Create Presentation
Download Presentation

Download Presentation

Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8

Download Presentation
## Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**Test Monday 3/2, 2:30 – when you’re finished Lecture Wed 3/4 and Friday 3/6: 2:30 – 3:20 123AH See schedule for topics Lab reports due next Wednesday 3/4**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**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 All include heat transfer and are dictated by thermal properties of material Generally diffusion of water in or out is involved**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**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 moves similar to heat by conduction Moisture diffusivity Volume change due to moisture content change**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**Terms: Specific heat Thermal conductivity Thermal diffusivity Thermal expansion coefficient Surface heat transfer coefficient Sensible and Latent heat Enthalpy**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**Specific heat: Amount of heat required to raise the temp. of one unit of mass one degree. Cp = specific heat at constant pressure Cp =4.18 kJ/kg-K = 1.00 BTU/lb-R=1.00 cal/g-K for water (unfrozen) oils and fats: ½ H2O See Table 8.1 pg. 219 grains, powders: ¼ - 1/3 H2O ice: ½ H2O Good list: http://www.engineeringtoolbox.com/specific-heat-capacity-food-d_295.html Q = quantity of heat required to change temperature of a mass Q = Mcp(T2-T1) M = mass or weight**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**For liquid H2O Cp = 0.837 + 3.348 M above freezing For solid H2O Cp = 0.837 + 1.256 M below freezing**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**Thermal Conductivity: measure of ability to transmit heat dQ/dt = -kA (dT/dx) K = coefficient of thermal conductivity W/m°K, Btu/h ft°F, 1 Btu/h ft °F = 1.731 W/m °K Greater the water content, the greater the thermal conductivity Tables 8.2 and 8.3**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**If we don’t know t-conductivity, approximate using... K = VwKw + VsKs K = KwXw + Ks(1-Xw) where X = decimal fraction so K = f(all the constituent volumes) Example 8.1 pg 224**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**Thermal Diffusivity, α, (m2/sec or ft2/sec) Material’s ability to conduct heat relative to its ability to store heat α = k/(ρcp) Estimate the thermal diffusivity of a peach at 22 C.**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**Surface Heat Transfer Coefficient, h: Placed in a flowing stream of liquid or gas, the solid’s T will change until it eventually reaches equilibrium with the fluid Q/T = hA(T2 – T1) “h” is determined experimentally Look for research that matches your needs. (bottom of pg 227)**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**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**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**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 0C and 4x that amount to freeze food) 420 kJ to raise T of water from 0C to 100C, 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…)**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**Latent Heat, L, (kJ/kg or BTU/lb) 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**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**Enthalpy, h, (kJ/kg or BTU/lb) Heat content of a material. Combines latent heat and sensible heat changes ΔQ = M(h2-h1)…amount of heat to raise a product from T1 to T2 ASHRAE Handbook of Fundamentals When data is not available use eqtn. 8.15 pg 230. Δh = M cp(T2 – T1) + MXw L**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**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.**Lecture 12: Thermal Properties, Moisture DiffusivityChpt 8**Problem 1: Determine the amount of heat removed from 2 kg of sour cherries when cooled from 28C to -7C. Assume MC of 92.3% and at -7C, 27% won’t be frozen. Problem 2: Estimate the thermal diffusivity of cheddar cheese at 22°C.