# Thermal Properties and Moisture Diffusivity BAE 2023 Thermal

Thermal Properties and Moisture Diffusivity BAE 2023

Thermal Properties, Moisture Diffusivity Processing and Storage of Ag Products ◦ Heating ◦ Cooling ◦ Combination of heating and cooling Ø Ø Ø Ø Grain dried for storage Noodles dried Fruits/Vegetables rapidly cooled Vegetables are blanched, maybe cooked and canned Powders such as spices and milk: dehydrated Cooking, cooling, baking, pasteurization, freezing, dehydration: all involve heat transfer Design of such processes require knowledge of thermal properties of material

Continue…. Heat is transferred by Ø Conduction: Temperature gradient exists within a body…heat transfer within the body Ø Convection: Heat transfer from one body to another by virtue that one body is moving relative to the other Ø Radiation: Transfer of heat from one body to another that are separated in space in a vacuum. (blackbody heat transfer) We’ll consider Ø Ø Conduction w/in the product Convection: transfer by forced convection from product to moving fluid Moisture movement through agricultural product is similar to movement of heat by conduction Ø Ø Moisture diffusivity Volume change due to moisture content change

Continue…. Terms used to define thermal properties Ø Ø Ø Ø Specific heat Thermal conductivity Thermal diffusivity Thermal expansion coefficient Surface heat transfer coefficient Sensible and Latent heat Enthalpy

Specific Heat �

Specific Heat �

Specific heat Eq. for calculating Cp based on moisture Content For liquid H 2 O �Cp = 0. 837 + 3. 348 M above freezing For solid H 2 O �Cp = 0. 837 + 1. 256 M below freezing Eq. based on composition Cp=4. 18 Xw+1. 711 Xp+1. 928 Xf+1. 547 Xc+0. 908 Xa X is the mass or weight fraction of each component The subscript denote following components: w=water, p= protein, f=fat, c= carbohydrate, a=ash

Thermal Conductivity (k) �

Thermal Conductivity (k) k water =0. 566 at 0°C = 0. 602 at 20°C = 0. 654 at 60°C Ø At room temp. value of k for endosperm of cereal grains, flesh of fruits and veg. , dairy products, fats and oil and sugar are less than that of water. Ø Higher the moisture content higher will be thermal conductivity of food product Ø Another factor is porosity e. g. freeze dried products and porous fruits like apple have low thermal conductivity.

Thermal Conductivity (k) �

Thermal Diffusivity (α) �

Surface heat transfer coefficient (h) �

Sensible and Latent heat Sensible heat: Temperature that can be sensed by touch or measured with a thermometer. Temperature change due to heat transfer into or out of product Latent heat: Transfer of heat energy with no accompanying change in temperature. Happens during a phase change. . . solid to liquid. . . liquid to gas. . . solid to gas

Latent heat (L) Latent Heat, L, (k. J/kg or BTU/lb) Ø Heat that is exchanged during a change in phase Ø Dominated by the moisture content of foods Ø Requires more energy to freeze foods than to cool foods (90 k. J removed to lower 1 kg of water from room T to 0 °C and 4 x that amount to freeze food) Ø 420 k. J to raise T of water from 0 ° C to 100 ° C, 5 x that to evaporate 1 kg of water Heat of vaporization is about 7 x greater than heat of fusion (freezing) Ø Ø Therefore, evaporation of water is energy intensive (concentrating juices, dehydrating foods…)

Latent heat (L) Determine L experimentally when possible. When data is not available (no tables, etc) use…. L = 335 Xw where Xw is weight fraction of water Many fruits, vegetables, dairy products, meats and nuts are given in ASHRAE Handbook of Fundamentals

Enthalpy (h) Units: (k. J/kg or BTU/lb) Ø Heat content of a material. Ø Used frequently to evaluate changes in heat content of steam or moist air Combines latent heat and sensible heat changes ΔQ = M(h 2 -h 1) Where, ΔQ = amount of heat needed to raise temperature from T 1 to T 2 M = mass of product h 2= enthalpy at temp T 1 h 1 = enthalpy at temp T 2

Enthalpy (h) Approach useful when one of the temperatures is below freezing Ø Measurements based on zero values of enthalpy at a specified temperature e. g. at -40°C, -18°C or 0°C. Ø Enthalpy changes rapidly near the freezing point Change in enthalpy of a frozen food can be calculated from eq. below: Δh = M cp(T 2 – T 1) + MXw L Xw is the mass fraction of water that undergoes phase change(frozen fraction) L is the latent heat of fusion of water M is the mass of product Δh = Change in enthalpy of frozen food

Example 8. 3: Calculate the amount of heat which must be removed from 1 kg of raspberries when their temperature is reduced from 25 C to -5 C. Assume that the specific heat of raspberries above freezing is 3. 7 k. J/kg. C and their specific heat below freezing is 1. 86 k. J/kg. C. The moisture content of the raspberries is 81% and the ASHRAE tables for freezing of fruits and vegs. Indicate that at -5 C, 27% will not yet be frozen.

Homework Assignment Due February 20 th Problem 1: Determine the amount of heat removed from 1. 5 kg of bologna (sausage) when cooled from 24 C to -7 C. Assume MC of 59% and at -7 C, 22% won’t be frozen. Problem 2: Estimate thermal diffusivity of butter at 20°C.

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