ABE 313 Thermal Properties PDF

ABE 313 Properties of AB Materials Topic 8 Thermal Properties Topic 8 THERMAL PROPERTIES FOURIER’S LAW OF HEAT CONDUCTION We need a driving force to overcome a resistance in order to transfer a property. For any kind of molecular transport processes (mo...

ABE 313 Properties of AB Materials Topic 8 Thermal Properties Topic 8 THERMAL PROPERTIES FOURIER’S LAW OF HEAT CONDUCTION We need a driving force to overcome a resistance in order to transfer a property. For any kind of molecular transport processes (momentum, heat or thermal energy, and mass) the general equation can be written as follows: Driving force Rate of a transfer process = 8.1 Resistance Consider a wall of thickness 𝑋 and surface area 𝐴. Imagine that the wall is initially uniform at a temperature 𝑇0. At time 𝑡 = 0, one side of the wall is suddenly brought to a slightly higher temperature 𝑇1 and maintained at that temperature. Heat is conducted through the wall as a result of the temperature difference, and as time proceeds, the temperature profile in the wall changes. Finally, linear steady-state temperature distribution is achieved as shown in Fig. 8.1. Figure 8. 1. Steady-state heat transfer through the wall. The driving force for the heat transfer to occur is the temperature difference: Driving force = 𝑇1 − 𝑇0 8.2 While the rate of heat conduction through the wall is proportional to the heat transfer area (𝐴), the thickness of the wall (𝑋) provides resistance to heat transfer. In addition, the ability of the wall material to conduct heat should be considered. Each material has a different ability to conduct heat. The responses of steel and wood to heating are not the same when they are exposed to the same amount of heat. This material property is named thermal conductivity (𝑘). Considering all these parameters, the resistance to heat transfer can be written as: 𝑋 Resistance = 8.3 𝑘𝐴  When the steady-state condition has been reached, the rate of heat flow (𝑄) through the wall can be written by substituting Eqs. (8.2) and (8.3) into Eq. (8.1): 𝑇1 − 𝑇0 𝑄 = 𝑘𝐴 8.4 𝑋 Equation (8.4) in differential form gives Fourier’s law of heat conduction: 𝑑𝑇 𝑄𝑥 = −𝑘𝐴 8.5 𝑑𝑥 where 𝑄𝑥 is the rate of heat flow in the 𝑥-direction. Heat is conducted in the direction of decreasing temperature and the temperature gradient becomes negative when temperature decreases with increasing 𝑥. Therefore, a negative sign is added to Eq. (8.5). THERMAL CONDUCTIVITY The thermal conductivity of a material is defined as a measure of its ability to conduct heat. It has a unit of W/m K in the SI system. A solid may be comprised of free electrons and atoms bound in a periodic arrangement called a lattice. Thermal energy is transported through the molecules as a result of two effects: lattice waves and free electrons. These two effects are additive: 𝑘 = 𝑘𝑒 + 𝑘𝑙 8.6 In pure metals, heat conduction is based mainly on the flow of free electrons and the effect of lattice vibrations is negligible. In alloys and nonmetallic solids, which have few free electrons, heat conduction from molecule to molecule is due to lattice vibrations. Therefore, metals have higher thermal conductivities than alloys and nonmetallic solids. The regularity of the lattice arrangement has an important effect on the lattice component of thermal conductivity. For example, diamond has very high thermal conductivity because of its well-ordered structure (Fig. 8.2). As temperature increases, lattice vibrations increase. Therefore, thermal conductivities of alloys increase with an increase in temperature while the opposite trend is observed in metals because the increase in lattice vibrations impedes the motion of free electrons. In porous solids such as foods, thermal conductivity depends mostly on composition but also on many factors that affect the heat flow paths through the material, such as void fraction, shape, size and arrangement of void spaces, the fluid contained in the pores, and homogeneity. Thermal conductivity in foods having fibrous structures such as meat cannot be the same in different directions (anisotropy) because heat flow paths through the material change with respect to direction. Thermal conductivity increases with moisture content. Thermal conductivities of food materials vary between that of water (𝑘water = 0.614 W/m°C at 27°C) and that of air (𝑘air = 0.026 W/m°C at 27°C), which are the most and the least conductive components in foods, respectively (Fig. 8.2). The thermal conductivity values of the other food components fall between these limits. Dry porous solids are very poor heat conductors because the pores are occupied by air. For porous materials, the measured thermal conductivity is an apparent one, called the effective thermal conductivity. It is an overall thermal transport property assuming that heat is transferred by conduction through the solid and the porous phase of the material. Figure 8. 2. Thermal conductivities of various materials at 27◦C. Thermal conductivity of ice is nearly four times greater than that of water (𝑘ice = 2.24 W/m°C at 0°C). This partly accounts for the difference in freezing and thawing rates of food materials. During freezing, the outer layer is in the frozen state and the ice propagates toward the center replacing nonflowing water by a better conductor. Heat is conducted through the ice layer, which has higher thermal conductivity than that of water. However, in thawing, the frozen part is at the center and heat is conducted through the thawed region, which has lower thermal conductivity or higher resistance than that of ice. As a result, the thawing process is slower than freezing process. Above their freezing point, thermal conductivity of foods increases slightly with increasing temperature. However, around their freezing point, temperature has a strong impact on the thermal conductivity as dramatic changes in the physical nature of the foods take place during freezing or thawing. In the case of liquids and gases, heat conduction occurs as a result of molecular collisions. Since the intermolecular spacing is much larger and the motion of the molecules is more random in fluids as compared to that in solids, thermal energy transport is less effective. Therefore, thermal conductivities of fluids are lower than those of solids.  Gases have very low thermal conductivity values (Fig. 8.2). Thermal conductivities of dilute monatomic gases can be predicted by kinetic theory, which is analogous to the prediction of viscosity of gases. Consider a pure gas composed of rigid, nonattracting spherical molecules of diameter 𝑑 and mass 𝑚 present in a concentration of 𝑁 molecules per unit volume. 𝑁 is considered to be small enough so that the average distance between molecules is many times their diameter 𝑑. The specific heat of the gas is 𝑐𝑣. Assume that the gas molecules are arranged in layers normal to the 𝑧-axis and the upper layer is warmer. Then, the heat is transferred in the 𝑧-direction only. When steady state is reached, the equation of energy becomes: 𝑑2𝑇 =0 8.7 𝑑𝑧 2 Integration states that: 𝑑𝑇 = Constant 8.8 𝑑𝑧 Therefore, at any point in the gas: 𝑑𝑇 𝑇 = 𝑇0 + 𝑧 8.9 𝑑𝑧 One third of the molecules are moving along the 𝑧-axis and carry heat from one layer to the other or one third of all the velocity components lie along 𝑧. According to kinetic theory, it is assumed that an average molecule traverses a distance equal to the mean free path between impacts. If mean free path is 𝜆, one may consider that the length of this path is the thickness of the layer of gas. On the two sides of a gas layer having a 𝑑𝑇 thickness of 𝜆, the average temperature difference of the molecules is expressed as 𝜆. 