1. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
Test Monday 3/2, 2:30 – when you’re finished
Lecture Wed 3/4 and Friday 3/6: 2:30 – 3:20 123AH
See schedule for topics
Lab reports due next Wednesday 3/4

2. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
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
All include heat transfer and are dictated by thermal properties of material
Generally diffusion of water in or out is involved

3. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
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 moves similar to heat by conduction
Moisture diffusivity
Volume change due to moisture content change

4. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
Terms:
Specific heat
Thermal conductivity
Thermal diffusivity
Thermal expansion coefficient
Surface heat transfer coefficient
Sensible and Latent heat
Enthalpy

5. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
Specific heat: Amount of heat required to raise the temp. of one unit of mass one degree.
Cp = specific heat at constant pressure
Cp =4.18 kJ/kg-K = 1.00 BTU/lb-R=1.00 cal/g-K for water (unfrozen)
oils and fats: ½ H2O See Table 8.1 pg. 219
grains, powders: ¼ - 1/3 H2O
ice: ½ H2O
Good list:
Q = quantity of heat required to change temperature of a mass
Q = Mcp(T2-T1)
M = mass or weight

6. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
For liquid H2O
Cp = M above freezing
For solid H2O
Cp = M below freezing

7. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
Thermal Conductivity:
measure of ability to transmit heat
dQ/dt = -kA (dT/dx)
K = coefficient of thermal conductivity
W/m°K, Btu/h ft°F,
1 Btu/h ft °F = W/m °K
Greater the water content, the greater the thermal conductivity
Tables 8.2 and 8.3

8. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
If we don’t know t-conductivity, approximate using...
K = VwKw + VsKs
K = KwXw + Ks(1-Xw) where X = decimal fraction
so K = f(all the constituent volumes)
Example 8.1 pg 224

9. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
Thermal Diffusivity, α, (m2/sec or ft2/sec)
Material’s ability to conduct heat relative to its ability to store heat
α = k/(ρcp)
Estimate the thermal diffusivity of a peach at 22 C.

10. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
Surface Heat Transfer Coefficient, h:
Placed in a flowing stream of liquid or gas, the solid’s T will change until it eventually reaches equilibrium with the fluid
Q/T = hA(T2 – T1)
“h” is determined experimentally
Look for research that matches your needs. (bottom of pg 227)

11. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
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

12. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
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…)

13. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
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 Fundamentals

14. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
Enthalpy, h, (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 pg Δh = M cp(T2 – T1) + MXw L

15. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
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.

16. Lecture 12: Thermal Properties, Moisture Diffusivity Chpt 8
Problem 1: Determine the amount of heat removed from 2 kg of sour cherries when cooled from 28C to -7C. Assume MC of 92.3% and at -7C, 27% won’t be frozen.
Problem 2: Estimate the thermal diffusivity of cheddar cheese at 22°C.