirculation. Energy savings of 50% or ore can be achieved with recirculation.

268 Food & Process Engineering Technology Figure 10.04. Sketch of a cabinet dryer with recirculation. (Food Engineering Fundaentals, J. C. Batty and S. L. Folkan, 1983. by John Wiley & Sons, Inc. Reprinted by perission of John Wiley & Sons, Inc. ) 10.6.3 Tunnel Drying Tunnel dryers are a large-scale odification of the cabinet dryer concept. The drying chaber is a tunnel with ultiple carts containing trays of the product being dried. New carts of undried product are loaded at one end of the tunnel as carts of dried product are reoved fro the other end. Air flow in these dryers ay be parallel or counter to the oveent of carts in the tunnel. Figure 10.05 shows exaples of parallel and counter flow tunnel dryers. 10.6.4 Dru Drying Large rotating drus are used for drying slurries (liquids with a high solids content). A thin fil of the slurry is deposited on the botto of a rotating dru as it passes through the slurry. The slowly rotating dru is heated and soeties held under a vacuu. The dried product is scraped fro the dru before the rotating surface reenters the slurry. Figure 10.06 shows four different configurations of dru dryers.

Chapter 10 Drying and Dehydration 269 Figure 10.05. Tunnel dryer sketches showing (a) parallel and (b) counter flow operation (fro Brennan et al. 1990). Figure 10.06. Different dru dryer configurations (fro Brennan et al. 1990).

270 Food & Process Engineering Technology 10.6.5 Other Drying Nuerous other air drying techniques are also used. The ajor function of all such systes is to ove air over the product in such a anner that the product is dried as econoically as possible without daage. 10.7 Special Drying Methods 10.7.1 Spray Drying Spray drying is used to dry liquid products. The product to be dried is sprayed into a strea of heated air. Water evaporates into the air leaving the dry particles to be collected. The two ajor operations of concern in spray drying are droplet atoization and powder collection. To optiize drying, droplets should be sall and unifor in size. Thus special procedures ust be used to insure that atoization is satisfactory. Figure 10.07. Spray dryer illustration. (Food Engineering Fundaentals, J. C. Batty and S. L. Folkan, 1983. by John Wiley & Sons, Inc. Reprinted by perission of John Wiley & Sons, Inc. )

Chapter 10 Drying and Dehydration 271 Collection of the dried powder also requires special techniques. The powder particles are sall and ove easily within an air strea. Collection chabers or special filters are norally used. Figure 10.07 shows a siple representation of a spray dryer. Actual operations are ore coplex, and any different configurations are available. 10.7.2 Vacuu Drying In vacuu drying, the product is placed inside a chaber where the pressure is reduced to produce a vacuu. Since the total pressure in the chaber is very low, the partial pressure of the water vapor in the chaber is also very low. This low partial pressure causes a large partial pressure difference between the water in the product and the surroundings. Thus water oves ore readily fro the product to the surrounding environent in the chaber. Drying under vacuu conditions perits drying at a lower teperature. This characteristic of vacuu drying is very iportant for products that ay suffer significant flavor changes at higher teperatures. 10.7.3 Freeze Drying Freeze drying involves the reoval of oisture fro a frozen product without thawing that product. The teperature ust be below freezing for that product (to insure that the product reains frozen) and the vapor pressure ust be aintained at a very low level to perit oisture reoval by subliation. Because of the low teperature, low pressure, and low drying rate, freeze drying is quite expensive copared to any other drying ethods. However, freeze drying can produce high quality dried products. Thus, it is the preferred drying ethod for soe high value aterials. Figure 10.08 shows a representation of a freeze dryer. Freezer burn, soeties seen in frozen foods, is an exaple of undesirable freeze drying. This very slow type of freeze drying can occur when inadequately protected food is stored in a freezer for extended periods. Figure 10.08. Representation of freeze drying coponents. Low teperature and pressure are required. (Fro Brennan et al. 1990.)