𝑑𝑧 Molecules coming from upper or warmer layer to the lower or cooler layer carry an excess energy 𝑑𝑇 of 𝑚𝑐𝑣 𝜆 from upper to the lower side. It can be said that on average one third of the molecules 𝑑𝑧 are moving along 𝑧, and one half are moving downward. Thus, the number of molecules having average velocity of 𝑐̅, moving downward through the layer 𝜆, per unit area per second will be one sixth of 𝑁𝑐̅. The energy transferred across this layer by the molecules can be expressed as, 1 𝑑𝑇 1 1 𝑑𝑇 𝑁𝑐̅𝑚𝑐𝑣 𝜆. Similarly, 𝑁𝑐̅ molecules pass upward and they carry − 𝑁𝑐̅𝑚𝑐𝑣 𝜆 energy units 6 𝑑𝑧 6 6 𝑑𝑧 across unit area per second. The net energy transfer is the difference of energy carried up and down. Then, the energy transferred through unit area per second is: 1 𝑑𝑇 𝑞𝑧 = − 𝑁𝑐̅𝑚𝑐𝑣 𝜆 8.10 3 𝑑𝑧 From Fourier’s law of heat conduction: 𝑑𝑇 𝑞𝑧 = −𝑘 8.11 𝑑𝑧 Equating Eqs. (8.10) and (8.11): 1 𝑘= 𝑁𝑐̅𝑚𝑐𝑣 𝜆 8.12 3 As discussed in Rheological properties, the mean free path is given by the following equation: 1 𝜆= 7.7 √2𝜋𝑑 2 𝑁 Inserting Eq. (7.7) into (8.12) gives: 𝑚𝑐𝑣 𝑐̅ 𝑘= 8.13 3√2𝜋𝑑 2 According to kinetic theory, molecular velocities relative to fluid velocity have an average magnitude given by the following equation: 8𝑅𝑇 𝑐̅ = √ 7.9 𝜋𝑁𝐴 𝑚 where 𝑁𝐴 is the Avogadro number, 𝑚 is the mass of the molecule, 𝑅 is the gas constant, and 𝑇 is the absolute temperature. Inserting Eq. (7.9) into Eq. (8.13) gives: 2𝑐𝑣 𝑘= √𝑚𝐾𝑇 8.14 3𝜋 3/2 𝑑 2 where 𝐾 = 𝑅/𝑁𝐴 and it is the Boltzmann constant. This equation predicts that thermal conductivity should increase with the square root of temperature. Thermal conductivities of gases are independent of pressure over a wide range. The mechanism of heat conduction in liquids is qualitatively the same as in gases. However, the situation is more complicated since the molecules are more closely spaced and they exert a stronger intermolecular force field. The thermal conductivities of liquids usually lie between those of solids and gases. Unlike those of gases, thermal conductivities of nonmetallic liquids usually decrease as temperature increases. There are some exceptions to this behavior such as water and glycerine. Polar or associated liquids such as water may exhibit a maximum in the thermal conductivity versus temperature curve. Generally, thermal conductivity decreases with increasing molecular weight. Prediction of Thermal Conductivity Predictive models have been used to estimate the effective thermal conductivity of foods. A number of models for thermal conductivity exist in the literature; however, many of them contain empirical factors and product-specific information, which restrict their applicability.  Modeling of thermal conductivity based on composition has been a subject of considerable interest. It is important to include the effect of air in the porous foods and ice in the case of frozen foods. Temperature dependence of thermal conductivities of major food components has been studied. Thermal conductivities of pure water, carbohydrate (CHO), protein, fat, ash, and ice at different temperatures can be empirically expressed according to Choi and Okos (1986) as follows: 𝑘water = 0.57109 + 1.7625 × 10−3 𝑇 − 6.7036 × 10−6 𝑇 2 8.15 𝑘CHO = 0.20141 + 1.3874 × 10−3 𝑇 − 4.3312 × 10−6 𝑇 2 8.16 𝑘protein = 0.17881 + 1.1958 × 10−3 𝑇 − 2.7178 × 10−6 𝑇 2 8.17 𝑘fat = 0.18071 − 2.7604 × 10−3 𝑇 − 1.7749 × 10−7 𝑇 2 8.18 𝑘ash = 0.32961 + 1.4011 × 10−3 𝑇 − 2.9069 × 10−6 𝑇 2 8.19 𝑘ice = 2.2196 − 6.2489 × 10−3 𝑇 + 1.0154 × 10−4 𝑇 2 8.20 where thermal conductivities (𝑘) are in W/m°C; temperature (T) is in °C and varies between 0 and 90◦C in these equations. Rahman (1995) correlated thermal conductivity of moist air at different temperatures from the data of Luikov (1964) as: 𝑘air = 0.0076 + 7.85 × 10−4 𝑇 + 0.0156𝑅𝐻 8.21 where 𝑅𝐻 is the relative humidity, changing from 0 to 1, and temperature varies from 20 to 60◦C. Generally, in multiphase systems, the effect of geometric distribution of phases is taken into account by using structural models. There are also some models that assume an isotropic physical structure. The most well-known models in literature are parallel and series models; the Krischer model, which is obtained by combining the parallel and series models; and the Maxwell- Eucken and Kopelman models used for two-component food systems consisting of a continuous and a dispersed phase. Parallel Model In the parallel model, components are assumed to be placed parallel to the direction of heat flow (Fig. 8.3a). The effective thermal conductivity of a food material made of n components can be calculated using volume fractions (𝑋𝑖𝑣 ) and thermal conductivities (𝑘𝑖 ) of each component (𝑖) from the following equation: 𝑛 𝑘pa = ∑ 𝑘𝑖 𝑋𝑖𝑣 8.22 𝑖=1 where  𝑋𝑖𝑤⁄ 𝜌𝑖 𝑋𝑖𝑣 = 𝑤 8.23 𝑋 ∑𝑛𝑖=1 ( 𝑖 ⁄𝜌𝑖 ) where 𝑋𝑖𝑣 = volume fraction of the 𝑖th constituent, 𝑋𝑖𝑤 = mass fraction of the 𝑖th constituent, 𝜌𝑖 = density of the 𝑖th constituent (kg/m3). Figure 8. 3. (a) Parallel, (b) Series, and (c) Krischer models. The parallel distribution results in maximum thermal conductivity value. If the food material is assumed to be composed of three components (water, solid, and air), effective thermal conductivity can be calculated from: 𝑣 + 𝑘 𝑋𝑣 + 𝑘 𝑋𝑣 𝑘pa = 𝑘w 𝑋w s s a a 8.24 𝑣 , 𝑋 𝑣 , and 𝑋 𝑣 are the volume fractions of moisture, solid, and air, respectively and 𝑘 , where 𝑋w s a w 𝑘s , and 𝑘a are the corresponding thermal conductivities. Series (Perpendicular) Model In the perpendicular model, components are assumed to be placed perpendicular to the direction of heat flow (Fig. 8.3b). The effective thermal conductivity of a food material can be calculated from the following equation: 𝑛 1 𝑋𝑖𝑣 =∑ 8.25 𝑘se 𝑘𝑖 𝑖=1  The perpendicular distribution results in a minimum thermal conductivity value. If the food material is assumed to be composed of three components (water, solid, and air), effective thermal conductivity can be calculated from: 𝑣 1 𝑋w 𝑋s𝑣 𝑋a𝑣 = + + 8.26 𝑘se 𝑘w 𝑘s 𝑘a Krischer Model The real value of thermal conductivity should be somewhat between the thermal conductivities derived from the parallel model and series model. The series and parallel models do not take into account the natural distribution of component phases. Thus, Krischer proposed a generalized model by combining the parallel and series models using a phase distribution factor. The distribution factor, 𝑓𝑘 , is a weighing factor between these extreme cases (Fig. 8.3c). Krischer’s model is described by the following equation: 1 𝑘= 1 − 𝑓𝑘 𝑓𝑘 8.27 + 𝑘pa 𝑘se where 𝑘, 𝑘pa , and 𝑘se are the effective thermal conductivity by the Krischer, parallel, and series models, respectively. The disadvantage of this model is that it is not possible to find the values of 𝑓𝑘 without making experiments because 𝑓𝑘 depends on moisture content, porosity, and temperature of foods. Maxwell-Eucken Model Krischer’s model is one of the most widely used one in food engineering literature. However, for isotropic materials, it is not very useful. For a two-component food system consisting of a continuous and a dispersed phase, a formula based on the dielectric theory had been developed by Maxwell. The model was then adapted by Eucken and is known as the Maxwell-Eucken model. The effective thermal conductivity of a food by this model is defined as: 2𝑘𝑐 + 𝑘𝑑 − 2𝑋𝑑𝑣 (𝑘𝑐 − 𝑘𝑑 ) 𝑘 = 𝑘𝑐 ( ) 8.28 2𝑘𝑐 + 𝑘𝑑 + 𝑋𝑑𝑣 (𝑘𝑐 − 𝑘𝑑 ) where 𝑘𝑐 and 𝑘𝑑 are the thermal conductivities of continuous and dispersed phases, respectively. A three-step modification of the Maxwell-Eucken model was developed for porous foods by Hamdami, Monteau, and Le Bail (2003). In the first step, water was the continuous and ice was the discontinuous phase. In the second step, the solid was the continuous and water-ice was the discontinuous phase. In the third step, solid-water-ice was the continuous and air the discontinuous phase.  Carson, Lovatt, Tanner, and Cleland (2005) recommended a modified version of the Maxwell-Eucken model. They proposed that porous materials should be divided into two types according to porosity (external or internal porosity) because the mechanism of heat conduction in these two types differs. Kopelman Model The isotropic model of Kopelman (1966) describes the thermal conductivity of a composite material as a combination of continuous and discontinuous phases. 