Presentation on theme: "Thermal Properties and Moisture Diffusivity"— Presentation transcript:
1Thermal Properties and Moisture Diffusivity BAE2023
2Thermal Properties, Moisture Diffusivity Processing and Storage of Ag ProductsHeatingCoolingCombination of heating and coolingGrain dried for storageNoodles driedFruits/Vegetables rapidly cooledVegetables are blanched, maybe cooked and cannedPowders such as spices and milk: dehydratedCooking, cooling, baking, pasteurization, freezing, dehydration: all involve heat transferDesign of such processes require knowledge of thermal properties of material
3Continue…. Heat is transferred by Conduction: Temperature gradient exists within a body…heat transfer within the bodyConvection: Heat transfer from one body to another by virtue that one body is moving relative to the otherRadiation: Transfer of heat from one body to another that are separated in space in a vacuum. (blackbody heat transfer)We’ll considerConduction w/in the productConvection: transfer by forced convection from product to moving fluidMoisture movement through agricultural product is similar to movement of heat by conductionMoisture diffusivityVolume change due to moisture content change
4Continue…. Terms used to define thermal properties Specific heat Thermal conductivityThermal diffusivityThermal expansion coefficientSurface heat transfer coefficientSensible and Latent heatEnthalpy
5Specific HeatSpecific Heat: Amount of heat required to raise temperature of unit amount of substance by one degreeUnits:𝐾𝐽 𝐾𝑔°𝐾 (SI System)𝐵𝑇𝑈 lb°𝐹 (English System)𝐶𝑎𝑙. 𝑔°𝐾 (if calories are used)Conversion of units: 1 𝐾𝐽 𝐾𝑔°𝐾 = 𝐵𝑇𝑈 lb°𝐹 =0.239 𝐶𝑎𝑙. 𝑔°𝐾
6Specific Heat Q = M Cp (T2-T1) Once Specific heat of material is known, then the amount of heat (Q) needed to increase temp. from T1 to T2 is calculated by:Q = M Cp (T2-T1)Where,Q = quantity of heat required to change temperature of a massCp = Specific heat at constant pressureM = mass or weightWater is a major component of all agricultural productsCp of water = = 1Cp of oils and fats = ½ of Cp of water ………See Table 8.1 pg. 219Cp of grains, powders = ¼ to 1/3 of Cp of waterCp of ice = ½ Cp of H2O ( therefore, less heat required to raise temp. of frozen product then the same product when it is thawed)
7Specific heat Eq. for calculating Cp based on moisture Content For liquid H2OCp = M above freezingFor solid H2OCp = M below freezingEq. based on compositionCp=4.18Xw+1.711Xp+1.928Xf Xc+0.908XaX is the mass or weight fraction of each componentThe subscript denote following components: w=water, p= protein, f=fat, c= carbohydrate, a=ash
8Thermal Conductivity (k) Measure of ability to transmit heat= -k AK= coefficient of thermal conductivityFor 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 directionUnits:(SI system)(English System) 1 = 1.731
9Thermal Conductivity (k) k water =0.566 at 0°C= at 20°C= at 60°CAt 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 productAnother factor is porosity e.g. freeze dried products and porous fruits like apple have low thermal conductivity.
10Thermal Conductivity (k) If we don’t know thermal conductivity, approximate using... K = kw Xw + ks(1-Xw) Where, Kw =Thermal conductivity of water Xw= Weight fraction of water Ks = Thermal conductivity of solids = 𝑊 m°𝐾 If the moisture in product is more than 50%, then K = Xw
11Thermal 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 pressureExample : Estimate the thermal diffusivity of a peach at 22 C.
12Surface heat transfer coefficient (h) When a body is placed in flowing stream of liquid or gas, the body’s temperature will change until it eventually reaches in equilibrium with the fluid. In eq. form also known as Newton’s Law of cooling𝑑𝑄 𝑑𝑡 = h A (Tf- Ts)Where,h = surface heat transfer coefficient and has same units as k i.e. 𝑊 m°𝐾Tf = temp. of fluidTs = temp of solid bodyh depends on fluid velocity, surface characteristics of solids, size and shape of solid and fluid properties ( density and viscosity)Difficult to tabulate value of h, therefore experimentally determined
13Sensible 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
14Latent heat (L) Latent Heat, L, (kJ/kg or BTU/lb) Heat that is exchanged during a change in phaseDominated by the moisture content of foodsRequires 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 waterHeat of vaporization is about 7x greater than heat offusion (freezing)Therefore, evaporation of water is energy intensive (concentrating juices, dehydrating foods…)
15Latent heat (L)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
16Enthalpy (h) Units: (kJ/kg or BTU/lb) Heat content of a material. Used frequently to evaluate changes in heat content of steam or moist airCombines latent heat and sensible heat changesΔQ = M(h2-h1)Where,ΔQ = amount of heat needed to raise temperature from T1 to T2M = mass of producth2= enthalpy at temp T1h1 = enthalpy at temp T2
17Enthalpy (h)Approach useful when one of the temperatures is below freezingMeasurements 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 pointChange in enthalpy of a frozen food can be calculated from eq. below:Δh = M cp(T2 – T1) + MXw LXw is the mass fraction of water that undergoes phase change(frozen fraction)L is the latent heat of fusion of waterM is the mass of productΔh = Change in enthalpy of frozen food
18ExampleExample 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.
19Homework 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.