272 Food & Process Engineering Technology Exaple 10.3 A counter flow dryer unit uses heated air to dry apple slices. The slices enter at a rate of 200 kg/h and a oisture content of 1 = 0.9. The dried slices have a oisture content of M 2 = 0.10. The drying air enters at 50 C and exits at 25 C and 90% relative huidity. (1) Find the water reoved, w re, kg/h. (2) Find the entering air flow rate, 3 /in. Solution: Analysis of this proble requires calculation of ass balances for dry atter, water, and dry air. We begin by calculating the aount of dry atter (solids) and water entering and leaving with the slices. Since we know the oisture content and the input rate, we can deterine the aount of water and dry atter in the entering slices: w1 w1 1 = 0.9 = = giving w1 = 0.90 = 200 T 1 kg H2O 200 180 h Next solve for the entering dry atter. Since only water is reoved, the dry atter in the dried product is the sae as that entering in the undried slices: kg DM D1 = T1 w1 = 200 180 = 20 = D2 h Knowing the oisture content and the aount of dry atter in the dried slices, we can now calculate the output rate of the product: w2 w2 kg H2O M 2 = = 0.10 = giving w2 = 20 0.10 = 2.0 20 h D2 The water reoved fro the apple slices is then: kg H 2O wre = w1 w2 = 180 2 = 178 ANSWER 1 h We know that the drying air follows a constant wet-bulb process as it gains oisture fro the apple sliced. Using the teperature and huidity of the exit air, we can go to a psychroetric chart, find the wet-bulb teperature, and follow the constant wet-bulb line to the inlet air teperature. Using this process we find the following air properties: Location t db ( C) t wb ( C) φ (%) v ( 3 /kg DA) W (g H 2 O/kg DA) Entering 50 23.7 0.925 7.7 Exiting 25 23.7 90 18

Chapter 10 Drying and Dehydration 273 For use in our solution, we need to convert the huidity ratio to units of kg H 2 O/kg DA (divide the table values by 1000). We can now calculate the oisture gained by each unit ass of dry air: W kg H2O = W1 W2 = 0.018 0.007 = 0.0103 kg DA Knowing the total water gain by the air and the gain by each kilogra of the air, we can now calculate the air flow rate: kg H2O 178 h DA = = 17 280 kg H2O 0.0103 kg DA kg DA h = 288 kg DA in Using the calculated ass flow rate and the entering air specific volue fro the psychroetric chart, and listed in the table above, we can deterine the voluetric air flow rate: 3 3 kg DA Q = 288 0. 925 = 266 ANSWER 2 in kg DA in Exaple 10.4 Repeat Exaple 10.3 with all paraeter values the sae except that the dried slices exit at a oisture content of 2 = 0.10 (wet basis instead of dry basis). Solution: We follow a solution process very siilar to that of the previous exaple. Input paraeters are exactly the sae as before. Thus: kg DM 1 = D D2 = 200 180 = 20 h Solving for the water in the dried slices: 2 w2 = 0.10 = T 2 = w2 w2 + D2 w2 = w2 + 20 kg H2O 0.1w2 + 2 = w2 w2 = 2.22 h The oisture reoved is then:

