PHYSICS

Continuation of Table 13.2
Hydrogen 0.172 20
Nitrogen 0.0234 20
Oxygen 0.0238 20
Nonmetals
Water 0.565 0
» 0.599 20
» 0.627 40
Concrete 2.43 20
Wood 0.126 20
Plastic 0.04 20
Biological objects
Fat 0.205 20
Skin 0.502 20
Muscles 0.4 20
Fur 0.036 - 0.063 20
Example. When a pig lies on concrete, the animal's belly and the
concrete on which it lies are in contact. The temperature of the concrete's
surface approximates that of the pig's belly surface. Assume that
the 8-cm-thick concrete slab is on ground with a temperature of O°C,
that belly-floor contact area is 3000 ern", the body temperature of pig is
38'C, and the thermal conductivity of the concrete is 2.43 W·m-i.K-I.
Estimate the conductive heat transfer under steady-state conditions.
Solution. Conductive heat transfer flux from the belly to the concrete
can be calculated as:
2(0
Qc =-«.s t1.T = -2.43~. 3000 .10- 4 m - 38) K K8 .10- 4 m =
t1.x
m·K
=- 2.43 ·3 ·10 -I ( - 38) / 8·10 -2 = 346.27 J .s:'
13.3.2. Convection
Heat transferred by the movement of a heated substance is said
to have been transferred by convection. When the movement results
from differences in density, it is referred to as natural convection.
When the heated substance is forced to move by a fan or
pump, the process is called forced convection.
Consider the convective heat exchange of objects with various
shapes. The amount of heat conducted across the boundary layer
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and convected away from the tJdL i-'~di.e (e.g., the surface of a leaf
per unit time and area) by forced convection is:
J = _ Zlc: (~eal - Talr )
e air 8
(13.2)
where k air
is the thermal conductivity coefficient of air, T'eal
the leaf temperature, and T alr
the temperature of the air outside
a boundary layer of thickness 8. This can be expressed as:
8 (mm) = 4.0 I L(m) (13.3)
, Vtm/s)
where L is the mean length of the leaf, V is the ambient wind
speed, and the factor 4.0 has dimensions in m/s!".
In the case of a cylinder (e.g., an animal), the heat flux
density is:
J = _ 2k . (Tsuif - Talr )
e air .' + 8] (13.4)
rln--,
r )
where r is the cylinder radius, T,urf surface temperature, and 8 is
calculated as:
D(m)
8 (mm) =5.8,/V(m/s) (13.5)
where D is the cylinder diameter.
It is possible to use the .following relation for objects of
irregular shape which is known as Newton's law of cooling. It describes
the rate of heat loss (Jc) in W1m 2 per unit surface area of a
body in a cool air stream as:
J e = k/T,uif - To) (13.6)
where k, is the convection coefficient with units W·m- 2·deg- J •
Example. Calculate the heat flux density conducted across the
boundary layer and convected away from the surface of a sheep if the
body of the animal approximates a cylinder with a radius of 60 ern,
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