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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