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Thermal Properties and Moisture Diffusivity BAE2023.

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Presentation on theme: "Thermal Properties and Moisture Diffusivity BAE2023."— Presentation transcript:

1 Thermal Properties and Moisture Diffusivity BAE2023

2 Thermal Properties, Moisture Diffusivity Processing and Storage of Ag Products ◦ Heating ◦ Cooling ◦ Combination of heating and cooling  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

3 Continue…. Heat is transferred by  Conduction: Temperature gradient exists within a body…heat transfer within the body  Convection: Heat transfer from one body to another by virtue that one body is moving relative to the other  Radiation: Transfer of heat from one body to another that are separated in space in a vacuum. (blackbody heat transfer) We’ll consider  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  Moisture diffusivity  Volume change due to moisture content change

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

5 Specific Heat

6 Once Specific heat of material is known, then the amount of heat (Q) needed to increase temp. from T 1 to T 2 is calculated by: Q = M C p (T 2 -T 1 ) Where, Q = quantity of heat required to change temperature of a mass C p = Specific heat at constant pressure M = mass or weight Water is a major component of all agricultural products C p of water = 4.18 = 1 C p of oils and fats = ½ of C p of water ………See Table 8.1 pg. 219 C p of grains, powders = ¼ to 1/3 of C p of water C p of ice = ½ C p of H 2 O ( therefore, less heat required to raise temp. of frozen product then the same product when it is thawed)

7 Specific heat Eq. for calculating C p based on moisture Content For liquid H 2 O  C p = M above freezing For solid H 2 O  C p = M below freezing Eq. based on composition C p =4.18X w X p X f X c X a 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

8 Thermal Conductivity (k) Measure of ability to transmit heat = -k A K= coefficient of thermal conductivity For one dimensional heat flow in x direction, k is numerically equal to the quantity of heat (Q) that will flow across a unit cross sectional area (A) per unit of time (t) in response to a temperature gradient () of 1 degree per unit distance in x direction Units: (SI system) (English System)1 = 1.731

9 Thermal Conductivity (k) k water =0.566 at 0°C = at 20°C = at 60°C  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.  Higher the moisture content higher will be thermal conductivity of food product  Another factor is porosity e.g. freeze dried products and porous fruits like apple have low thermal conductivity.

10 Thermal Conductivity (k)

11 Thermal Diffusivity ( α ) α quantifies the materials ability to conduct heat relative to its ability to store heat. α = Where, α = Thermal Diffusivity, Units () or () k = Thermal conductivity = density of material = Specific heat at constant pressure Example : Estimate the thermal diffusivity of a peach at 22 C.

12 Surface heat transfer coefficient (h)

13 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

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

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

16 Enthalpy (h) Units: (kJ/kg or BTU/lb)  Heat content of a material.  Used frequently to evaluate changes in heat content of steam or moist air Combines latent heat and sensible heat changes Δ Q = M(h 2 -h 1 ) Where, Δ Q = amount of heat needed to raise temperature from T1 to T2 M = mass of product h 2 = enthalpy at temp T 1 h 1 = enthalpy at temp T 2

17 Enthalpy (h) Approach useful when one of the temperatures is below freezing  Measurements based on zero values of enthalpy at a specified temperature e.g. at - 40°C, -18°C or 0°C.  Enthalpy changes rapidly near the freezing point Change in enthalpy of a frozen food can be calculated from eq. below: Δ h = M c p (T 2 – T 1 ) + MX w L X w 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 Δ h = Change in enthalpy of frozen food

18 Example Example 8.3: Calculate the amount of heat which must be removed from 1 kg of raspberries when their temperature is reduced from 25C to -5C. Assume that the specific heat of raspberries above freezing is 3.7 kJ/kgC and their specific heat below freezing is 1.86 kJ/kgC. The moisture content of the raspberries is 81% and the ASHRAE tables for freezing of fruits and vegs. Indicate that at -5C, 27% will not yet be frozen.

19 Homework Assignment Due February 20th Problem 1: Determine the amount of heat removed from 1.5 kg of bologna (sausage) when cooled from 24C to -7C. Assume MC of 59% and at -7C, 22% won’t be frozen. Problem 2: Estimate the thermal diffusivity of butter at 20°C.

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