𝑘𝑐[1 − 𝑄] 𝑘= 8.29 1 − 𝑄[1 − (𝑋𝑑𝑣 )1/3 ] where 𝑘𝑑 𝑄 = (𝑋𝑑𝑣 )2⁄3 (1 − ) 8.30 𝑘𝑐 This equation is useful for two-component models, but needs modification for multicomponent systems. In particular, the definition of continuous and dispersed phases needs specification. Since foods are multicomponent systems, and the phases are associated with each other in complex ways, a stepwise procedure in which the mean thermal conductivity of pairs of components is found at each step in the Kopelman and Maxwell equations. Moreira, Palau, Sweat, and Sun (1995) and Sahin, Sastry, and Bayındırlı (1999) used the isotropic model of Kopelman to predict the effective thermal conductivity of tortilla chips and french fries, respectively. The approach chosen in Kopelman equation is to successively determine the thermal conductivity of two-component systems, starting with water continuous and carbohydrate dispersed phases. Then, the water–carbohydrate phase is taken as continuous and protein as dispersed phases and the iterative procedure is continued through all phases using the order: water (phase 1), carbohydrate (2), protein (3), fat (4), ice (5), ash (6), and air (7). The following iterative algorithm was obtained for the thermal conductivity of a system of 𝑖 + 1 components: 𝑘𝑖 [1 − 𝑄𝑖+1 ] 𝑘comp,𝑖+1 = 1⁄3 8.31 𝑣 1 − 𝑄𝑖+1 [1 − (𝑋𝑑,𝑖+1 ) ] where the following definitions apply: 𝑣 2⁄3 𝑘𝑖+1 𝑄𝑖+1 = (𝑋𝑑,𝑖+1 ) (1 − ) 8.32 𝑘𝑖 𝑣 𝑉𝑖+1 𝑋𝑑,𝑖+1 = 8.33 ∑𝑖+1 𝑖 𝑉𝑖 Improved Thermal Conductivity Prediction Models  Many food products are porous and the thermal conductivity strongly depends on porosity since thermal conductivity of air in pores is an order of magnitude lower than the thermal conductivities of other food components. In general, the effective thermal conductivity of heterogeneous materials cannot be predicted with simple additive models. Heat is transported in pores of food materials not only by conduction but also by latent heat. Therefore, for porous food materials, the two kinds of heat transport should be considered in the modeling of effective thermal conductivity. Effective thermal conductivity of a highly porous model food, at above and below freezing temperatures, was studied by Hamdami et al. (2003). The Krischer model was in good agreement with the experimental data. The model, including the effect of evaporation–condensation phenomena in addition to heat conduction, was also useful in predicting the effective thermal conductivity of porous model food. If there is heat transfer in porous food material, moisture migrates as vapor in the pore space as a result of the vapor pressure gradient caused by the temperature gradient. The effective thermal conductivity in pores can be expressed by the following equation by considering the effect of latent heat transport: 𝑘pores = 𝑘air + 𝑘eva−con 𝑓eva−con 8.34 where 𝑘pores = effective thermal conductivity in pores (W/m K), 𝑘air = thermal conductivity of air (W/m K), 𝑘eva−con = equivalent thermal conductivity due to evaporation–condensation phenomena (W/m K), 𝑓eva−con = resistance factor against vapor transport. Equivalent thermal conductivity due to evaporation–condensation phenomena (𝑘eva−con ) can be calculated from 𝐷 𝑃 𝑑𝑃sat 𝑘eva−con = 𝜆𝑎𝑤 8.35 𝑅𝑇 𝑃 − 𝑎𝑤 𝑃sat 𝑑𝑇 where 𝐷 = diffusivity of water vapor in air (m2/s), 𝑅 = gas constant (8.3145 J/mol K), 𝑇 = temperature (K), 𝑃 = total pressure (Pa), 𝑃sat = saturated pressure of water vapor (Pa), 𝑎𝑤 = water activity,  𝜆 = latent heat of evaporation (J/mol). Thermal conductivities of foods change during processing with variation in composition. For example, thermal conductivity decreases with decreasing moisture content during drying. The formation of pores (air phase) in foods further decreases the thermal conductivity. Rahman (1992) developed a thermal conductivity model for beef, apple, potato, pear, and squid during drying by introducing a porosity term: 𝑤 𝑘 1 𝑋𝑤 ( )( ) = [1.82 − 1.66 exp (−0.85 𝑤 )] 8.36 𝑘0 1 − 𝜀 𝑋𝑤0 where 𝑘 = effective thermal conductivity (W/m K), 𝑘0 = initial thermal conductivity (W/m K), 𝜀 = volume fraction of air or porosity, 𝑤 𝑋𝑤 = mass fraction of water (wet basis), 𝑤 𝑋𝑤0 = mass fraction of initial water (wet basis). The above correlation is an extension of the parallel model, which is realistic in the case of homogeneous food materials. It is valid for moisture content from 5% to 88% (wet basis), porosity from 0 to 0.5, and temperature from 20 to 25◦C. However, this correlation has some disadvantages: When porosity is 1.0, the left side of the correlation becomes infinity, which is physically incorrect. Moreover, the effect of temperature on thermal conductivity is not taken into account. In addition, the thermal conductivity values of fresh foods (i.e., before processing) are required. Therefore, an improved general thermal conductivity prediction model has been developed for fruits and vegetables during drying as a function of moisture content, porosity, and temperature. 𝜑 0.713 = 0.996 (𝑇⁄𝑇 ) 𝑤 )0.285 (𝑋𝑤 𝑘air ref 8.37 1 − 𝜀 + [ ⁄(𝑘 ) ] 𝑤 ref where 𝑇ref is the reference temperature which is 0°C and (𝑘𝑤 )𝑟𝑒𝑓 is the thermal conductivity of water evaluated at reference temperature. 𝜑 is the Rahman-Chen structural factor related to the phase distribution of components of food material and expressed as a function of temperature, moisture content, and porosity as: 𝑘 − 𝜀𝑘air 𝜑= 𝑤 )𝑘 + 𝑋 𝑤 𝑘 8.38 (1 − 𝜀 − 𝑋𝑤 𝑠 𝑤 𝑤 In the development of this model, the moisture content varied from 14% to 88% (wet basis), porosity varied from 0 to 0.56, and temperature varied from 5 to 100°C.  Empirical models, relating thermal conductivity to temperature and moisture content of the material, are fitted to literature data for different food. The thermal conductivities of the materials with intermediate moisture were estimated using the two-phase parallel model assuming that the material is a uniform mixture of two components: a dried material and a wet material with infinite moisture. 𝑤 )𝑘 𝑤 𝑘 = (1 − 𝑋𝑤 dry + 𝑋𝑤 𝑘wet 8.39 where 𝑘 = effective thermal conductivity (W/m K), 𝑘dry = thermal conductivity of the dried material (phase a) (W/m K), 𝑘wet = thermal conductivity of the wet material (phase b) (W/m K), 𝑤 𝑋𝑤 = mass fraction of moisture in the material (w/w in db). The temperature dependence of the thermal conductivities of both phases is expressed by an Arrhenius type equation: 𝐸0 1 1 𝑘dry = 𝑘0 exp [− ( − )] 8.40 𝑅 𝑇 𝑇ref 𝐸𝑖 1 1 𝑘wet = 𝑘𝑖 exp [− ( − )] 8.41 𝑅 𝑇 𝑇ref where 𝑇ref = reference temperature (60◦C) which is a typical temperature of air drying of foods, 𝑅 = the ideal gas constant (8.3143 J/mol K), 𝑘0 = thermal conductivity at moisture 𝑋 = 0 and temperature 𝑇 = 𝑇ref (W/m K), 𝑘𝑖 = thermal conductivity at moisture 𝑋 = ∞ and temperature 𝑇 = 𝑇ref (W/m K), 𝐸0 = activation energy for heat conduction in dry material at 𝑋 = 0 (J/mol), 𝐸𝑖 = activation energy for heat conduction in wet material at 𝑋 = ∞ (J/mol). Example 8.1. The composition of date fruit (Phoenix dactylifera) and the densities of food components are given in Table E.8.1.1. Determine the thermal conductivity of the fruit at 25◦C, using parallel, series, and isotropic Kopelman models. Table E.8.1.1. Composition of the Date Fruit and Densities of Food Components at 25◦C Component Weight (%) Density (kg/m3) Water 22.5 995.7 Carbohydrate 72.9 1592.9 Protein 2.2 1319.6 Fat 0.5 917.15 Ash 1.9 2418.2 Solution:  To calculate the thermal conductivity of the fruit using predictive models, thermal conductivity values of food components at 25°C are required. They can be calculated using Eqs. (8.15)–(8.19) (Table E.8.1.2). Table E.8.1.2. Thermal Conductivity Values of Components at 25°C Using the composition and density of components data given in the question, the specific volume of each component is calculated: Mass fraction of component 𝑖 Specific volume of component 𝑖 = Density of component 𝑖 Table E.8.1.3. Specific Volume and Volume Fraction of Components in Date Fruit Thermal conductivity of date fruit is calculated by using volume fractions and thermal conductivity values of components using the parallel model (Eq. 8.22): 𝑛 𝑘pa = ∑ 𝑘𝑖 𝑋𝑖𝑣 8.22 𝑖=1 𝑘pa = (0.61)(0.32) + (0.233)(0.64) + (0.207)(0.023) + (0.112)(0.0076) +(0.363)(0.011) = 0.353 W⁄mK Using the series model (Eq. 8.25): 𝑛 1 𝑋𝑖𝑣 =∑ 8.25 𝑘se 𝑘𝑖 𝑖=1 0.32 0.64 0.023 0.0076 0.011 = + + + + 0.61 0.233 0.207 0.112 0.363 = 3.48 mk/W 1 𝑘se = = 0.287 W/mk 3.48  In the case of the isotropic model of Kopelman (Eqs. 8.31–8.33), the 𝑘 value of date fruit is calculated using the order of water (1), carbohydrate (2), protein (3), fat (4), and ash (5) in the iteration. 𝑘𝑖 [1 − 𝑄𝑖+1 ] 𝑘comp,𝑖+1 = 1⁄3 8.31 𝑣 1 − 𝑄𝑖+1 [1 − (𝑋𝑑,𝑖+1 ) ] where 𝑣 2⁄3 𝑘𝑖+1 𝑄𝑖+1 = (𝑋𝑑,𝑖+1 ) (1 − ) 8.32 𝑘𝑖 𝑣 𝑉𝑖+1 𝑋𝑑,𝑖+1 = 8.33 ∑𝑖+1 𝑖 𝑉𝑖 Starting with the water continuous and carbohydrate (CHO) dispersed phase, the thermal conductivity of water–CHO system is calculated as: 𝑣 𝑉CHO 4.58 × 10−4 𝑋𝑑,CHO = = = 0.669 𝑉water + 𝑉CHO 2.26 × 10−4 + 4.58 × 10−4 𝑣 2⁄3 𝑘CHO 𝑄CHO = (𝑋𝑑,CHO ) (1 − ) 𝑘water 0.233 𝑄CHO = (0.699)2⁄3 (1 − ) = 0.473 0.610 𝑘water [1 − 𝑄CHO ] 𝑘water−CHO = 1⁄3 𝑣 1 − 𝑄CHO [1 − (𝑋𝑑,CHO ) ] 0.61[1 − 0.473] 𝑘water−CHO = = 0.342 W/mK 1 − 0.473[1 − (0.669)1⁄3 ] Then, the water–carbohydrate phase is taken as the continuous phase and protein as the dispersed phase. 