1、20.1CHAPTER 20COOLING AND FREEZING TIMES OF FOODSThermodynamics of Cooling and Freezing . 20.1Cooling Times of Foods and Beverages 20.1Sample Problems for Estimating Cooling Time 20.5Freezing Times of Foods and Beverages 20.7Sample Problems for Estimating Freezing Time 20.13Symbols . 20.14RESERVATIO
2、N of food is one of the most significant applica-Ptions of refrigeration. Cooling and freezing food effectivelyreduces the activity of microorganisms and enzymes, thus retard-ing deterioration. In addition, crystallization of water reduces theamount of liquid water in food and inhibits microbial gro
3、wth (Hel-dman 1975).Most commercial food and beverage cooling and freezing oper-ations use air-blast convection heat transfer; only a limited numberof products are cooled or frozen by conduction heat transfer in platefreezers. Thus, this chapter focuses on convective heat transfer.For air-blast conv
4、ective cooling and freezing operations to be cost-effective, refrigeration equipment should fit the specific requirementsof the particular cooling or freezing application. The design of suchrefrigeration equipment requires estimation of the cooling and freez-ing times of foods and beverages, as well
5、 as the corresponding refrig-eration loads.Numerous methods for predicting the cooling and freezing timesof foods and beverages have been proposed, based on numerical,analytical, and empirical analysis. Selecting an appropriate estima-tion method from the many available methods can be challenging.Th
6、is chapter reviews selected procedures available for estimatingthe air-blast convective cooling and freezing times of foods and bev-erages, and presents examples of these procedures. These proce-dures use the thermal properties of foods, discussed in Chapter 19.THERMODYNAMICS OF COOLING AND FREEZING
7、Cooling and freezing food is a complex process. Before freezing,sensible heat must be removed from the food to decrease its temper-ature to the initial freezing point of the food. This initial freezingpoint is somewhat lower than the freezing point of pure water becauseof dissolved substances in the
8、 moisture within the food. At the initialfreezing point, a portion of the water within the food crystallizes andthe remaining solution becomes more concentrated, reducing thefreezing point of the unfrozen portion of the food further. As the tem-perature decreases, ice crystal formation increases the
9、 concentrationof the solutes in solution and depresses the freezing point further.Thus, the ice and water fractions in the frozen food, and consequentlythe foods thermophysical properties, depend on temperature.Because most foods are irregularly shaped and have temperature-dependent thermophysical p
10、roperties, exact analytical solutions fortheir cooling and freezing times cannot be derived. Most researchhas focused on developing semianalytical/empirical cooling andfreezing time prediction methods that use simplifying assumptions.COOLING TIMES OF FOODS AND BEVERAGESBefore a food can be frozen, i
11、ts temperature must be reduced to itsinitial freezing point. This cooling process, also known as precoolingor chilling, removes only sensible heat and, thus, no phase changeoccurs.Air-blast convective cooling of foods and beverages is influ-enced by the ratio of the external heat transfer resistance
12、 to the inter-nal heat transfer resistance. This ratio (the Biot number) isBi = hL/k (1)where h is the convective heat transfer coefficient, L is the character-istic dimension of the food, and k is the thermal conductivity of thefood (see Chapter 19). In cooling time calculations, the characteristic
13、dimension L is taken to be the shortest distance from the thermal cen-ter of the food to its surface. Thus, in cooling time calculations, L ishalf the thickness of a slab, or the radius of a cylinder or a sphere.When the Biot number approaches zero (Bi 40), internal resistance to heat transfer is mu
14、ch greater than ex-ternal resistance, and the foods surface temperature can be assumedto equal the temperature of the cooling medium. For this latter situ-ation, series solutions of the Fourier heat conduction equation areavailable for simple geometric shapes. When 0.1 100Source: Lacroix and Castaig
15、ne (1987a)Fig. 5 Relationship Between jsValue for Surface Tempera-ture and Biot Number for Various ShapesFig. 5 Relationship Between jsValue for Surface Temperature and Biot Number for Various ShapesfL2-10lnBi-=jc1.0=fL2-10lnu2-=jc2 usinuuucossin+-=u 0.860972 0.312133 Biln+=0.007986 Biln20.016192 Bi
