Thermal Properties, Moisture Diffusivity Chpt 8

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1 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 1

2 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 2

3 Terms: Specific heat Thermal conductivity Thermal diffusivity Thermal expansion coefficient Surface heat transfer coefficient Sensible and Latent heat Enthalpy 3

4 Specific heat: Amount of heat required to raise the temp. of one unit of mass one degree. C p = specific heat at constant pressure C p =4.18 kj/kg-k = 1.00 BTU/lb-R=1.00 cal/g-k for water (unfrozen) oils and fats: ½ H 2 O See Table 8.1 pg. 219 grains, powders: ¼ - 1/3 H 2 O ice: ½ H 2 O Good list: Q = quantity of heat required to change temperature of a mass Q = Mc p (T 2 -T 1 ) M = mass or weight 4

5 For liquid H 2 O C p = M above freezing For solid H 2 O C p = M below freezing 5

6 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 6

7 If we don t know t-conductivity, approximate using... K = V w K w + V s K s K = K w X w + K s (1-X w ) where X = decimal fraction so K = f(all the constituent volumes) Example 8.1 pg 224 7

8 Thermal Diffusivity, α, (m 2 /sec or ft 2 /sec) Material s ability to conduct heat relative to its ability to store heat α = k/(ρc p ) Estimate the thermal diffusivity of a peach at 22 C. 8

9 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) 9

10 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 10

11 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 ) 11

12 Latent Heat, L, (kj/kg or BTU/lb) Determine L experimentally when possible. When data is not available (no tables, etc) use. L = 335 X w where X w is weight fraction of water Many fruits, vegetables, dairy products, meats and nuts are given in ASHRAE Handbook of Fundamentals 12

13 Enthalpy, h, (kj/kg or BTU/lb) Heat content of a material. Combines latent heat and sensible heat changes Q = M(h 2 -h 1 ) amount of heat to raise a product from T 1 to T 2 ASHRAE Handbook of Fundamentals When data is not available use eqtn pg 230. h = M c p (T 2 T 1 ) + MX w L 13

14 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. 14

15 Lecture 12: Thermal Properties, Moisture Diffusivity Due March 5th 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. 15