Thermal Properties and
Moisture Diffusivity
24th February, 2010
Thermal Properties, Moisture Diffusivity
Processing and Storage of Ag Products
◦ Heating
◦ Cooling
◦ Combination of heating and cooling
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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
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Conduction: Temperature gradient exists within a body…heat transfer
within the body
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Convection: Heat transfer from one body to another by virtue that one
body is moving relative to the other
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Radiation: Transfer of heat from one body to another that are
separated in space in a vacuum. (blackbody heat transfer)
We’ll consider
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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
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Moisture diffusivity
Volume change due to moisture content change
Continue….
Terms used to define thermal
properties
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Specific heat
Thermal conductivity
Thermal diffusivity
Thermal expansion coefficient
Surface heat transfer coefficient
Sensible and Latent heat
Enthalpy
Specific Heat
Specific Heat: Amount of heat required to raise
temperature of unit amount of substance by one
degree
Units:
◦
◦
◦
𝐾𝐽
𝐾𝑔°𝐾
𝐵𝑇𝑈
lb°𝐹
𝐶𝑎𝑙.
𝑔°𝐾
(SI System)
(English System)
(if calories are used)
Conversion of units: 1
𝐾𝐽
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
𝐾𝐽
𝐵𝑇𝑈
𝐶𝑎𝑙.
Cp of water = 4.18
=1
=1
𝐾𝑔°𝐾
lb°𝐹
𝑔°𝐾
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 heat
Eq. 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 composition
Cp=4.18Xw+1.711Xp+1.928Xf+1.547 Xc+0.908Xa
X is the mass or weight fraction of each component
The subscript denote following components: w=water, p=
protein, f=fat, c= carbohydrate, a=ash
Thermal Conductivity (k)
Measure of ability to transmit heat
𝑑𝑄
𝑑𝑡
= -k A
𝑑𝑇
𝑑𝑥
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
𝑑𝑇
( ) of 1 degree per unit distance in x direction
𝑑𝑥
Units:
𝑊
m°𝐾
𝐵𝑇𝑈
h ft °𝐾
(SI system)
(English System)
1
𝐵𝑇𝑈
h ft °𝐾
= 1.731
𝑊
m°𝐾
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 pdts, 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,
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
If the moisture in product is more than 50%, then
K = 0.056 + 0.57Xw
Thermal Diffusivity (α)
α quantifies the materials ability to conduct
heat relative to its ability to store heat.
α=
𝑘
𝞺 𝐶𝑝
Where,
𝑚2
𝑓𝑡2
α = Thermal Diffusivity, Units ( ) or ( )
𝑠𝑒𝑐
𝑠𝑒𝑐
k = Thermal conductivity
𝞺 = density of material
𝐶𝑝 = 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
Where,
𝑑𝑄
𝑑𝑡
= h A (Tf- Ts)
𝑊
h = surface heat transfer coefficient and has same units as k i.e.
m°𝐾
Tf = temp. of fluid
Ts = temp of solid body
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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 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)
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Heat that is exchanged during a change in phase
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Dominated by the moisture content of foods
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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)
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420 kJ to raise T of water from 0 ° C to 100 ° C, 5x that to evaporate 1
kg of water
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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: (kJ/kg or BTU/lb)
 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 temperature is below freezing
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Measurements based on zero values of enthalpy at a specified
temperature e.g. at -40°C, -18°C or 0°C.
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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 March
7th
Problem 1: Determine the amount of heat
removed 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 20°C.