𝑣 𝑉prot 4.58 × 10−4 𝑋𝑑,prot = = = 0.024 𝑉water + 𝑉CHO + 𝑉prot 2.26 × 10−4 + 4.58 × 10−4 + 1.67 × 10−5 𝑣 2⁄3 𝑘prot 𝑄prot = (𝑋𝑑,prot ) (1 − ) 𝑘water−CHO 0.207 𝑄CHO = (0.024)2⁄3 (1 − ) = 0.033 0.342 𝑘water−CHO [1 − 𝑄prot ] 𝑘water−CHO−prot = 1⁄3 𝑣 1 − 𝑄prot [1 − (𝑋𝑑,prot ) ] 0.342[1 − 0.033] 𝑘water−CHO−prot = = 0.338 W/mK 1 − 0.033[1 − (0.024)1⁄3 ]  The same procedure is followed throughout all phases to find the thermal conductivity of date fruit. The calculated values in the iterative procedure are shown in Table E.8.1.4. The thermal conductivity of date fruit is found to be 0.337 W/m K using the isotropic model of Kopelman. Table E.8.1.4. Results Obtained in the Iterative Procedure Measurement of Thermal Conductivity Measurement of thermal conductivity can be done by either steady-state or transient-state methods. There are a number of experimental measurement techniques under each of these two categories. The advantages of steady-state methods are the simplicity in the mathematical processing of the results, the ease of control of the experimental conditions, and often quite high precision in the results. However, a long time is required for temperature equilibration. The moisture migration and the necessity to prevent heat losses to the environment during this long measurement time are the disadvantages of steady-state methods. In addition, these methods require definite geometry of the sample and relatively large sample size. On the other hand, the transient methods are faster and more versatile than the steady- state methods and are preferable for extensive experimental measurements. Transient methods are preferred over steady-state methods because of the short experimental duration and minimization of moisture migration problems. Steady State Methods In steady-state methods, two sides of a flat object are maintained at constant but different temperatures and the heat flux through the sample is measured. Steady-state methods are longitudinal heat flow, radial heat flow, heat of vaporization, heat flux, and differential scanning calorimeter methods. (a) Longitudinal Heat Flow Method. The most common method in this group is guarded hot plate method. This method is suitable mostly for determination of thermal conductivity of dry homogeneous materials in slab forms. It is the most widely used and the most accurate method for the measurement of thermal conductivity of materials, which are poor heat conductors.  In this method, the heat source (𝑇1 ), the sample, and the heat sink (𝑇2 ) are placed in contact with each other and with a thermal guard heated electrically. The thermal guard plates are kept at the same temperature as the adjacent surfaces, in a way that no heat leakage takes place from source, sample, or sink boundaries. Thermal conductivity is measured after the sample has reached steady state condition. However, achieving steady-state conditions may take several hours. It is assumed that all of the measured heat input is transferred across the sample. The thermal conductivity is calculated by measuring the amount of heat input required to maintain the unidirectional steady-state temperature profile across the test sample. 𝑄𝐿 𝑘= 8.42 𝐴(𝑇1 − 𝑇2 ) where 𝑘 = thermal conductivity of sample (W/m K), 𝑄 = measured rate of heat input (W), 𝐿 = sample thickness (m), 𝐴 = area of sample (m2). (b) Radial Heat Flow Methods. These methods are suitable mostly for loose, powdered, or granular materials. (i) Concentric Cylinder Method. In this method, the sample is placed between two concentric cylinders. This method is preferable for liquid samples. The heater is usually located at the outer cylinder. A coolant fluid flows through the inner cylinder. The heat that the coolant absorbed is assumed to be equal to the heat transferred through the sample. Thermal conductivity can be calculated from the unidirectional radial steady-state heat transfer equation as: 𝑄 ln(𝑟2 ⁄𝑟1 ) 𝑘= 8.43 2𝜋𝐿(𝑇1 − 𝑇2 ) where 𝑄 = power used by central heater (W), 𝐿 = length of the cylinder (m), 𝑇1 = temperature of the sample at the outer surface of radius 𝑟1 (K), 𝑇2 = temperature of the sample at the inner surface of radius 𝑟2 (K). The length-to-diameter ratio of the cylinder must allow the radial heat flow assumption. End guard heaters may be used to minimize the error due to axial heat flow. (ii) Concentric Cylinder Comparative Method. This method uses a central heater followed by a cylindrical sample and a cylindrical standard. The temperatures 𝑇1 and 𝑇2 at radii 𝑟1 and 𝑟2 of the sample, respectively, and temperature 𝑇3 and 𝑇4 at radii 𝑟3 and 𝑟4 of the standard, respectively, are measured. Assuming radial heat flow, the thermal conductivity can be determined from Eq. (8.44): 𝑘ref (𝑇3 − 𝑇4 ) ln(𝑟2 ⁄𝑟1 ) 𝑘= 8.44 (𝑇1 − 𝑇2 ) ln(𝑟4 ⁄𝑟3 ) where 𝑘ref is the thermal conductivity of the standard. (iii) Sphere with Central Heating Source. In this method, the sample is placed between the central heater which has a radius 𝑟1 and the outer radius of sphere, 𝑟2. The sample completely encloses the heating source so that end losses are eliminated. Assuming that the inner and outer surfaces of the sample are 𝑇1 and 𝑇2 , respectively, after the steady state has been established heat flow will essentially be radial and Eq. (8.45) can be used to determine thermal conductivity: 𝑄(1⁄𝑟1 − 1⁄𝑟2 ) 𝑘= 8.45 4𝜋(𝑇1 − 𝑇2 ) This is the most sensitive method among the steady-state methods because the error due to heat losses can be practically eliminated. However, it cannot be widely used because of the difficulty in obtaining suitably shaped food samples. This method has been used mainly for granular materials. Samples should be filled in a vacuum environment because air bubbles trapped inside the sphere could increase contact resistance. (c) Heat of Vaporization Method. In this method, a small test sample is put between two silver plates, one of which is in contact with a liquid 𝐴 at its boiling point and the other one is in contact with liquid 𝐵. Heat transferred through the sample vaporizes some of the liquid 𝐵, which has a lower boiling point. Since the time necessary to vaporize a unit mass of liquid 𝐵 is known, the thermal conductivity of the sample is calculated using Eq. (8.46): 𝜆𝐿 𝑘= 8.46 𝜃𝐴(𝑇𝐴 − 𝑇𝐵 ) where 𝜃 = time necessary to vaporize a unit mass of liquid 𝐵 (s/kg), 𝜆 = heat of vaporization of the liquid at lower boiling point (J/kg), 𝑇𝐴 , 𝑇𝐵 = boiling points of liquids A and B, respectively (K), 𝐿 = thickness of sample (m), 𝐴 = area of sample (m2). (d) Heat Flux Method. The heat flow meter is a device for measuring heat flux. It is suitable for materials with conductance (𝑘/𝐿) less than 11.3 W/m2K. Tong and Sheen (1992) proposed a technique based on heat flux to determine effective thermal conductivity of multilayered constructions. In this technique, a heat flux sensor is attached to the inner surface of the wall with a very thin layer of high thermal conductivity adhesive. A temperature difference of 5 to 7°C is maintained within the system and the thermal conductivity is evaluated at the arithmetic mean temperature. At steady state, the heat flux is: 𝑞 = 𝑈∆𝑇 8.47 where 𝑞 is the heat flux in W/m2 and 𝑈 is the overall heat transfer coefficient in W/m2K. The overall heat transfer coefficient can be written in terms of convective and conductive resistances: 𝑁 1 1 1 𝐿𝑖 = + +∑ 8.48 𝑈 ℎ𝑖 ℎ 𝑜 𝑘𝑖 𝑖=1 where ℎ𝑖 = internal heat transfer coefficient (W/m2K), ℎ𝑜 = external heat transfer coefficient (W/m2K), 𝑁 = number of layers, 𝐿𝑖 = thickness of layer 𝑖 (m), 𝑘𝑖 = thermal conductivity of layer 𝑖 (W/mK). If ℎ𝑖 and ℎ𝑜 are very large, convective resistances are negligible and Eq. (8.48) becomes: 𝑁 1 𝐿𝑖 =∑ 8.49 𝑈 𝑘𝑖 𝑖=1 Since it is not possible to measure the thermal conductivity of each layer, an effective thermal conductivity (𝑘eff ) can be used: 𝑁 𝐿𝑖 ∑𝑁𝑖=1 𝐿𝑖 ∑ = 8.50 𝑘𝑖 𝑘eff 𝑖=1 Combining Eqs. (8.49) and (8.50) gives: 𝑁 𝑘eff = 𝑈 ∑ 𝐿𝑖 8.51 𝑖=1 Then, substituting Eq. (8.47) into Eq. (8.51): 𝑞 ∑𝑁 𝑖=1 𝐿𝑖 𝑘eff = 8.52 ∆𝑇 (e) Differential Scanning Calorimeter (DSC). An attachment to a differential scanning calorimeter was designed to measure thermal conductivity of foods as shown in Fig. 8.4. The sample of uniform cross section (possibly cylindrical) is placed in the sample pan, the opposite end of which is in contact with a heat sink at constant temperature. Initially, the sample is maintained at a constant temperature. At a predetermined time, the pan temperature is immediately increased to a predetermined higher value. A new steady state is reached in a few minutes and the heat flow into the DSC pan levels off. A typical DSC response curve is shown in Fig. 8.5. The difference in heat flow (∆𝑄) between the two states is recorded from the thermogram. Then, thermal conductivity of the sample can be calculated using Fourier’s heat conduction equation: 𝐿∆𝑄 𝑘= 8.53 𝐴(∆𝑇2 − ∆𝑇1 ) where 𝐿 is the sample thickness, 𝐴 is the cross-sectional area, ∆𝑇1 is the initial temperature difference, and ∆𝑇2 is the final temperature difference. Figure 8. 