U. S. Army Corps of Engineers

Figure 4-9. Effect of heated structure size on depth and rate of thaw.

building floor temperature and 32 F.

Rv =

d

=

8

+

4

+

6

(a) Example: Estimate the depth of

k

(12) (1.0) (12) (0.033)

(12) (1.0)

thaw after a period of one year for a building floor

= 11.2 ft hr F/Btu

2

consisting of 8 inches of concrete, 4 inches of cellular

The average volumetric heat capacity of the floor system

glass insulation, and 6 inches of concrete, placed directly

is

on a 5-feetthick sand pad overlying permanently frozen

silt for the following conditions:

(30) (8) + (1.5) (4) + (30) (6)

Cf

Mean annual temperature (MAT), 20 F.

8+4+6

Building floor temperature, 65 F.

3

Sand pad: Dry unit weight γ d = 72 lb/ft , w = 45

3

=23.7 Btu/ft F

percent.

The solution to this problem (table 4-2) predicts a total

Concrete: Coefficient of thermal conductivity, K =

thaw depth of 7.8 feet. This solution did not consider

1.0 Btu/ft hr F; Volumetric heat capacity, C = 30

3

edge effects; i.e., a long narrow building will have less

Btu/ft F.

Insulation: K = 0.033 Btu/ft hr F, C = 1.5

depth of thaw than a square building with the same floor

area because of the difference in lateral heat flow.

Btu/ft F.

3

(b) Figures 4-3, 4-10, 4-11 and 4-12a

The resistances of the three floor layers are in

show measured rates of thaw beneath buildings placed

series, and the floor resistance Rf is the sum of

direct-

the three layer resistances (d = thickness of

layer in feet).