1 lecture 12: thermal properties, moisture diffusivity chpt 8 test monday 3/2, 2:30 – when...

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Lecture 12: Thermal Properties, Moisture DiffusivityChpt 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

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• 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

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• 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

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

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Specific heat: Amount of heat required to raise the temp. of one unit of mass one degree.

Cp = specific heat at constant pressureCp =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. 219grains, powders: ¼ - 1/3 H2Oice: ½ H2O

Good list:http://www.engineeringtoolbox.com/specific-heat-capacity-food-d_295.html

Q = quantity of heat required to change temperature of a mass

Q = Mcp(T2-T1)M = mass or weight

http://www.engineeringtoolbox.com/specific-heat-capacity-food-d_295.html

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For liquid H2O

Cp = 0.837 + 3.348 M above freezing

For solid H2O

Cp = 0.837 + 1.256 M below freezing

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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 = 1.731 W/m °K

Greater the water content, the greater the thermal conductivity

Tables 8.2 and 8.3

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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

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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.

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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)

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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

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• 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…)

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• 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

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• 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. 8.15 pg

230. Δh = M cp(T2 – T1) + MXw L

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• 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.

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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.

1 lecture 12: thermal properties, moisture diffusivity chpt 8 test monday 3/2, 2:30 – when...

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