16、ln3+0.001190 Biln40.000581 Biln5+fL2- 0 . 9 3 3 2=jc1.273=1fcomp-1fi-i=Table 2 Expressions for Estimating f and jcFactors for Thermal Center Temperature of Infinite CylindersBiot Number Range Equations for f and j factorsBi 0.10.1 100Source: Lacroix and Castaigne (1987a)Table 3 Expressions for Estim
17、ating f and jcFactors for Thermal Center Temperature of SpheresBiot Number Range Equations for f and j factorsBi 0.10.1 100Source: Lacroix and Castaigne (1987a)fL2-10ln2 Bi-=jc1.0=fL2-10lnv2-=jc2J1vvJ02v J12v-=v 1.257493 0.487941 Biln+=0.025322 Biln20.026568 Biln3+0.002888 Biln40.001078 Biln5+fL2- 0
18、 . 3 9 8 2=jc1.6015=fL2-10ln3 Bi-=jc1.0=fL2-10lnw2-=jc2 wsin wwcoswwwcossin-=w 1.573729 0.642906 Biln+=0.047859 Biln20.03553 Biln3+0.004907 Biln40.001563 Biln5+fL2- 0 . 2 3 3 3=jc2.0=jiiG 0.2538B12-38B22-+=20.4 2010 ASHRAE HandbookRefrigerationwhere B1and B2are related to the cross-sectional areas o
19、f the food:(8)where L is the shortest distance between the thermal center of thefood and its surface, A1is the minimum cross-sectional area con-taining L, and A2is the cross-sectional area containing L that isorthogonal to A1.G is used in conjunction with the inverse of the Biot number m anda nomogr
20、aph (shown in Figure 6) to obtain the characteristic valueM12. Smith et al. showed that the characteristic value M12can be re-lated to the f factor by(9)where is the thermal diffusivity of the food. In addition, an expres-sion for estimating a jmfactor used to determine the mass averagetemperature i
21、s given asjm= 0.892e0.0388M21 (10)As an alternative to estimating M12from the nomograph devel-oped by Smith et al. (1968), Hayakawa and Villalobos (1989) ob-tained regression formulas for estimating M12. For Biot numbersapproaching infinity, their regression formula isln (M12) = 2.2893825 + 0.353305
22、39Xg 3.8044156Xg2 9.6821811Xg3 12.0321827Xg4 7.1542411Xg5 1.6301018Xg6(11)where Xg= ln(G). Equation (11) is applicable for 0.25 G 1.0.For finite Biot numbers, Hayakawa and Villalobos (1989) gave thefollowing:ln (M12) = 0.92083090 + 0.83409615Xg 0.78765739Xb 0.04821784XgXb 0.04088987Xg2 0.10045526Xb2
23、+ 0.01521388Xg3+ 0.00119941XgXb3+ 0.00129982Xb4(12)where Xg= ln(G) and Xb= ln(1/Bi). Equation (12) is applicable for0.25 G 1.0 and 0.01 1/Bi 100.Cooling Time Estimation Methods Based on Equivalent Heat Transfer DimensionalityProduct geometry can also be considered using a shape factorcalled the equi
24、valent heat transfer dimensionality (Cleland andEarle 1982a), which compares total heat transfer to heat transferthrough the shortest dimension. Cleland and Earle developed anexpression for estimating the equivalent heat transfer dimensional-ity of irregularly shaped foods as a function of Biot numb
25、er. Thisovercomes the limitation of the geometry index G, which wasderived for the case of Biot number approaching infinity. However,the cooling time estimation method developed by Cleland and Earlerequires the use of a nomograph. Lin et al. (1993, 1996a, 1996b)expanded on this method to eliminate t
26、he need for a nomograph.In the method of Lin et al., the cooling time of a food or beverageis estimated by a first term approximation to the analytical solutionfor convective cooling of a sphere: = (13)Equation (13) is applicable for center temperature if Yc1, 2 =)21.01 0 1 11Squat cylinder(1 = 2, 1
27、 1)3 1.01 0.75 1 1.22511.22521Short cylinder(1 = 1, 2 1)31.010.751 11.521Sphere(1 = 2 = 1)31.011.24 0 1 1 1Ellipsoid(1 1, 2 1)31.011.24 1 121Source: Lin et al. (1996b)20.6 2010 ASHRAE HandbookRefrigerationFor cooling time problems, the characteristic dimension is the short-est distance from the ther
28、mal center of a food to its surface. Assumingthat the thermal center of the ham coincides with its geometric center,the characteristic dimension becomesL = (4/12 ft)/2 = 0.1667 ftThe dimensional ratios then become Equations (16) and (17)1= = 1.6252= = 2.75Step 4: Calculate the Biot number.Bi = hL/k
29、= (8.5)(0.1667)/0.22 = 6.44 Step 5: Calculate the heat transfer dimensionality.Using Equation (19), E0becomesAssuming the ham to be ellipsoidal, the geometric factors can beobtained from Table 4:p1= 1.01 p2= 1.24 p3= 1From Equation (22),f (1) = = 0.4114f (2) = = 0.1766From Equation (21),E= 0.75 + (1
30、.01)(0.4114) + (1.24)(0.1766) = 1.38Thus, using Equation (15), the equivalent heat transfer dimension-ality becomesE = = 1.44Step 6: Calculate the lag factor applicable to the mass average tempera-ture.From Table 4, = 1, 1= 1, and 2= 2. Using Equation (24), jbecomesj= 1.271 + 0.305 exp(0.172)(1.625)