274 Food & Process Engineering Technology kg H 2O w re = w1 w2 = 180 2. 22 = 177. 78 ANSWER h Continuing as in the previous exaple, we find: kg DA 3 DA = 287. 7 and Q = 266. 1 ANSWER in in Note that the results differ little between these two exaples. That is because, at low oisture content, there is little difference between the wet basis and dry basis easureents. The difference would be uch greater if an exit oisture content of 0.20 had been used in the exaples. 10.8 Introduction to Evaporation An iportant step in the production of any food products is the reoval of water to produce a ore concentrated liquid. The ter used in the food industry to define this operation is evaporation. Exaples of evaporated food products are evaporated ilk, syrups, and various fruit juice concentrates. The basic principle involved in the evaporation process is the application of heat to evaporate free water present in the product. The configuration of the equipent used to accoplish this concentration defines the evaporator type. 10.8.1 Open (Atospheric) and Vacuu Evaporators The earliest evaporators were open kettles or pans placed over an open flae. Maple syrup, sweet sorghu syrup, and sugar cane syrup were produced using open evaporators during the early history of the United States. Figure 10.09 shows a sketch of an open pan evaporator used for sorghu syrup evaporation during the early 1900s (Walton et al., 1938). Evaporators very siilar to that shown in Figure 10.09 continue to be used by any sorghu syrup producers in the United States. The priary difference is that few processors use wood as a heat source. Most now use gas (natural or LP) or stea as the direct heat source for their evaporators. Two ajor disadvantages of open evaporators are higher operating teperatures and higher energy cost in coparison to vacuu evaporation systes. While acceptable for syrup production, the teperatures required to boil a product, thus evaporating water, are sufficient to produce undesirable flavors for any food products. In addition, such evaporation requires huge aounts of energy ore than 2200 kj for every kilogra of water evaporated. The probles noted above for open evaporators can be avoided by using vacuu evaporators. By evaporating at a lower pressure, the boiling point is lowered. This reduces the possibility of undesirable flavors. In addition, vacuu evaporators can be placed in series. These ultiple effect evaporators use the vapor fro one evaporator to provide energy for evaporation in the next evaporator in the series. This cobination results in a significant reduction in energy use, although initial cost is greater.

Chapter 10 Drying and Dehydration 275 Figure 10.09. Open pan evaporator used for sorghu syrup evaporation (odified fro Walton et al., 1938). Exaple 10.5 A sall open pan evaporator is used to produce sorghu syrup. Juice ( = 0.87 wb) enters at the rate of 800 kg/h. The finished syrup leaving the evaporator contains 80% solids. Five percent of the entering solids are reoved as ipurities during the evaporation process. (1) Find the production rate of the syrup (kg/h). (2) Find the ass of water reoved per unit ass of syrup. (3) Find the energy required per hour to evaporate the water reoved. (4) Find the stea flow required for evaporation if 20% of the heat is lost to surroundings. The stea enters at 350 kpa (saturated) and exits at a quality of 0.10. Solution: As with dryer analysis, the solution requires ass and energy balances. However, one unique difference fro previous exaples is that the dry atter ass balance ust account for the solids reoved. For the solids balance: Solids (dry atter) entering ( DI ):

276 Food & Process Engineering Technology w1 = 0.87 800 = 696 kg H 2 O/h D1 = T1 w1 = 800 696 = 104 kg DM/h (or D1 = 0.13 800 = 104) Ipurities reoved are 5% of the entering solids, thus: 1 = 0.05 D1 = 0.05 104 = 5.2 kg DM/h The solids reaining in the product are: D2 = D1 1 = 104 5.2 = 98.8 kg DM/h or D2 = D1 0.5 D1 = 0.95 D1 = 0.95 (104) = 98.8 We can now deterine the syrup output. D2 For 80% solids we can write the relationship as 0.80 = T 2 D2 98. 8 kg product Thus, T 2 = = = 123. 5 ANSWER 1 0. 80 0. 80 h The aount of water in the product can now be deterined as: kg H2O w2 = T 2 D2 = 123.5 98.8 = 24.7 h Knowing the water input ( w1 ) and water output ( w2 ), we can now deterine the water reoved: kg H2O w re = w1 w2 = 696 24.7 = 671.3 h Converting to water reoved per unit ass of syrup, we have: w re 671. 3 kg H 2O = = 5. 44 ANSWER 2 123. 5 kg syrup T 2 We can use the latent heat of the evaporated water to deterine the energy needed for evaporation. Assuing an average evaporation teperature of 105 C we find that h fg = 2244 kj/kg. Thus, Q = 2244 kj/kg 671.3 kg/h = 1 506 400 kj/h = 418.4 kw ANSWER 3 We can now copute the required stea flow. The energy output per unit ass (kg) of stea is: h s = h g [h f + x h fg ] = 2731.6 [584.3 + 0.1 2147.7] = 2731.6 799. = 1932.6 kj/kg stea Knowing the energy required for evaporation ( Q ) and the energy per unit ass

