TM 5-852-5/AFR 88-19, Volume 5
b. Estimate the ice thickness on the reservoir when
--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.
--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
d1 = 3/12 = 0.25 feet. The moisture content in the
soil = (0.06)(125 pcf) = 7.5 lb water/ft3 soil.
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
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.
d. Determine the rate of heat loss per linear foot
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.
the insulation material is 0.0133 BTU/EF ft hr.
c. The Stefan equation (equation 12-6) can also
-- 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
Outside radius = rout = 2.5 + 0.5 = 3.0 inch