1. Topics in Processing
Dr. C. L. Jones
Biosystems and Ag. Engineering

2. Thermal Properties and Moisture Diffusivity
Assignment:
Read Chapter 8 in Stroshine Book
Dr. C. L. Jones
Biosystems and Ag. Engineering

3. Thermal Properties, Moisture Diffusivity
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)
Dr. C. L. Jones
Biosystems and Ag. Engineering

4. Thermal Properties, Moisture Diffusivity
Specific heat: Amount of heat required to raise the temp. of one unit of mass one degree.
Q = Mcp(T2-T1)
Q = quantity of heat,
M = mass or weight
Cp = specific heat at constant pressure
Cp =4.18 kJ/kg-K = 1.00 BTU/lb-R=1.00 cal/g-K for water
Dr. C. L. Jones
Biosystems and Ag. Engineering

5. Thermal Properties, Moisture Diffusivity
For liquid H2O
Cp = M above freezing
For solid H2O
Cp = M below freezing
Dr. C. L. Jones
Biosystems and Ag. Engineering

6. Thermal Properties, Moisture Diffusivity
Thermal Conductivity:
dQ/dt = -kA (dT/dx)
K = coefficient of thermal conductivity
Dr. C. L. Jones
Biosystems and Ag. Engineering

7. Thermal Properties, Moisture Diffusivity
K = VwKw + VsKs
K = KwXw + Ks(1-Xw) where X = decimal fraction
so K = f(all the constituent volumes)
Dr. C. L. Jones
Biosystems and Ag. Engineering

8. Thermal Properties, Moisture Diffusivity
Power: Pwatt = -KA(ΔT/ ΔX)
P = -KA (T2 – T1)/x
Kfats&oils<<Kwater
Kdrymaterials<<Kwater
Dr. C. L. Jones
Biosystems and Ag. Engineering

9. Thermal Properties, Moisture Diffusivity
Force Convection:
Q/T = hA(T2 – T1)
Dr. C. L. Jones
Biosystems and Ag. Engineering

10. Thermal Properties, Moisture Diffusivity
Thermal Diffusivity, α, (m2/sec)
Material’s ability to conduct heat relative to its ability to store heat
Estimate the thermal diffusivity of a peach at 22 C.
α = k/(ρcp)
Dr. C. L. Jones
Biosystems and Ag. Engineering

11. Thermal Properties, Moisture Diffusivity
Water Phase Diagram
Dr. C. L. Jones
Biosystems and Ag. Engineering

12. Thermal Properties, Moisture Diffusivity
Water Phase Diagram
Dr. C. L. Jones
Biosystems and Ag. Engineering

13. Thermal Properties, Moisture Diffusivity
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…)
Dr. C. L. Jones
Biosystems and Ag. Engineering

14. Thermal Properties, Moisture Diffusivity
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
Dr. C. L. Jones
Biosystems and Ag. Engineering

15. Thermal Properties, Moisture Diffusivity
Enthalpy, L, (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 230.
Dr. C. L. Jones
Biosystems and Ag. Engineering

16. Thermal Properties, Moisture Diffusivity
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.
Dr. C. L. Jones
Biosystems and Ag. Engineering

17. Thermal Properties, Moisture Diffusivity
Homework Due 4/20
Problem 1: do 8.1 in your book (see example of a solution
Problem 2: Determine the amount of heat removed from 2 kg of blueberries when cooled from 28C to -7C. Assume MC of 83% and at -7C, 27% won’t be frozen.
Dr. C. L. Jones
Biosystems and Ag. Engineering

18. Thermal Properties, Moisture Diffusivity
Sample solutions for problem #1
A thawed turkey (whole) is placed in a turkey fryer. The turkey is at 3 C while the oil is at 400 C.
The turkey is at 3 C while the oil is at 400 C. Assuming the oil temperature is maintained at 400 C, the temp. at the surface of the turkey (boundary condition) will remain at 400C throughout the frying process. This assumes that the heat transfer coefficient across the oil-turkey fry boundary does not limit the heat transfer process. The thermal conductivity of the turkey can be estimated from the perpendicular model since the meat is fibrous and the thermal conductivity will be directionally dependent. The turkey can be modeled as a finite cylinder.
Dr. C. L. Jones
Biosystems and Ag. Engineering