Chapter 10 Drying and Dehydration 277 of stea ( h s ), we can solve for the required stea flow s hs s = s : s[ 1932.6 kj/kg stea] = Q = 1 506 400 kj/h = 1 506 400 /1932.6 = 779.5 kg stea/h (with no losses) Taking the 20% loss into account the above value represents only 80% of the required flow rate. Thus the required total stea flow rate is: st = s 779. 5 = = 0. 80 0. 80 974. 3 kg stea h ANSWER 4 10.8.2 Controlling Factors for Evaporation Analysis of evaporator systes is relatively coplex. In addition to the noral ass and energy balances that ust be satisfied, other therodynaic paraeters ust also be considered. These include boiling point and latent heat of vaporization of the raw and finished products. The boiling point of any liquid increases as pressure increases. Thus, vacuu evaporators, which operate below atospheric pressure, require lower operating teperatures than those required for the sae product in an open evaporator. The presence of sugars and/or other coponents in the liquid being concentrated increases the boiling point of that aterial at any given pressure. As the concentration of the liquid increases, so does the boiling point. This increase in boiling point is coonly referred to as boiling point rise. It is a function of the aount and type of constituents in the liquid. The cobination of evaporator pressure and boiling point rise ust be considered in deterining the actual boiling point in the evaporator. The latent heat of vaporization (h fg ) is the energy required to evaporate a unit ass of liquid at its boiling point. The value of h fg for water at atospheric pressure is 2257 kj/kg; however, it does not reain constant as the boiling point changes. It decreases slightly as the boiling point increases. At any evaporator pressure (and boiling point), h fg is quite large. Thus, substantial energy is required for evaporation processes. Because h fg of food products is large, evaporators require substantial aounts of energy input. This energy ust be transferred fro the energy source (coonly stea) to the liquid being evaporated. Thus, the heat transfer rate ust be as great as possible in the heat exchanger area of the evaporator. This is a ajor consideration in evaporator design. In addition, foaing of the product and fouling of the heat exchanger surface can have significant adverse effects upon the heat transfer rate. Foaing occurs during concentration of soe food products. This can substantially reduce the surface heat transfer coefficient (and the rate of heat transfer fro the heat exchanger surface to the fluid.) Fouling of the heat transfer surface can create added resistance to heat flow, thus reducing the heat transfer rate. Properties of raw aterial and finished product are ajor factors that enter into evaporator design and selection. Maxiu teperature and length of tie exposed to that teperature are significant in deterining product quality for soe products. Evaporators ust therefore operate within specified teperature liitations and yet

278 Food & Process Engineering Technology allow for rapid evaporation. Viscosity is another iportant paraeter since ore viscous fluids tend to have higher boiling points and lower heat transfer rates. 10.8.3 Multiple Effect Evaporators To reduce overall energy costs, evaporators ay be placed in series such that the vapor fro one evaporator is the heat source to evaporate water fro the next unit in the series. Figure 10.10 shows how evaporators could be connected to provide a ultiple effect syste. In this syste stea for evaporation is required for the first unit only. The latent heat of vaporization of the condensing stea provides energy to evaporate water in the first stage of the evaporator. Vapor fro this stage is condensed in the next effect to vaporize the water there. This continues through all effects. Because we recycle the vapor into subsequent evaporator stages, we could ake the following gross estiate of energy requireents: in a ultiple effect syste, one kilogra of stea input will evaporate as any kilogras of water as there are effects. This would iply that for a triple effect syste, one kilogra of stea would produce three kilogras of vapor. This siplification is an aid in showing the advantage of ultiple effects, but it gives a isleading (high) indication of the vapor produced. Aside fro inefficiencies and heat losses, the latent heat of vaporization of water is not constant. It increases as the boiling point decreases. Thus, at every effect in the syste, less vapor is evaporated in the current effect than was condensed fro the previous effect. The syste shown in Figure 10.10 is called a forward feed syste. The heat source and the product being concentrated flow in parallel. In this syste, the pressure and teperature decrease with each effect. This decrease is necessary to perit use of vapor fro one effect as the heat source for the next effect. The vapor can only be used if the boiling point in the next effect is lower. The boiling point can only be lower if the pressure is lower; thus, P 1 > P 2 > P 3 and T 1 > T 2 > T 3. An advantage of this syste is that flow between effects occurs because of the pressure differences. v1 v2 v3 t 1, p 1 t 2, p 2 t 3, p 3 f1 s1 p1 v1 p2 v2 s1 v1 v2 p1 p2 p3 Figure 10.10. Multiple effect (three effects) forward feed evaporator syste.

