--Assume: sandy soil, dry density 125 pcf, mois-

ture content 6% and a freezing index (Ig) = 3000 EF

there is an 8-inch snow cover on top of the ice and

Ig = 3000 EF d.

d

--Find depth of frost penetration under 3-inch-

--From equation 12-3, and Table 12-4:

thick polystyrene board. From table 12-2:

k1 = 0.020 BTU/ft EF hr (for polystyrene),

k2 = 1.0 BTU/ft EF hr (for sand) and thus

X=

m(Ig)

=

0.7(3000)

d1 = 3/12 = 0.25 feet. The moisture content in the

soil = (0.06)(125 pcf) = 7.5 lb water/ft3 soil.

=

38 inches

=

3.2 ft.

Latent heat of water = 144 BTU/lb

L2 = (144 BTU/lb)(7.5 lb/ft3) = 1080 BTU/ft3 of

--Or, use the Stefan equation (equation 12-6) for

soil

a two-layer system:

L1 = 0

--The first layer is snow, d1 = 8 inches = 0.667

feet, assumed to be drifted and compact. From

table 12-2, k1 = 0.4 BTU/ft hr EF. Since no

phase change occurs in the snow, L1 = 0.

--Ice: k2 = 1.28 BTU/ft hr EF

L2 = (144)(62.4) = 8986 BTU/ft3

--The depth of frost penetration would be 11.5

feet in the same soil, under the same conditions, if

the insulation board were not in place.

of above-ground pipe from a 5-inch-ID (wall

thickness inch) plastic pipe encased in a 2-inch

thickness of polyurethane insulation. Water inside

the pipe is maintained at 40EF, ambient air

= 3.5 ft (includes the 8 in. of snow).

temperature is -40EF, and wind speed is 15 mph.

Thermal conductivity of pipe material is

0.208BTU/EF ft hr and thermal conductivity q of

the insulation material is 0.0133 BTU/EF ft hr.

-- Use equations a and b from figure 12-4:

be used to estimate the depth of frost penetration

beneath a gravel pad or an insulation board. The L1

in either case would be zero. The L2 in this example

would be the latent heat of fusion for the soil and

Inside radius = rin = 2.5 inch

would be dependent on the moisture content in the

soil.

Outside radius = rout = 2.5 + 0.5 = 3.0 inch

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