Thermal Properties and Moisture Diffusivity

BAE2023

Thermal Properties, Moisture DiffusivityProcessing 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

Continue….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 movement through agricultural product is similar to movement of heat by conduction

Moisture diffusivity Volume change due to moisture content change

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

Specific HeatSpecific Heat: Amount of heat required to raise temperature of unit amount of substance by one degree

Units:◦ (SI System)◦ (English System)◦ (if calories are used)Conversion of units: = 0.239

Specific HeatOnce 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 mass

Cp = Specific heat at constant pressure

M = mass or weight

Water is a major component of all agricultural products

Cp of water = 4.18 = 1

Cp of oils and fats = ½ of Cp of water ………See Table 8.1 pg. 219

Cp of grains, powders = ¼ to 1/3 of Cp of water

Cp of ice = ½ Cp of H2O ( therefore, less heat required to raise temp. of frozen product then the same product when it is thawed)

Specific heatEq. for calculating Cp based on moisture Content

For liquid H2O Cp = 0.837 + 3.348 M above freezing

For solid H2O Cp = 0.837 + 1.256 M below freezing

Eq. based on compositionCp=4.18Xw+1.711Xp+1.928Xf+1.547 Xc+0.908Xa

X is the mass or weight fraction of each componentThe subscript denote following components: w=water, p= protein, f=fat, c= carbohydrate, a=ash

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

Thermal Conductivity (k)

k water =0.566 at 0°C

= 0.602 at 20°C = 0.654 at 60°C

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.

Thermal Conductivity (k)If we don’t know thermal conductivity, approximateusing...

K = kw Xw + ks(1-Xw)

Where,Kw =Thermal conductivity of water

Xw= Weight fraction of water

Ks = Thermal conductivity of solids = 0.259

If the moisture in product is more than 50%, thenK = 0.056 + 0.57Xw

Thermal 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 pressure

Example : Estimate the thermal diffusivity of a peach at 22 C.

Surface 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 cooling

= h A (Tf- Ts)

Where,h = surface heat transfer coefficient and has same units as k i.e. Tf = temp. of fluid

Ts = temp of solid body

h 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

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 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 gas

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…)

Latent 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, meatsand nuts are given in ASHRAE Handbook ofFundamentals

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 airCombines latent heat and sensible heat changes

ΔQ = M(h2-h1)

Where,ΔQ = amount of heat needed to raise temperature from T1 to T2M = mass of producth2= enthalpy at temp T1

h1 = enthalpy at temp T2

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 -40°C, -18°C or 0°C.

Enthalpy changes rapidly near the freezing point

Change in enthalpy of a frozen food can be calculated from eq. below:

Δh = M cp(T2 – T1) + MXw L

Xw 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 Δh = Change in enthalpy of frozen food

ExampleExample 8.3:Calculate the amount of heat which must be removedfrom 1 kg of raspberries when their temperature isreduced from 25C to -5C. Assume that the specific heat of raspberries abovefreezing is 3.7 kJ/kgC and their specific heat belowfreezing is 1.86 kJ/kgC.The moisture content of the raspberries is 81% and theASHRAE tables for freezing of fruits and vegs. Indicatethat at -5C, 27% will not yet be frozen.

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% won’t be frozen.

Problem 2: Estimate the thermal diffusivityof butter at 20°C.