4. Illustration of sample placement in DSC method for determination of thermal conductivity. Figure 8. 5. Typical DSC thermogram for measurement of thermal conductivity  This method is simple and suitable for small size samples, for both low- and high-moisture foods. Time to achieve the new steady state is small enough (10 to 15 min) to prevent moisture migration since the sample is small. This approach may be modified to measure thermal conductivity as a function of temperature by using small thermal perturbations. However, measurement of thermal conductivity under ultrahigh temperature (UHT) conditions may require extensive equipment modification. Thus, it is more expensive than either the line heat source or modified Fitch methods, which are the most commonly used unsteady-state thermal conductivity measurement methods in food systems. Unsteady-State Methods The most important transient methods are the thermal conductivity probe method, transient hot wire method, modified Fitch method, point heat source method, and comparative method. (a) Thermal Conductivity Probe Method. The theory of the thermal conductivity probe or line heat source method has been reviewed by many authors. This method is the most popular method for determining thermal conductivity of food materials because of its relative simplicity and speed of measurement. In addition, this method requires relatively small sample sizes. On the other hand, it requires a fairly sophisticated data acquisition system. In this method, a constant heat source is applied to an infinite solid along a line with infinitesimal diameter, such as a thin resistant wire. The electrical wire must have a low resistance so that the voltage drop across it is negligible compared to the voltage drop across the heater. The cross section of the line heat source probe and the experimental apparatus for measurement of the effective thermal conductivity using the line heat probe are shown in Figs. 8.6 and 8.7, respectively. Figure 8. 6. Cross section of thermal conductivity probe. For measurement of thermal conductivity, the container is filled with sample and the line heat source probe is inserted at the center of the container. The container is placed in a constant temperature bath and equilibrated at room temperature. After the initial temperature is recorded, the probe heater is activated and heated at a constant rate of energy input. Then, the time versus temperature adjacent to the line heat source is recorded. Figure 8. 7. Experimental apparatus for measurement of the effective thermal conductivity using line heat probe. Theoretically, the line heat source technique is based on unsteady state heat conduction in an infinite medium and is expressed by: 𝜕𝑇 = 𝛼∇2 𝑇 8.54 𝜕𝑡 where 𝛼 is the thermal diffusivity of the material (𝛼 = 𝑘⁄𝜌𝑐𝑝 ). The initial condition is: at 𝑡 = 0 ∆𝑇(𝑟, 𝑡) = 0 8.55 The line heat source theory is based on a line heat source of infinite length with negligible axial heat flow. Thermal conductivity is measured only in a direction radial to the probe because the heat flow is only in the radial direction. Therefore, the following boundary conditions can be written: 𝜕𝑇 𝑄 B. C. 1 at 𝑟 = 0 (𝑟 )=− 8.56 𝜕𝑟 2𝜋𝑘 B. C. 2 at 𝑟 = ∞ ∆𝑇(𝑟, 𝑡) = 0 8.57 It is required to assume that the power input per unit probe length, 𝑄 and the thermal properties are constant and the thermal mass of the heater is negligible. The solution of Eq. (8.54) is: ∞ −𝑢 𝑄 𝑒 ∆𝑇 = ∫ 𝑑𝑢 8.58 4𝜋𝑘 𝛽2 𝑢 where  𝑟2 𝑢= 8.59 4𝛼(𝑡 − 𝑡0 ) 𝑟 𝛽= 8.60 2√𝛼𝑡 The analytical solution of Eq. (8.58) gives: 𝑄 ∆𝑇 = − 𝐸 [−𝛽 2 ] 8.61 4𝜋𝑘 1 where 𝐸1 is the first-order exponential integral function, which can be evaluated using the following equation: ∞ (−1)𝑛 𝛽 2𝑛 𝐸1 [−𝛽 2 ] = 𝐶𝑒 + ln(−𝛽 2 ) +∑[ ] 8.62 𝑛𝑛! 𝑛=1 The Euler constant, 𝐶𝑒 , is equal to 0.57721. Then, Eq. (8.61) can be written as: 𝑄 2) 𝛽2 𝛽4 ∆𝑇 = [−0.57721 − ln(𝛽 + − + ⋯] 8.63 4𝜋𝑘 2! 4(2!) For small values of 𝛽 (large time values), the higher order terms in Eq. (8.63) become negligible. This can be achieved either by making the probe radius as small as possible or by making the test duration as long as possible (Eq. 8.60). The simplified equation for the line heat source method is obtained if the temperature versus time data are collected within a specific time interval (𝑡 − 𝑡0 ) as: 𝑄 𝑡 ∆𝑇 − ∆𝑇0 = ln ( ) 8.64 4𝜋𝑘 𝑡0 The rate of rise in temperature of the sample is a function of thermal conductivity of the material. A plot of (∆𝑇 − ∆𝑇0 ) versus 𝑙𝑛(𝑡/𝑡0 ) should yield a straight line and thermal conductivity can be found from the slope. The initial time (𝑡0 ) is arbitrarily chosen but it will simplify the data analysis if it is set equal to the time when the time–temperature plot in a semilogarithmic axis starts to become linear. Based on visual inspection, the initial data points are discarded to eliminate any transient initial effects. The thermal conductivity can be expressed as: 𝑄 ln(𝑡⁄𝑡0 ) 𝑘= 8.65 4𝜋 (∆𝑇 − ∆𝑇0 ) The sources of error with the line heat source probe method can be summarized as follows: 1. One source of error is due to finite probe length since the line heat source theory is based on a line heat source of infinite length. Heat flow in axial direction must be considered if the probe length is not long enough as compared to its diameter. Murakami, Sweat, Sastry, Kolbe,  Hayakawa, and Datta (1996b) have shown that error reduces to ≤0.1% if the ratio of the probe length to diameter is greater than 30. 2. Finite probe diameter and the fact that the probe has different thermal properties than the samples may cause error. Van der Held and van Drunen (1949) introduced the time correction factor, 𝑡𝑐 , which they subtracted from each time observation to correct for the effect of finite heater diameter and any resistance to heat transfer between the heat source and sample. Nix et al. (1967) reported that the time correction factor will be greater for larger probes because the deviation from the line heat source theory is greater for larger probes. The time correction factor is not necessary for proper probe design and also for food samples at temperatures above freezing because the specific heat of the probe is about the same as that of the food sample. To determine the time correction factor, 𝑡𝑐 , first temperature data are plotted with respect to time using arithmetic scales. Then, the instantaneous slope 𝑑𝑇/𝑑𝑡 is taken at several different times from this plot. Next, 𝑑𝑇/𝑑𝑡 values against time are plotted on an arithmetic scale. The intercept on the time axis gives the time correction factor 𝑡𝑐 , at which the rate of change of temperature 𝑑𝑇/𝑑𝑡 becomes zero. If a time correction factor is required, the thermal conductivity is expressed as: 𝑄 ln[(𝑡 − 𝑡𝑐 )⁄(𝑡0 − 𝑡𝑐 )] 𝑘= 8.66 4𝜋(∆𝑇 − ∆𝑇0 ) where 𝑘 = effective thermal conductivity (W/m °C), 𝑡 = time since the probe heater is energized (s), 𝑡0 = initial time (s), 𝑡𝑐 = time correction factor (s), 𝑇, 𝑇0 = temperatures of probe thermocouple at time 𝑡 and 𝑡0 , respectively (°C). 3. This method cannot be used for thin samples because the probe must be surrounded by a sufficient layer of sample. Finite sample size can cause errors if the sample boundaries experience a temperature change during measurement. The measurement time can be shortened and sample diameter can be increased to minimize the error arising from this situation. Excessive test duration also creates error as a result of convection and moisture transfer. If the plot of temperature versus the natural logarithm of time is linear (𝑟 2 > 0.99), it means the sample diameter is sufficient. The ratio of the probe diameter to sample (container) diameter must be less than 1/30 to minimize the error. In addition, if the 4𝛼𝑡/𝑑^2 < 0.6 (in  which 𝛼 is the thermal diffusivity, 𝑡 is the duration of experiment, and 𝑑 is the diameter of the cylinder), error is negligible according to Vos (1955). 