Thermal Properties, Moisture DiffusivityChpt 8• Processing and Storage of Ag Products– Heating– Cooling– Combination of heating and cooling•Grain dried for storage1•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 involvedThermal 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 2–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 changeThermal Properties, Moisture DiffusivityChpt 8• Terms:–Specific heat–Thermal conductivity3–Thermal conductivity–Thermal diffusivity–Thermal expansion coefficient–Surface heat transfer coefficient–Sensible and Latent heat–EnthalpyThermal Properties, Moisture DiffusivityChpt 8Specific heat: Amount of heat required to raise the temp. of one unit of mass one degree.Cp= specific heat at constant pressureCp=4.18 kJ/kg-K = 1.00 BTU/lb-R=1.00 cal/g-K for water (unfrozen)4(unfrozen)oils and fats: ½ H2O See Table 8.1 pg. 219grains, powders: ¼ - 1/3 H2Oice: ½ H2OGood list:http://www.engineeringtoolbox.com/specific-heat-capacity-food-d_295.htmlQ = quantity of heat required to change temperature of a mass Q = Mcp(T2-T1)M = mass or weightThermal Properties, Moisture DiffusivityChpt 8For liquid H2OCp= 0.837 + 3.348 M above freezing5For solid H2OCp= 0.837 + 1.256 M below freezingThermal Properties, Moisture DiffusivityChpt 8Thermal Conductivity:measure of ability to transmit heatdQ/dt = -kA (dT/dx)K = coefficient of thermal conductivity6K = coefficient of thermal conductivityW/m°K, Btu/h ft°F, 1 Btu/h ft °F = 1.731 W/m °KGreater the water content, the greater the thermal conductivityTables 8.2 and 8.3Thermal Properties, Moisture DiffusivityChpt 8If we don’t know t-conductivity, approximate using...K = VwKw + VsKs7K = VwKw + VsKsK = KwXw + Ks(1-Xw) where X = decimal fractionso K = f(all the constituent volumes)Example 8.1 pg 224Thermal Properties, Moisture DiffusivityChpt 8Thermal Diffusivity, α, (m2/sec or ft2/sec)Material’s ability to conduct heat relative to its ability to store heat8its ability to store heatα = k/(ρcp)Estimate the thermal diffusivity of a peach at 22 C.Thermal Properties, Moisture DiffusivityChpt 8Surface Heat Transfer Coefficient, h:Placed in a flowing stream of liquid or gas, the solid’s T will change until it eventually 9the solid’s T will change until it eventually reaches equilibrium with the fluidQ/T = hA(T2 – T1)“h” is determined experimentallyLook for research that matches your needs. (bottom of pg 227)Thermal Properties, Moisture DiffusivityChpt 8Sensible heat: Temperature that can be sensed by touch or measured with a thermometer. Temperature change due to 10thermometer. Temperature change due to heat transfer into or out of productLatent heat: transfer of heat energy with no accompanying change in temperature. Happens during a phase change...solid to liquid...liquid to gas...solid to gasThermal 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 11(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…)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) 12•When data is not available (no tables, etc) use….• L = 335 Xw where Xwis weight fraction of water • Many fruits, vegetables, dairy products, meats and nuts are given in ASHRAE Handbook of FundamentalsThermal Properties, Moisture DiffusivityChpt 8• Enthalpy, h, (kJ/kg or BTU/lb)• Heat content of a material. •Combines latent heat and sensible heat changes13•Combines latent heat and sensible heat changes• ∆Q = M(h2-h1)…amount of heat to raise a product from T1to T2• ASHRAE Handbook of Fundamentals• When data is not available use eqtn. 8.15 pg 230. ∆h = M cp(T2– T1) + MXwLThermal 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.14reduced 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 8Due March 5thProblem 1: Determine the amount of heat removed from 3 kg of bologna (sausage) 15removed from 3 kg of bologna (sausage) when cooled from 23C to -7C. Assume MC of 59% and at -7C, 22% won’t be frozen. Problem 2: Estimate the thermal diffusivity of butter at
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