Chapter 10 Drying and Dehydration 279 Vapor Vapor Vapor Feed Stea Concentrated Product Condensate Figure 10.11. Backward feed ultiple effect evaporator. A different ultiple effect arrangeent, a backward flow syste, is shown in Figure 10.11. Here, the product being concentrated flows counter to the flow of vapor used as the heat source. The nuber of effects (evaporators in the series) ay vary depending upon the application. Econoy is usually the controlling factor. Stea costs are a ajor coponent of operating costs and ultiple effects significantly reduce stea requireents. However, the increased efficiency of ore effects is eventually offset by the increased initial and aintenance costs of the added effects. 10.8.4 Mass and Energy Balances Mass and energy balances are key coponents in analyses of any evaporator syste. For ultiple effect systes, this results in ultiple equations that ust be solved. Such an analysis is beyond the scope of our study. We will only exaine a siplified analysis of a single effect syste following the notation of Figure 10.12. The feed, product, vapor, and stea flow rates are represented by subscripts as indicated. Siilarly, ass ratios of solids in the feed and product streas are X f and X p, respectively. v t f f s t bp s Figure 10.12. Notation for analysis of a single effect evaporator. p

280 Food & Process Engineering Technology A total ass balance for this syste gives the equation: A siilar analysis of the solids balance gives: f = p + v (10.08) f X f = p X p (10.09) And a water balance gives: f ( X f ) = p ( 1 X p ) + v 1 (10.10) Entering and exiting stea flow rates will be equal. The required agnitude of this flow is a function of the flow rate of feed aterial and the aount of water evaporated. An energy analysis is necessary to deterine the stea flow required. The energy balance analysis is soewhat ore coplicated. We will siplify the analysis by assuing that the evaporator is perfectly insulated. Thus, all energy fro the stea is used to heat the feedstock to the boiling point and then evaporate the desired aount of water. An overall energy balance then becoes: ( tbp t f ) + v h fs s hs = s cpf (10.11) Exaple 10.6 A single-effect evaporator is used to concentrate syrup fro 10% to 70% solids. The production rate is 50 kg/h. The evaporation occurs at a pressure of 50 kpa (a boiling point of 81.3 C for pure water). The feed syrup enters at a teperature of 20 C. Energy for evaporation is provided by stea entering as saturated vapor at 200 kpa and exiting as saturated liquid at the sae pressure. What stea flow is required to accoplish the evaporation if we assue no energy losses? Solution: We ust first perfor a ass balance to deterine the aount of water evaporated. Perforing a solids balance (Equation 10.09) we have: 0.1 f = 0.7 p = 0.7 f = 350 kg/h Then, solving for v using Equation 10.10: ( 50 kg/h) = 35 kg/h v = 350 = or by using Equation 10.08: ( 1 0.1) 50( 1 0.7) = 350( 0.9) 50( 0.3) 300 kg/h

