Topics in ProcessingThermal Properties and Moisture DiffusivityThermal Properties, Moisture DiffusivitySlide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 181Dr. C. L. JonesBiosystems and Ag. EngineeringTopics in Processing2Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties and Moisture Diffusivity •Assignment: –Read Chapter 8 in Stroshine Book3Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•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)4Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•Specific heat: Amount of heat required to raise the temp. of one unit of mass one degree.•Q = Mcp(T2-T1)•Q = quantity of heat,•M = mass or weight•Cp = specific heat at constant pressure•Cp =4.18 kJ/kg-K = 1.00 BTU/lb-R=1.00 cal/g-K for water5Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•For liquid H2O–Cp = 0.837 + 3.348 M above freezing•For solid H2O–Cp = 0.837 + 1.256 M below freezing6Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•Thermal Conductivity:•dQ/dt = -kA (dT/dx)•K = coefficient of thermal conductivity7Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•K = VwKw + VsKs•K = KwXw + Ks(1-Xw) where X = decimal fraction•so K = f(all the constituent volumes)8Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•Power: Pwatt = -KA(ΔT/ ΔX)•P = -KA (T2 – T1)/x•Kfats&oils<<Kwater•Kdrymaterials<<Kwater9Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•Force Convection:•Q/T = hA(T2 – T1)10Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•Thermal Diffusivity, α, (m2/sec)•Material’s ability to conduct heat relative to its ability to store heat•Estimate the thermal diffusivity of a peach at 22 C.•α = k/(ρcp)11Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•Water Phase Diagram12Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•Water Phase Diagram13Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•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…)14Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•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 Fundamentals15Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•Enthalpy, L, (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 230.16Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•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.17Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•Homework Due 4/20•Problem 1: do 8.1 in your book (see example of a solution•Problem 2: Determine the amount of heat removed from 2 kg of blueberries when cooled from 28C to -7C. Assume MC of 83% and at -7C, 27% won’t be frozen.18Dr. C. L. JonesBiosystems and Ag. EngineeringThermal Properties, Moisture Diffusivity•Sample solutions for problem #1•A thawed turkey (whole) is placed in a turkey fryer. The turkey is at 3 C while the oil is at 400 C. –The turkey is at 3 C while the oil is at 400 C. Assuming the oil temperature is maintained at 400 C, the temp. at the surface of the turkey (boundary condition) will remain at 400C throughout the frying process. This assumes that the heat transfer coefficient across the oil-turkey fry boundary does not limit the heat transfer process. The thermal conductivity of the turkey can be estimated from the perpendicular model since the meat is fibrous and the thermal conductivity will be directionally dependent. The turkey can be modeled as a finite
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