4. Power input should be low. This is particularly important for frozen foods. Materials having higher thermal conductivity require higher power levels to obtain sufficient increase in temperature. High power inputs tend to cause localized fusion, resulting in highly variable properties over the experimental period. In addition, it creates a high temperature gradient, which may also cause moisture migration and heat convection. Sweat (1995) recommended that the power level should be selected depending on the temperature rise, which is necessary to get a high correlation between 𝑇 and ln 𝑡. 5. Truncation of infinite series of solution may also cause error. This error is minimized if the probe size is minimized and the experimental duration is optimized (the error is maximum at 𝑡 = 𝑡0 but decreases thereafter). 6. Contact resistance between sample and probe may cause error. This error shifts the time– temperature plot but does not change the slope. Therefore, no correction is required. 7. Convection effects may occur with liquid samples, but confining the analysis to the linear portion of the curve can eliminate it. It cannot be used with nonviscous liquids owing to convection currents which develop around the heated probe. Convection becomes important if PrGr > 100 – 1200 at which a deviation of straight line (𝑇 vs. ln 𝑡) starts. Agar may be used in the samples to form a gel to minimize convection effects. The probe method is not appropriate for determination of thermal conductivity at temperatures slightly below the initial freezing temperatures because of the large variation caused by ice formation. A longer probe and larger sample diameter may be required for measuring the thermal conductivity of ice using the line heat source probe method because of the high thermal diffusivity of ice. Although it is simple and fast, the line heat source probe method performs local measurements and therefore relies on the assumption that the food is homogeneous. Therefore, this method is not suitable for porous foods. Example 8.2. The line heat source probe method was used to determine the thermal conductivity of Red Delicious apples. The sample container was filled with the sample with the probe inserted at the center and it was placed in a constant temperature bath at 21°C for equilibration. After equilibrium was reached, the probe heater was activated. The electrical resistance of heated wire was 223.1 Ω and the electrical current through the  heated wire was measured as 0.14 A. Calculate the thermal conductivity of Red Delicious apples from the time–temperature data recorded (Table E.8.2.1). Table E.8.2.1. Time–Temperature Data Recorded Using Line Heat Source Probe Method for Red Delicious Apple Time (s) Temperature (°C) 5 21.00 10 21.51 15 21.72 20 21.97 25 22.15 30 22.29 35 22.33 40 22.44 45 22.57 50 22.63 55 22.72 60 22.77 Solution: The equation for the line heat source method is: 𝑄 𝑡 ∆𝑇 − ∆𝑇0 = ln ( ) 8.64 4𝜋𝑘 𝑡0 From the linear regression of (∆𝑇– ∆𝑇0 ) versus ln(𝑡/𝑡0 ), the slope of 𝑄/4𝜋𝑘 is obtained as 0.718 K (𝑟 2 = 0.995). The heat supplied per unit length (𝑄) is calculated from electrical resistance of heated source and the electrical current: 𝑄 = 𝐼2𝑅 𝑄 = (0.14)2 (223.1) = 4.37 W/m Then, from the slope, thermal conductivity of Red Delicious apples is calculated as: 𝑄 Slope = = 0.718 K 4𝜋𝑘 ⇒ 𝑘 = 0.499 W/m ∙ K (b) Transient Hot Wire Method. This method involves a thin heater wire similar to the line heat source method. However, in this case, the hot wire is located at the interface between the sample and a reference of known thermal conductivity (Fig. 8.8). The heater and temperature sensor in the hot wire thermal conductivity apparatus consist of a single wire that is exposed to the material, while in the thermal conductivity probe there are separate wires that are usually sealed in a tube.  Figure 8. 8. Transient hot wire method. When electrical power is applied to the heater wire, the temperature rise ∆𝑇 at a point located on the interface between the two materials at a distance 𝑥 from the heater wire may be described as: 2 2 𝑄𝜅𝜎 1 𝐸1 {−𝜅 𝑥 ⁄ } 4𝛼1 𝑡(𝜅 2 𝑢 + 1 + 𝑢) ∆𝑇 = − ∫ 2 𝑑𝑢 8.67 4𝜋𝑘1 (𝜅 𝑢 + 1 − 𝑢)1⁄2 (1 − 𝑢 + 𝜎 2 𝑢)3⁄2 0 where 𝐸1 {} is an integral exponential function and: 𝛼1 𝜅 = √( ) 8.68 𝛼2 𝑘2 √𝜎1 𝜎= 8.69 𝑘1 √𝜎2 where 𝑘 is the thermal conductivity and 𝛼 is the thermal diffusivity. The subscripts 1 and 2 refer to the thermal properties of sample and reference, respectively. If 𝑥 is small, Eq. (8.67) may be simplified to: 𝛼 𝑑∆𝑇 𝑄 (𝑘1 1 + 𝑘2 ) 𝑥 2 𝛼2 = {1 − } 8.70 𝑑 ln 𝑡 2𝜋(𝑘1 + 𝑘2 ) 𝑘1 + 𝑘2 4𝛼2 𝑡 𝑥2 If the thermal diffusivities of sample and reference are the same (𝛼1 ≈ 𝛼2 ), and < 10−2 , the 4𝛼2 𝑡 thermal conductivity of the sample can be determined from: 𝑄 𝑑 ln 𝑡 𝑘1 = − 𝑘2 8.71 2𝜋 𝑑𝑡 Note that if the two materials are the same (i.e., 𝑘1 = 𝑘2 ), the equation reduces to the thermal conductivity probe equation (Eq. 8.65). Sharity-Nissar et al. (2000) used a ribbon hot wire of width ℎ and an infinitesimal thickness instead of a line heat source. The solution is then:  𝛼1 𝑑∆𝑇 𝑄 ℎ2 (𝑘1 𝛼2 + 𝑘2 ) 𝑥 2 = {1 − (1 + ) } 8.72 𝑑 ln 𝑡 2𝜋(𝑘1 + 𝑘2 ) 12𝑥 2 𝑘1 + 𝑘2 4𝛼2 𝑡 Sharity-Nissar et al. (2000) indicated that the method is more convenient for measurements under high pressures than the thermal conductivity probe method. One obvious advantage is that probe insertion is no longer an issue, because the heater wire is sandwiched between two different materials. The method works best if the thermal diffusivities of the two materials are nearly equal. Thus, it may be necessary to change the reference for each sample. Because of the similarity of this method to the line heat source method, many of the error sources would be expected to be similar. (c) Modified Fitch Method. One of the most common transient methods used to measure the thermal conductivity of low conductivity materials is the Fitch method. The Fitch method consists of a heat source or a sink in the form of a vessel filled with constant temperature liquid, and a sink or a source in the form of a copper plug insulated on all sides except one face through which heat transfer occurs. The sample is sandwiched between the vessel and the open face of the plug. Then, the temperature of the plug varies with time depending on the heat flow rate through the sample. Copper may be considered as lumped system since its thermal conductivity is high enough, and its temperature history may be used together with its mass and physical properties for calculation of the sample thermal conductivity. Bennett, Chace, and Cubbedge (1962) suggested a modified version of the commercially available Fitch apparatus for measuring thermal conductivity of soft materials such as fruits and vegetables. In the modification, a screw mechanism was provided to control the pressure on the sample. Zuritz et al. (1989) modified the Fitch apparatus to make the device suitable for measuring thermal conductivity of small food particles that can be formed into slabs. The cross section of the modified Fitch apparatus is shown in Fig. 8.9. In the Fitch apparatus, the sample is placed on the copper plug and equilibrated to room temperature. The initial sample temperature is recorded prior to the test. Then, the copper rod, the temperature of which is kept constant at a higher value than that of the sample and copper plug by circulating a fluid, is lowered and contacted with the sample. Time versus temperature data are recorded at the same time. One-dimensional heat transfer occurs through the slab. Assuming constant properties, the governing differential equation for the temperature field within the sample is expressed as:  𝜕𝑇 𝜕 2𝑇 =𝛼 8.73 𝜕𝑡 𝜕𝑥 2 where 𝛼 is the thermal diffusivity of the sample. Figure 8. 