Chapter 10 Drying and Dehydration 281 v = f + p = 350 50 = 300 kg/h Knowing the aount of water evaporated in the concentration process, we can now deterine the energy required to heat the feedstock to the boiling point and evaporate the water. The boiling point of a sugar solution is a function of pressure, sugar content, and type(s) of sugar present. The boiling point rise is less than 1 C at relatively low sugar concentrations and increases rapidly at concentration above 70% (Pancoast and Junk, 1980). At a concentration of 70% the boiling point rise would be near 5 C for a sucrose solution. For siplicity, we will ignore the increase in boiling point teperature as the syrup is concentrated and use 82 C as the approxiate boiling point teperature of a 10% sugar solution at 50 kpa. This is a 0.7 C increase above the boiling point of pure water. We analyze the energy balance using Equation 10.11, which states that: Stea Energy = Energy for Heating + Energy for Evaporation Fro stea tables: at 200 kpa, h fg = 2201.6 kj/kg (for condensing stea) at 82 C, h fg = 2303.8 kj/kg (for boiling solution) We now have all paraeters needed for a solution of Equation 10.11 except for the specific heat c pf. At 82 C water has a specific heat of 4.2 kj/kg K, and a 10% sucrose solution has a specific heat that is 95% of that for pure water. Thus: c pf = 4.2(0.95) = 3.99 kj/kg K. Substituting into Equation 10.11, we find: s 2201.6 = 350 3.99 82 20 + 300 2303.8 = 86 583 + 691140 = 777 ( ) 723 s = 353.2 kg/h ANSWER Note that the energy for evaporation is alost eight ties larger than the energy required to heat the solution fro 20 C to 82 C. Thus, energy for evaporation is the priary energy factor in evaporator systes; and our act of ignoring the boiling point rise had little effect upon the answer obtained.

282 Food & Process Engineering Technology List of Sybols a w water activity, decial c p specific heat capacity at constant pressure, J/(kg K) or kj/(kg K) C, N, K equilibriu oisture equation constants (see Tables 10.01 and 10.02). EMC equilibriu oisture content, decial or percent ERH equilibriu relative huidity, decial or percent h fg (h f h g ), latent heat of vaporization, kj/kg ass, kg oisture content wet basis (wb), decial or percent M oisture content dry basis (db), decial or percent equilibriu oisture content (db), decial or percent M e ass flow rate, kg/s P pressure, kpa rh relative huidity, decial or percent t teperature, C T absolute teperature, K X ass ratio (ass fraction), decial φ relative huidity, decial or percent Subscripts bp boiling point d dry f feed p product re reoved s stea, saturation t total v vapor w water wb wet-bulb ws water at saturation References 1. AOAC. 1990. Official Methods of Analysis of the Association of Analytical Cheists, 15 th ed. Kenneth Helrich, ed. AOAC, Arlington, VA. 2. ASAE. 2000. ASAE Standards D245.5: Moisture relationships of grains. ASAE, St. Joseph, MI. 3. Batty, J. C., and S. L. Folkan. 1983. Food Engineering Fundaentals. John Wiley & Sons, New York, NY. 4. Brennan, J. G., J. R. Butters, N. D. Cowell, and A. E. V. Lilly. 1990. Food Engineering Operations, 3 rd ed. Elsevier, London, UK. 5. Brooker, D. B., F. W. Bakker-Arkea, and C. W. Hall. 1974. Drying Cereal Grains. AVI, Westport, CT. 6. Char, S. E. 1971. The Fundaentals of Food Engineering. AVI, Westport, CT.

