PHYSICS

the surface temperature (Tsar[) is 3SoC, the temperature of the air (T a)
is 20"C, and the ambient wind speed (V) is SO em-s:".
Solution. The average of the boundary layer (8 ) is:
8 = 5 8· {D = 5 8 . JO,6 = 5 mm = 5 .10- 3 m
, 'Ii~ ' 0,8
Using Equation (13.6) and k , ai
(0.0257 W·m-I·K-I at 20"C), calculate
the heat flux density conducted across the boundary layer.
(T,arj - Ta;r ) _ 0,0257· (38 - 20) = 93 W . m-2
J c = - 2k air [ r + 8 ) - [ 0,3 + 0,005 J
rln - 0,3·ln 03
r ,
13.3.3. Radiation
Two bodies at different temperatures will exchange heat even
when there is no possibility of exchange by conduction or
convection; the transfer of heat takes place by radiation.
The rate at which an object emits radiant energy is proportional
to the fourth power of its absolute temperature. This is
known as Stefan-Boltzmann's law which is expressed in equation
form as:
R = aSET4 (13.7)
where R is the power radiated by the body (W); (J =5,67051'10- 8
W·m- 2 •K-4 is a constant; S is the surface area of the object, EO is a
constant called the emissivity, and T is the temperature in Kelvin.
When an object is in equilibrium with its surrounding, it
radiates and absorbs energy at the same time:
R = Q - Q = Sa(aT4 - ET41 (13.8)
n a e s e/
where a is a constant called the absorptivity.
The values of constants a and EO are given in Table 13.3.
Example. If a brown cow has an effective radiant-surface area of 4 m'
and a radiant-surface temperature averaging 27"C, and the average
temperature of the environment is -3°C, calculate the net flux of
thermal radiant energy between an animal and its environment.
104
13.3. Radiation absorptivity and emissivity ratios for various
farm-animals and humans
Object I Absorptivity, a r Emissivity, EO
Cow
white 0.50 0.95
brown 0.80 0.95
black 0.90 0.95
Pig
white 0.50 0.95
black 0.90 0.95
Sheep
without wool 0.60 0.95
with wool 0.75 0.95
Human organism 0.65 - O.SO 0.95
Solution. A net heat flux from animal to environment via thermal
radiation can be determined from Equation ( 13.S ):
R n
= Q a
- Q, = Sa (aT/ - ET,4) =
= 4 m2·5,67051·10-8 W·m- 2. K-4(0.SO·300 4 - 0.95'270 4) = 632 W.
13.3.4. Evaporation
Evaporative loss of water is an excellent way for animals to
dissi pate heat. The rate of evaporation depends not only on the
surface temperature, but also on the difference in water vapor
density between the animal's surface and the environment, and
on the resistance to water loss from the surface. Typical values
for evaporative water loss are 7.0 - 20.9 W·m- 2 •
Chapter 14. THERMOBIOLOGY
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