Thermal Properties

Thermal Properties, Moisture Diffusivity

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

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 areseparated 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 Heat

Specific Heat: Specific heat is the quantity of

heat that is gained or lost by a unit mass of

product to accomplish a unit change in

temperature, without a change in state

Units:

◦𝐽

𝑘𝑔°𝐾(SI System)

Conversion of units: 1𝐾𝐽

𝐾𝑔°𝐾= 0.239

𝐵𝑇𝑈

lb°𝐹= 0.239

𝐶𝑎𝑙.

𝑔°𝐾

Specific Heat

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 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 kJ/kg0C

Cp of oils and fats = ½ of Cp of water ………

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 heat

Eq. for calculating Cp based on moisture

Content

SI unit- J/kg K

One calorie equals 4.187 joules

Siebel’s equation for fat free materials

Above freezing Cavg = 3349M + 837.36 J/kgK

Below freezing Cavg = 1256M + 837.36 J/kgK

M- Mass fraction of water (ie. 0.8 or moisture wet basis)

Modified Siebel’s

Above freezing Cavg = 1674F + 837.36 SNF+4186M J/kgK

Below freezing Cavg = 1674.72F + 837.36 SNF+ 2093.4M J/kgK

SNF- solids non fat (fraction)

F – Fat (Fraction)

Based on composition

Heldman and Singh (1981)

Calculate the specific heat for a model

food with the following composition:

carbohydrate 40%, protein 20%, fat 10%,

ash 5%, moisture 25%.

Thermal Conductivity (k)

Measure of ability to transmit heatQ= -k A (dT)

l

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) (W/m K)

Thermal Conductivity (k)

k water =0.566 at 0°C

= 0.597 at 20°C

= 0.654 at 60°C

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.

■ Metals: 50 - 400 W/(m0C)

■Alloys: 10 - 120 W/(m0C)

■Air: 0.0251 W/(m0C) (at 200C)

■ Insulating materials: 0.035 - 0.173 W/(m0C)

Thermal Conductivity (k)

If we don’t know the 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 = 0.259 𝑊

m°𝐾

For meats and fish, temperature 0–60C, water

content 60–80%, wet basis, Sweat (1975)

k = 0.08 + 0.52Xw

Estimate the thermal conductivity of hamburger

beef that contains 68.3% water at 20 C.

Given

X m = 0.683

One face of a stainless-steel plate 1 cm thick is

maintained at 110°C, and the

other face is at 90°C. Assuming steady-state

conditions, calculate the rate of heat transfer

per unit area through the plate. The thermal

conductivity of stainless steel is 17 W/(m °C).

Thermal Diffusivity (α)

α quantifies the materials ability to conduct heat relative to its ability to store heat.

α = 𝑘

ρ𝐶𝑝Where,

α = Thermal Diffusivity, Units or m²/s.

k = Thermal conductivity

ρ = density of material

Cp = Specific heat at constant pressure (J/(kg·K)

Example : Estimate the thermal diffusivity of a peach at 22 C. k= 0.57 W/m.K, Cp = 3.6 kJ/kg.K, ρ = 670 kg/m3

Surface heat transfer coefficient (h)

(convective heat-transfer coefficient)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

𝑑𝑄

𝑑𝑡= h A (Tf- Ts)

Where,

h = surface heat transfer coefficient and has same units as k i.e. 𝑊

m°𝐾Tf = 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

Tf

The rate of heat transfer per unit area

from a metal plate is 1000 W/m 2 . The

surface temperature of the plate is 120°C,

and ambient temperature is 20°C.

Estimate the convective heat transfer

coefficient.

Radiation Heat Transfer

Radiation heat transfer occurs between

two surfaces by the emission and later

absorption of electromagnetic waves (or

photons). In contrast to conduction and

convection, radiation requires no physical

medium for its propagation—it can even

occur in a perfect vacuum, moving at the

speed of light,

Liquids are strong absorbers of radiation.

Gases are transparent to radiation, (except

some gases absorb radiation of a particular wavelength (for example, ozone absorbs

ultraviolet radiation).

Solids are opaque to thermal radiation.

Solid materials, such as with solid foods,

our analysis is concerned primarily with

the surface of the material.

All objects at a temperature above 0

Absolute emit thermal radiation.

Thermal radiation emitted from an

object’s surface is proportional to the

absolute temperature raised to the fourth

power and the surface characteristics.

where σ is the Stefan–Boltzmann constant, equal to 5.669 x 10 -8 W/

(m2 K 4); T A is temperature, Absolute; A is the area (m 2); and ε is emissivity,

which describes the extent to which a surface is similar to a blackbody. For

a blackbody, the value of emissivity is 1.

where σ is the Stefan–Boltzmann constant,

equal to 5.669 x 10 -8 W/(m2 K 4);

T A is temperature, Absolute;

A is the area (m 2); and

ε is emissivity,

which describes the extent to which a

surface is similar to a blackbody. For a

blackbody, the value of emissivity is 1.

Calculate the rate of heat energy emitted

by 100 m 2 of a polished iron surface

(emissivity 0.06) as shown in Figure E4.5 .

The temperature of the surface is 37°C.

STEADY-STATE HEAT TRANSFER

In problems involving heat transfer, we

often deal with steady state and unsteady

state (or transient) conditions. Steady-

state conditions imply that time has no

influence on the temperature distribution

within an object, although temperature may

be different at different locations within the

object.

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 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

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, meats

and nuts are given in ASHRAE Handbook of

Fundamentals

Enthalpy (h)

Units: (J/kg)

Heat content of a material.

Used frequently to evaluate changes in heat content of steam or moist air

Combines latent heat and sensible heat changes

ΔQ = M(h2-h1)

Where,

ΔQ = amount of heat needed to raise temperature from T1 to T2

M = mass of product

h2= 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

Example

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.

Homework Assignment Due

February 20thProblem 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.