31、 (0.115)(1.625)2 + 0.425 exp(0.09)(2.75) (0.128)(2.75)2 = 1.78Using Equation (23), the lag factor applicable to the center tem-perature becomesjc= = 1.72Using Equations (25) and (26), the lag factor for the mass averagetemperature becomesjm= (1.72) = 0.721Step 7: Find the root of transcendental Equa
32、tion (14):cot + Bi 1 = 0cot + 6.44 1 = 0 = 2.68Step 8: Calculate cooling time.The unaccomplished temperature difference isY = = = 0.1538 Using Equation (13), the cooling time becomes = = 3.40 hExample 2. Repeat the cooling time calculation of Example 1, but useHayakawa and Villalobos (1989) estimati
33、on algorithm based on theuse of f and j factors.Solution.Step 1: Determine the thermal properties of the ham.The thermal properties of ham are given in Example 1.Step 2: Determine the heat transfer coefficient.From Example 1, h = 8.5 Btu/hft2F.Step 3: Determine the characteristic dimension L and the
34、 dimensionalratios 1and 2.From Example 1, L = 0.1667 ft, 1= 1.625, 2= 2.75.Step 4: Calculate the Biot number.From Example 1, Bi = 6.44.Step 5: Calculate the f and j factors using the method of Hayakawa andVillalobos (1989).For simplicity, assume the cross sections of the ham to be ellipsoi-dal. The
35、area of an ellipse is the product of times half the minor axistimes half the major axis, orA1= L21A2= L22Using Equations (7) and (8), calculate the geometry index G:B1= = 1= 1.625B2= = 2= 2.75G = 0.25 + = 0.442Using Equation (12), determine the characteristic value M12:Xg= ln(G) = ln(0.442) = 0.816X
36、b= ln(1/Bi) = ln(1/6.44) = 1.86 ln (M12) = 1.27M12= 3.56From Equation (9), the f factor becomesf = f = = 4.91 hFrom Equation (10), the j factor becomes6.54-114-E01.51.625 2.75 1.625212.75+2.7521 1.625+ +1.6252.751 1.625 2.75+-=1.625 2.7520.415-2.06=11.6252- 0 . 0 11 1.6251.62526-exp+12.752- 0 . 0 11
37、 2.752.7526-exp+6.44431.85+6.44431.38-1.852.06-+-6.441.35 11.625-+6.441.351.78-11.625-+-1.5 0.696.44+1.5 6.44+-3TmTTmTi-30 5030 160-367.5 0.89 0.166722.6820.22 1.44-0.7210.1538-lnA1L2-L21L2-=A2L2-L22L2-=381.6252-382.752-+0.92083090 0.83409615 0.8160.78765739 1.86+=0.04821784 0.8161.860.04088987 0.81
38、620.10045526 1.8620.01521388 0.8163+0.00119941 0.8161.8630.00129982 1.864+2.303L2M12-2.303L2cM12k-=2.3030.1667267.50.893.560.22-Cooling and Freezing Times of Foods 20.7jm= 0.892e(0.0388)(3.56)= 0.777Step 6: Calculate cooling time.From Example 1, the unaccomplished temperature difference wasfound to
39、be Y = 0.1538. Using Equation (4), the cooling time becomes = = 3.45 hFREEZING TIMES OF FOODS AND BEVERAGESAs discussed at the beginning of this chapter, freezing of foodsand beverages is not an isothermal process but rather occurs over arange of temperatures. This section discusses Planks basic fre
40、ezingtime estimation method and its modifications; methods that calcu-late freezing time as the sum of the precooling, phase change, andsubcooling times; and methods for irregularly shaped foods. Thesemethods are divided into three subgroups: (1) equivalent heat trans-fer dimensionality, (2) mean co
41、nducting path, and (3) equivalentsphere diameter. All of these freezing time estimation methods usethermal properties of foods (Chapter 19).Planks EquationOne of the most widely known simple methods for estimatingfreezing times of foods and beverages was developed by Plank(1913, 1941). Convective he
42、at transfer is assumed to occur betweenthe food and the surrounding cooling medium. The temperature ofthe food is assumed to be at its initial freezing temperature, which isconstant throughout the freezing process. Furthermore, constantthermal conductivity for the frozen region is assumed. Planksfre
43、ezing time estimation is as follows: = (27)where Lfis the volumetric latent heat of fusion (see Chapter 19), Tfis the initial freezing temperature of the food, Tmis the freezingmedium temperature, D is the thickness of the slab or diameter ofthe sphere or infinite cylinder, h is the convective heat
44、transfercoefficient, ksis the thermal conductivity of the fully frozen food,and P and R are geometric factors. For an infinite slab, P = 1/2 andR = 1/8. For a sphere, P = 1/6 and R = 1/24; for an infinite cylinder,P = 1/4 and R = 1/16.Planks geometric factors indicate that an infinite slab of thick-
45、ness D, an infinite cylinder of diameter D, and a sphere of diameterD, if exposed to the same conditions, would have freezing times inthe ratio of 6:3:2. Hence, a cylinder freezes in half the time of a slaband a sphere freezes in one-third the time of a slab.Modifications to Planks EquationVarious researchers have noted that Planks method does notaccurately predict freezing times of foods and beverages. This isbecause, in part, Planks method assumes that foods freeze a
