1. Psychrometrics – Lecture 3
Assignment:
Read Chapter 8 in Stroshine Book
Dr. C. L. Jones
Biosystems and Ag. Engineering

2. Lecture 1 and 2 Review
This is all about removing and adding water from materials and from air-water vapor
Knowing 2 of the properties of a state point, we can find the other 5 properties
Wet Bulb T, Dry Bulb T, Dew Point, RH, enthalpy h, humidity ratio H, specific volume v
Dr. C. L. Jones
Biosystems and Ag. Engineering

3. Lecture 1 and 2 Review
Cooling and heating without changing amount of moisture: moves along a horizontal line (humidity ratio H stays the same…everything else changes)
In a mixing process (2 air streams at different state points)…resultant state point will lay on a line between the two initial stages
Example of mixing: cooling and dehumidifying…ie. air conditioning
Dr. C. L. Jones
Biosystems and Ag. Engineering

4. Lecture 1 and 2 Review
Product drying process using air: adiabatic…follows the enthalpy line…wetbulb stays the same, enthalpy stays the same, everything else changes
Dr. C. L. Jones
Biosystems and Ag. Engineering

5. Psychrometric Processes: Addition of Water at any state
Changes enthalpy and humidity ratio…and everything else, Figure 9.12
Mass of water: (eqtn 9.100)
Energy balance: (eqtn 9.101)
Example 9.14
Dr. C. L. Jones
Biosystems and Ag. Engineering

6. Psychrometric Processes: Addition of Water at any state
Greenhouse plants like to be kept at 30°C and 95% rh.
To achieve this condition, steam at 160°C is injected into the ventilation air stream (20°C, 30% rh).
Mass flow rate of the air stream is 1800 kg/min and the mass flow rate of the steam is 5 kg/min.
Will this system achieve the ideal conditions for the plants?
Dr. C. L. Jones
Biosystems and Ag. Engineering

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

8. Thermal Properties, Moisture Diffusivity
Heat is transferred by conduction, convection or radiation
Heat may be conducted within the product
Heat transfer by forced convection between the product and a moving liquid surrounding the product
Dr. C. L. Jones
Biosystems and Ag. Engineering

9. Thermal Properties, Moisture Diffusivity
Basic thermal properties
Specific heat
Thermal conductivity
Thermal diffusivity
Thermal expansion coefficient
Surface heat transfer coefficient
Heat may be conducted within the product
Dr. C. L. Jones
Biosystems and Ag. Engineering

10. Thermal Properties, Moisture Diffusivity
Specific heat, cp , (kJ/kg-K, BTU/lb-F, cal/g-K)
Amount of heat (joules, BTU’s or calories) needed to increase the temperature of a unit of mass by one degree
Q = amount of heat added to mass to raise temp. from T2 to T1
Q = Mcp(T2-T1)
Dr. C. L. Jones
Biosystems and Ag. Engineering

11. Thermal Properties, Moisture Diffusivity
Table 8.1 gives cp for food and ag products
Materials containing mostly water have a cp close to water (4.18 kJ/kg-K, 1 BTU/lb-F, 1 cal/g-K)
Oils and fats are approx. ½ that of water
Ice is ½ of water…frozen foods take less heat to increase T than after it is thawed!!!
Cp = M above freezing
Cp = M below freezing
Equation 8.4 pg 220 takes into account the composition of the solids
Dr. C. L. Jones
Biosystems and Ag. Engineering

12. Thermal Properties, Moisture Diffusivity
Thermal Conductivity, k, (W/m-K, BTU/h ft-K)
Measure of material’s ability to transmit heat,…. quantity of heat Q that will flow across a unit cross sectional area A per unit of time, t
dQ/dt = -kA (dT/dx)
Table 8.2, pg. 221
The greater the proportion of water, the higher the thermal conductivity.
Porosity and fiber orientation make a difference, too.
See wood in Table 8.3 for fiber direction differences.
Dr. C. L. Jones
Biosystems and Ag. Engineering

13. 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.
Dr. C. L. Jones
Biosystems and Ag. Engineering

14. 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.
Dr. C. L. Jones
Biosystems and Ag. Engineering

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

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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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