TM 5-852-4/AFM 88-19, Chap. 4
Building temperature, 60 0F.
Sensible heat (10% of latent heat, based upon
Thermal conductivity of concrete, Kc = 1.0 Btu/ft
experience)
hrF.
= 495 Btu/ft2
Thermal conductivity of insulation, Ki = 0.033
Total heat content
Btu/ft hr F.
= 5,445 Btu/ft'.
Other data required for solution are obtainable
The ducts will be open during the freezing
from TM 5-852-6/AFM 88-19, Chapter 6" or are
season (215 days), and the average rate of heat flow
introduced later. The thickness of gravel pad required is
from the gravel during this season is equal to 5,445/(215
determined by the following equation (derived from the
x 24) = 1.0 Btu/ft' hr. The average thawing index at the
modified Berggren equation):
surface of the pad is
This thawing index must be compensated by an
equal freezing index at the duct outlet on the surface of
the pad to assure freezeback. The average pad surface
temperature at the outlet equals the ratio
The inlet air during the freezing season has an
average temperature of
(In the computations, the dead airspace is
assumed equivalent to the thermal resistance of
Therefore, the average permissible temperature
concrete of the same thickness.)
rise TR along the duct is (25.4 - 13.4) = 12.0 F.
The heat flowing from the floor surface to the
duct air during the winter is equal to the temperature
difference between the floor and duct air divided by the
thermal resistance between them.
The thermal
resistance
Thus, the total amount of heat to be removed
where
from the gravel pad by cold-air ventilation during the
(For practical design purposes hrc = 1.0 Btu/ft' hr
freezing season with ducts open is equal to the latent
F, and represents the combined effect of convection
and sensible heat contained in the thawed pad. The
and radiation. At much higher air velocities, this value
heat content per square foot of pad is determined as
will be slightly larger; however, using a value of 1.0 will
follows:
lead to conservative designs.)
Latent heat = (X) (L) = (11.0) (450) = 4,950
The average heat flow between the floor and
Btu/ft'.
inlet duct air is [(60 - 13.4)/12.3] = 3.8 Btu/ft: hr, and
between the floor and outlet duct air is [(60 -25.4)/12.3] =
*Rf = Reciprocal of time rate of heat flow through a unit
2.8 Btu/ft2 hr. Thus, the average rate of heat flow from
area of a given temperature difference per unit thickness
the floor to the duct air is [(3.8 + 2.8)/2] = 3.3 Btu/ft2 V
hr. As previously calculated the average heat flow from
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