# thermal properties and moisture diffusivity bae2023

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- Slide 1
- Thermal Properties and Moisture Diffusivity BAE2023
- Slide 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
- Slide 3
- Continue. Heat is transferred by Conduction: Temperature gradient exists within a bodyheat 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) Well 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
- Slide 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
- Slide 5
- Specific Heat
- Slide 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 = of C p of water See Table 8.1 pg. 219 C p of grains, powders = to 1/3 of C p of water C p of ice = C p of H 2 O ( therefore, less heat required to raise temp. of frozen product then the same product when it is thawed)
- Slide 7
- Specific heat Eq. for calculating C p based on moisture Content For liquid H 2 O C p = 0.837 + 3.348 M above freezing For solid H 2 O C p = 0.837 + 1.256 M below freezing Eq. based on composition C p =4.18X w +1.711X p +1.928X f +1.547 X c +0.908X 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
- Slide 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
- Slide 9
- Thermal Conductivity (k) k water =0.566 at 0C = 0.602 at 20C = 0.654 at 60C 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.
- Slide 10
- Thermal Conductivity (k)
- Slide 11
- Thermal Diffusivity ( ) quantifies the materials ability to conduct heat relative to its ability to store heat. = Where, = 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.
- Slide 12
- Surface heat transfer coefficient (h)
- Slide 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
- Slide 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)
- Slide 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
- Slide 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 Q = M(h 2 -h 1 ) Where, 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
- Slide 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 - 40C, -18C or 0C. Enthalpy changes rapidly near the freezing point Change in enthalpy of a frozen food can be calculated from eq. below: 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 h = Change in enthalpy of frozen food
- Slide 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.
- Slide 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% wont be frozen. Problem 2: Estimate the thermal diffusivity of butter at 20C.

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