9. Cross-section of the modified Fitch apparatus. The sample is considered to be at an initial temperature of 𝑇0. One face of the sample is maintained at constant temperature (in contact with the vessel) and the other in contact with a perfect conductor (copper plug). Therefore, the differential equation (Eq. 8.74) can be solved with the following initial and boundary conditions: I. C. at 𝑡 = 0, 𝑇 = 𝑇0 for 0 < 𝑥 < 𝐿 8.74 B. C. 1 at 𝑥 = 0, 𝑇 = 𝑇Cu−rod for 𝑡 > 0 8.75 𝜕𝑇 B. C. 2 at 𝑥 = 𝐿, −𝑘 = 𝑚Cu−plug 𝑐𝑝Cu−plug for 𝑡 > 0 8.76 𝜕𝑥 where 𝑐𝑝Cu−plug = specific heat of the cupper plug (J/kg K), 𝑘 = thermal conductivity of the sample (W/m K), 𝐿 = sample thickness (m), 𝑚Cu−plug = mass of cupper plug (kg),  𝑇0 = initial temperature of both sample and copper plug (°C), 𝑇Cu−rod = temperature of cupper rod (°C). The analytical solution that satisfies the above differential equation is given by Carlslaw and Jaeger (1959) as: ∞ 2(𝜆2𝑛 + ℎ2 ) exp(−𝛼𝜆2𝑛 𝑡) sin(𝜆𝑛 𝐿) 𝑇 = 𝑇Cu−rod + (𝑇0 − 𝑇Cu−rod ) ∑ 8.77 𝜆𝑛 [𝐿(𝜆2𝑛 + ℎ2 ) + ℎ] 𝑛=1 where 𝜌𝑐𝑝 ℎ= 8.78 𝑚Cu−plug 𝑐𝑝Cu−plug and 𝜆𝑛 are the roots of: 𝜆𝑛 tan(𝜆𝑛 𝐿) = ℎ 8.79 The assumption of quasi-steady conduction heat transfer through the sample, thereby ignoring the energy storage in the sample, yields the following simplified equation: 𝑘𝐴(𝑇 − 𝑇Cu−rod ) 𝑑𝑇 = 𝑚Cu−plug 𝑐𝑝Cu−plug 8.80 𝐿 𝑑𝑡 Integrating Eq. (8.80) from 𝑡 = 0 to any time 𝑡: 𝑡 𝑇 𝑘𝐴 𝑑𝑇 ∫ 𝑑𝑡 = ∫ 8.81 𝐿𝑚Cu−plug 𝑐𝑝Cu−plug 𝑇 − 𝑇Cu−rod 0 𝑇0 𝑇0 − 𝑇Cu−rod 𝑘𝐴𝑡 ⇒ ln ( )= 8.82 𝑇 − 𝑇Cu−rod 𝐿𝑚Cu−plug 𝑐𝑝Cu−plug where 𝐴 = heat transfer area (m2), 𝑇 = temperature of both copper plug and sample at time t (°C), 𝑇Cu−rod = temperature of copper rod (°C). 𝑇0 −𝑇Cu−rod As can be seen in Eq. (8.82), a plot of ln ( ) versus time is a straight line and 𝑇−𝑇Cu−rod thermal conductivity is calculated from the slope. Data must be analyzed in the linear temperature history region. Based on visual inspection, the initial data points are discarded to eliminate transient initial effects. A satisfactory fit was arbitrarily defined as a straight line with coefficient of determination (𝑟 2 ) ≥ 0.995. When 𝑟 2 is below this value, the topmost points were discarded, one point at a time until the 𝑟 2 is at least 0.995. Mohsenin (1980) suggested that each Fitch apparatus should be calibrated with standard equipment such as a heat flow meter. The correction factor (𝛽) for the apparatus can be calculated from the ratio between the thermal conductivity determined using standard apparatus and the thermal conductivity measured using the Fitch method. Then, the thermal conductivity of any material using the Fitch method is equal to the measured value multiplied by the correction factor. While the analytical solution is the more accurate and general one, the simplified model is more frequently chosen because of its simplicity and ease of use. Equation (4.80) is valid only if the following assumptions hold: 1. Contact resistance is negligible: This is the most important source of error in this method and it is especially important for high thermal conductivity materials. Good contact can be achieved if the contact surfaces are smooth and the sample faces are parallel. For rigid samples, pressure can be applied to provide good contact. However, for soft samples, it may not be possible. For these kinds of samples, application of a thin layer of nonwetting liquid to eliminate the air gaps between contact surface and porous materials is suggested. 2. Heat storage in the sample is negligible (i.e., heat transfer in the sample is quasi-steady state): For a given material, the sample thickness should be as small as possible to make a quasi- steady-state assumption. For the heat storage in the sample to be negligible with respect to that in the copper plug, the following relationship must hold: 𝑑𝑇𝑠⁄ 𝑚𝑠 𝑐𝑝𝑠 ( 𝑑𝑡 ) ≪1 8.83 𝑑𝑇 𝑚Cu−plug 𝑐𝑝Cu−plug ( 𝑐⁄𝑑𝑡) where 𝑚𝑠 , 𝑚Cu−plug = mass of sample and copper plug, respectively (kg), 𝑐𝑝𝑠 , 𝑐𝑝Cu−plug = specific heat of sample and copper plug, respectively (J/kg K). Since the time rates of temperature change for both sample and the copper plug are of the same order of magnitude, the following relationship is obtained: 𝑚𝑠 𝑐𝑝𝑠 ≪1 8.84 𝑚Cu−plug 𝑐𝑝Cu−plug The mass of the sample can be written as: 𝑚𝑠 = 𝜌𝑠 (𝜋𝑅 2 𝐿max ) 8.85 where 𝐿max = the upper limit of optimum sample thickness (m), 𝑅 = radius of sample (m), 𝜌𝑠 = density of sample (kg/m3). The upper limit of the optimum sample thickness, 𝐿max , can be estimated by substituting Eq. (8.85) into Eq. (8.84). 3. The heat transfer at the edges of the sample and copper plug is negligible: The error coming from this assumption is minimal if the temperature gradient is small, the sample is thin (area of heat transfer to the air is small), and thermal conductivity of the surroundings is low. 4. The temperature of the copper rod is constant: This can be achieved by contact with a large thermal mass at constant temperature and by using a vacuum flask. 5. The initial temperature of the copper plug and the sample are the same: This can be achieved by holding the sample and the copper plug at the same temperature environment prior to the experiment. 6. The temperature distribution in the copper plug is uniform: The high thermal conductivity and the small size of plug satisfy this condition. For the lumped system assumption to be valid, the internal to external heat transfer resistances of copper plug should be less than 0.1. 7. The sample is homogeneous across the heat transfer area: A larger apparatus size may be necessary for porous materials with pore sizes comparable to the heat transfer area. Example 8.3. Thermal conductivity of an apple sample is measured by using a Fitch apparatus consisting of a copper rod and a copper plug. The sample was shaped to obtain a disk with 7.5 mm diameter and 3.0 mm thickness. The initial temperature of both sample and copper plug was 25°C. After equilibrium was reached, the copper rod, which has a constant temperature at 35°C, was lowered, making good contact with the sample surface and the temperature variation of copper plug was recorded (Table E.8.3.1). Table E.8.3.1. Time–Temperature Data Recorded Using Fitch Method for Apple Time (s) Temperature (°C) 0 25.00 5 25.08 10 25.16 15 25.24 20 25.32 25 25.39 30 25.47 (a) Calculate the thermal conductivity of the apple sample if the mass and specific heat of copper plug are 12.0 g and 385 J/kg.°C, respectively. (b) Is it reasonable to assume that the heat storage in the sample is negligible? (The mass and specific heat values of the sample are 0.32 g and 4019 J/kg°C, respectively and assume that the time rate of temperature change for both the sample and the copper plug are the same.) Solution: (a) The equation for the modified Fitch method is:  𝑇0 − 𝑇Cu−rod 𝑘𝐴𝑡 ln ( )= 8.82 𝑇 − 𝑇Cu−rod 𝐿𝑚Cu−plug 𝑐𝑝Cu−plug Heat transfer area is: 2 7.5 × 10−3 m 𝐴 = 𝜋( ) = 4.4 × 10−5 m2 4 𝑇0 −𝑇Cu−rod Using the data, ln ( ) is calculated as a function of time (Table E.8.3.2). 𝑇−𝑇Cu−rod 𝑇0 −𝑇Cu−rod Table E.8.3.2. ln ( ) Values as a Function of Time. 𝑇−𝑇Cu−rod 𝑇0 − 𝑇Cu−rod Time (s) ln ( ) 𝑇 − 𝑇Cu−rod 0 0 5 0.008032 10 0.016129 15 0.024293 20 0.032523 25 0.039781 30 0.048140 𝑇0 −𝑇Cu−rod Plotting ln ( ) versus time gives a straight line and thermal conductivity is 𝑇−𝑇Cu−rod calculated from the slope (Fig. E.8.3.1): 0𝑇 −𝑇 Cu−rod Figure E.8 3.1 Plot of ln ( 𝑇−𝑇 ) versus time for calculation of thermal conductivity. Cu−rod 𝑘𝐴 Slope = = 0.0016 s −1 𝐿𝑚𝑐 𝑐𝑝𝑐 (0.0016)(3 × 10−3 )(12 × 10−3 )(385) ⇒𝑘= 4.4 × 10−5 𝑘 = 0.504 W/m℃  (b) If the heat storage in the sample is negligible: 𝑚𝑠 𝑐𝑝𝑠 (𝑑𝑇𝑠 ⁄𝑑𝑡) ≪1 8.83 𝑚Cu−plug 𝑐𝑝Cu−plug (𝑑𝑇𝑐 ⁄𝑑𝑡) 𝑑𝑇𝑠 𝑑𝑇𝑐 ≅ 𝑑𝑡 𝑑𝑡 𝑚𝑠 𝑐𝑝𝑠 ≪1 8.84 𝑚Cu−plug 𝑐𝑝Cu−plug (0.32)(4019) = 0.28 ≪ 1 (12)(385) Therefore, it is reasonable to assume that heat storage in the apple sample is negligible. (d) Point Heat Source Method. This method involves a point heat source, which is heated for a period of time followed by monitoring of its temperature as the heat dissipates through the sample. The typical device used for this purpose is a thermistor that serves as both a heating element and a temperature sensor. The analytical solution is given in the following equation: 𝑄 𝑟 𝑇 − 𝑇0 = erfc ( ) 8.86 4𝜋𝑘𝑟 √4𝛼𝑡 where 𝑥 2 2 2 2 12 4 erfc(𝑥) = 1 − erf(𝑥) = 1 − ∫ exp(−𝜂 2 ) 𝑑𝜂 = 1 − ∫ (1 − 𝜂 + 𝜂 − ⋯) √𝜋 √𝜋 2! 4! 