Chapter 10 Drying and Dehydration 283 7. Farrall, Arthur W. 1963. Engineering for Dairy and Food Products. John Wiley & Sons, Inc., New York, NY. 8. Iglesias, H. A., and J. Chirfe. 1982. Handbook of Food Isothers. Acadeic Press, New York, NY. 9. Labuza, I. P. 1984. Moisture Sorption: Practical Aspects of Isother Measureent and Use. Aerican Association of Cereal Cheists, St. Paul, MN. 10. Minton, Paul E. 1986. Handbook of Evaporation Technology. Noyes Publications, Park Ridge, NJ. 11. Pancoast, Harry M., and W. Ray Junk. 1980. Handbook of Sugars, 2 nd ed. AVI, Westport, CT. 12. Troller, J. A., and J. H. B. Christian. 1978. Water Activity and Food. Acadeic Press, New York, NY. 13. Van Arsdel, W. B., M. J. Copley, and A. I. Morgan, Jr. 1973. Food Dehydration Volue 1: Drying Methods and Phenoena, 2 nd ed. AVI, Westport, CT. 14. Van Arsdel, W. B., M. J. Copley, and A. I. Morgan, Jr. 1973. Food Dehydration Volue 2: Practices and Applications, 2 nd ed. AVI, Westport, CT. 15. Walton, C. F., E. K. Ventre, and S. Byall. 1938. Far Production of Sorgo Sirup. USDA Farers Bulletin 1791. GPO, Washington, DC. Probles 10.1. Show the relationship between oisture content wet basis () and dry basis (M) by plotting M as a function of for values of between 0 and 0.95. 10.2. Show how the equation used to solve for M e in Exaple 10.2 can be obtained fro Equation 10.06. 10.3. Air at 75 C (167 F) and a dew point teperature of 20 C (68 F) is used to dry apple leather in a counter flow arrangeent. The air leaves at 90% relative huidity. The leather enters at a rate of 100 kg/hr and 85% oisture (wb). The dried product contains 20% oisture (wb). a. How uch dried product is produced per hour? b. How uch water is reoved hourly? c. What air flow rate is needed to accoplish the drying? 10.4. Repeat proble 10.3, assuing a finished product of 25% oisture (wb). 10.5. Repeat proble 10.3, assuing a finished product with 0.2 kg H 2 O/kg of dry atter. 10.6. Mashed potatoes at 40% oisture content (wb) are to be air dried to 12% oisture. The output rate is 500 kg/hr of the 12% oisture content (wb) product. The drying air enters at 55 C and 10% relative huidity and is reoved at 80% relative huidity. a. Find the oisture content of the potatoes on a dry basis before and after drying. b. What is the water reoval rate, kg/hr? c. What is the teperature of the exiting air?

284 Food & Process Engineering Technology d. Find the voluetric flow rate of the inlet air. e. If the drying air was heated fro an initial teperature of 27 C, what was its initial relative huidity? 10.7. Sliced apples at 88% oisture (wb) are to be dried to 25% oisture in a counter flow dryer. Drying air enters at 90 C (194 F) and 34 C (93 F) wet bulb. If the air leaves at 60% relative huidity, what air flow rate is required to dry 1000 kg/hr (2200 lb/hr) of sliced apples? 10.8. A sall, stea heated, open pan evaporator is used to produce sorghu syrup. In the process, 8 kg of juice produces 1 kg of syrup. The production rate is 100 kg/hr of syrup that contains 76% solids. Five percent of the solids in the raw juice is reoved as ipurities during the process. a. How any kilogras of water are reoved per kilogra of syrup?. b. Find the juice supply rate needed to aintain the production rate. c. How uch energy is required per hour to evaporate the water? d. What stea flow is required if saturated stea is supplied at 350 kpa and leaves at 10% quality? (Assue that 20% of the energy fro the stea is lost.) 10.9. The equilibriu oisture content of soe aterials is related to relative huidity by the following equation: 1 φ = exp N [ K ( t + C) ] where C, K, and N are constants for a given aterial, t is the teperature ( C), Ν is the relative huidity (decial), and M e is the equilibriu oisture content (% dry basis). Values of constants for selected aterials are given in Table 10.01. Plot the equilibriu oisture content as a function of relative huidity for any one of these products at teperatures of 0 C, 10 C, 30 C, and 50 C. 10.10. An evaporator is used to concentrate juice of sugar cane to produce syrup. Coplete the table below for the various cobinations of juice input, juice sugar content, syrup production and syrup sugar content. The sugar content is expressed in Brix, which represents the percent sugars present in the solution. Juice Syrup Condition Input (kg/h) Brix Brix Output (kg/h) A 1000 15 70 B 1000 15 150 C 15 65 300 D 1000 70 100 M e