0 8.87 2𝑥 2𝑥 3 = 1− +3 −⋯ √𝜋 √√𝑥 For small values of 𝑟⁄ , the higher order terms of complementary error function (erfc) drop √4𝛼𝑡 out and Eq. (8.87) becomes: 𝑄 1 1 𝑇 − 𝑇0 = ( )( − ) 8.88 4𝜋𝑘 𝑟 √𝜋𝛼𝑡 Thus, if (𝑇 − 𝑇0 ) is plotted against 1/√𝑡 , the thermal conductivity can be found from the intercept first and then the thermal diffusivity can be determined from the slope. The effective value of 𝑟 must be determined for the point heat source by conducting experiments using reference materials with known thermal properties. The theoretical analysis of Voudouris and Hayakawa (1994) indicated that the smaller the thermistor size, the smaller the dimensions of the sample. Much smaller thermistors are necessary for a measurement that is competitive with the line heat source method.  (e) Comparative Method. The comparative method is simple and a range of sample sizes can be handled via this method. Overall thermal conductivity measurement is possible rather than local measurement. Therefore, it is suitable for porous foods such as cakes. It was successfully applied to the measurement of thermal conductivity of a variety of porous food analogues having a porosity between 0 and 0.65 and a mean pore size relative to the size of the sample container between 10−6 and 4 × 10−3 (volume basis). This method involves cooling of two spheres side by side in a well stirred ice/water bath. One sphere contains the sample and the other contains a reference of known thermal conductivity. The thermal conductivity of the sample is calculated from the time–temperature data of the cooling spheres. It is based on the analytical solution for the center temperature of a sphere being cooled with convection boundary conditions as described by: ∞ 𝑇 − 𝑇∞ 2Bi(𝛽𝑖2 + (Bi − 1)2 ) sin 𝛽𝑖 =∑ exp(−𝛽𝑖2 Fo) 8.89 𝑇0 − 𝑇∞ 2 𝛽 (𝛽 + Bi(Bi − 1)) 𝑖=1 𝑖 𝑖 where Fo and Bi are the Fourier and Biot numbers, respectively, and 𝑇0 = initial temperature (°C), 𝑇∞ T∞ = bulk fluid temperature (°C). The 𝛽𝑖 coefficients are the roots of 𝛽𝑖 cot 𝛽𝑖 + (Bi − 1) = 0. The solution involves infinite series which are difficult to deal with. However, the terms in the solution converge rapidly with increasing time. Approximating the infinite series in Eq. (8.89) by the first term alone for Fo > 0.2 and using the temperature–time data, the thermal conductivity of the sample can be determined. The temperature–time histories of the cooling spheres are plotted on a logarithmic scale and the slope of the linear portion of the graph is determined: 𝑇 − 𝑇∞ 𝛽𝑖2 𝛼 ln ( )=𝐵− 2 𝑡 8.90 𝑇0 − 𝑇∞ 𝑅 Calculations are made for the sample and reference spheres relative to each other. Then: 2 𝑘sample Slopesample 𝜌sample 𝑐𝑝sample 𝑅sample 𝛽12ref = 2 8.91 𝑘ref Sloperef 𝜌ref 𝑐𝑝ref 𝑅ref 𝛽12sample The end losses are eliminated by using spherical sample containers in measurements. However, the measurement takes a long time. The temperature dependence of the thermal conductivities of the materials involved is not determined and instead temperature averaged conductivity values are measured. It is an indirect method for thermal conductivity measurement. Density and specific heat data are required since the method directly gives thermal diffusivity. SPECIFIC HEAT Specific heat is the amount of heat required to increase the temperature of a unit mass of the substance by unit degree. Therefore, its unit is J/kg K in the SI system. The specific heat depends on the nature of the process of heat addition in terms of either a constant pressure process or a constant volume process. However, because specific heats of solids and liquids do not depend on pressure much, except extremely high pressures, and because pressure changes in heat transfer problems of agricultural materials are usually small, the specific heat at constant pressure is considered. The specific heats of foodstuffs depend very much on their composition. Knowing the specific heat of each component of a mixture is usually sufficient to predict the specific heat of the mixture. Prediction of Specific Heat The specific heat of high-moisture foods is largely dominated by water content. As early as 1892, Siebel argued that the specific heat of the food can never be greater than the sum of the specific heat of the solid matter and water since water in food materials exists side by side with the solid matter without any heat producing chemical reactions. In addition, the specific heat of food materials cannot be much smaller than that of water since water is the major component of food materials. Siebel (1892) proposed the following equation for aqueous solutions such as vegetable and fruit juices or pastes: 𝑤 𝑐𝑝 = 0.837 + 3.349𝑋𝑤 8.92 Siebel (1892) also suggested the following equation for food materials below their freezing point: 𝑤 𝑐𝑝 = 0.837 + 1.256𝑋𝑤 8.93 𝑤 is the mass fraction of moisture within the sample and specific heat, and 𝑐 is given in where 𝑋𝑤 𝑝 kJ/kg K. The reason for the lower values of 𝑐𝑝 below freezing is that the specific heat of ice is about one half of that of the liquid water. This also partly explains the higher thawing times of foods as compared to their freezing times. Heldman (1975) proposed the following equation to estimate the specific heat of foodstuffs using the mass fraction of its constituents (water, protein, fat, carbohydrate, and ash): 𝑤 𝑤 𝑤 𝑤 𝑤 𝑐𝑝 = 4.180𝑋water + 1.547𝑋prot + 1.672𝑋fat + 1.42𝑋CHO + 0.836𝑋ash 8.94 Choi and Okos (1986) suggested the following equation for products containing 𝑛 components:  𝑛 𝑐𝑝 = ∑ 𝑋𝑖𝑤 𝑐𝑝𝑖 8.95 𝑖=1 where 𝑋𝑖𝑤 = mass fraction of component 𝑖 , 𝑐𝑝𝑖 = specific heat of component 𝑖 (J/kg K). The temperature dependence of specific heat of major food components has been studied. The specific heat of pure water, carbohydrate (CHO), protein, fat, ash, and ice at different temperatures can be expressed empirically in J/kg°C as follows: 𝑐𝑝water = 4081.7 − 5.3062𝑇 + 0.99516𝑇 2 (for − 40 to 0℃) 8.96 𝑐𝑝water = 4176.2 − 0.0909𝑇 + 5.4731 × 10−3 𝑇 2 (for 0 to 150℃) 8.97 𝑐𝑝CHO = 1548.8 + 1.9625𝑇 − 5.9399 × 10−3 𝑇 2 (for − 40 to 150℃) 8.98 𝑐𝑝protein = 2008.2 + 1.2089𝑇 − 1.3129 × 10−3 𝑇 2 (for − 40 to 150℃) 8.99 𝑐𝑝fat = 1984.2 + 1.4373𝑇 − 4.8008 × 10−3 𝑇 2 (for − 40 to 150℃) 8.100 𝑐𝑝ash = 1092.6 + 1.8896𝑇 − 3.6817 × 10−3 𝑇 2 (for − 40 to 150℃) 8.101 𝑐𝑝ice = 2062.3 + 6.0769𝑇 8.102 where temperature (𝑇) is in (°C) in these equations. Specific heat of moist air can be expressed as a function of relative humidity (RH) of air: 𝑐𝑝moist air = 𝑐𝑝dry air (1 + 0.837RH) 8.103 Generally, experimentally determined specific heat is higher than the predicted value. The reason may be the presence of bound water, variation of specific heat of the component phases with the source and interaction of the component phases. Rahman (1993) considered the excess specific heat, 𝑐ex , due to the interaction of the component phases and proposed the following equation: 𝑛 𝑐𝑝 = [∑ 𝑋𝑖𝑤 𝑐𝑝𝑖 ] − 𝑐ex 8.104 𝑖=1 Rahman (1993) correlated the excess specific heat for fresh seafood as: 𝑤 ) − 53.76(𝑋 𝑤 )2 𝑐ex = −33.77 + 85.58(𝑋𝑤 𝑤 8.105 These equations are valid in a temperature change when there is no phase change. If there is a phase change, latent heat must be incorporated. This is accomplished by using a new term, apparent specific heat, which includes both latent and sensible heat: 𝑑𝐻 𝑐𝑝app = 8.106 𝑑𝑇 Example 8.4. Estimate the specific heat of potatoes containing 85% water. Data: 𝑐𝑝water = 4186.80 J⁄kgK 𝑐𝑝nonfat solids = 837.36 J⁄kgK Solution: Using Eq. (8.95) as suggested by Choi and Okos (1986): 𝑛 𝑐𝑝 = ∑ 𝑋𝑖𝑤 𝑐𝑝𝑖 8.95 𝑖=1 𝑐𝑝 = (0.85)(4186.8) + (0.15)(837.36) = 3684.38 J⁄kg K Example 8.5. Calculate the specific heat of wild rice grain at 20°C with the approximate composition data given in Table E.8.5.1. Table E.8.5.1. Approximate Composition of Rice Component Weight (%) Water 8.5 Carbohydrate 75.3 Protein 14.1 Fat 0.7 Ash 1.4 Solution: Using the specific heat correlations, specific heat of each component at 20°C can be calculated. The results are tabulated in Table E.8.5.2. Table E.8.5.2. Predicted Specific Heat Values of Components at 20°C Introducing these values into Eq. (8.95) and using the composition of rice, 𝑐𝑝 of rice grain is determined. 𝑛 𝑐𝑝 = ∑ 𝑋𝑖𝑤 𝑐𝑝𝑖 8.95 𝑖=1 𝑐𝑝 = (0.085)(4176.6) + (0.753)(1585.7) + (0.141)(2031.9) + (0.007)(2011) + (0.014)(1128.9) = 1865.4 J⁄kg K Measurement of Specific Heat  Methods of mixture, guarded plate, comparison calorimeter, adiabatic agricultural calorimeter, differential scanning calorimeter (DSC), and calculated specific heat are some of the methods used for the determination of specific heat. Method of Mixture The method of mixture is the most widely used system for measuring specif