above ground or at a depth of 4 feet. Assume the

= 5.79 BTU/hrlinear foot (LF) of pipe.

minimum air temperature is -58EF, and the minimum

mean daily soil temperature at a depth of 4 ft is

--This is about one-third the heat loss rate

1.4EF. Thermal conductivity of soil is 1.2

calculated for an above-ground installation in the

BTU/fthrEF.

same location. The responsible factor is the

(1) *Above-ground installation. *Assume a 5-inch

attenuation of the extreme surface temperature at

ID plastic pipe, with 2 inches of polyurethane

the 4-foot depth.

insulation:

Rc = 6.306 hrftEF/BTU (from previous

the average rate of heat loss from a 6-inch steel pipe

example).

buried 4 feet below the surface in a clay soil, where

--The water inside pipe will be maintained at

the soil thermal conductivities are kf (thawed) =

40EF, so that the maximum rate of heat loss

0.60 BTU/hrftEF and kf (frozen) = 1.0

BTU/hrftEF. Mean soil temperature at the ground

surface is 27.5EF and the water in the pipe is

maintained at 45EF. (See figure 12-sb, for

schematic, symbols and equations.) A bare steel pipe

has negligible thermal resistance so

Rp = 0.

= 15.5 BTU/hrLF pipe.

--Outer pipe radius rp = 6 inch/(2)(12) = 0.25 ft

(2) *Buried installation. *Assume that the top of the

pipe is 4 feet below the surface, the radius to the

--Depth to center of pipe Hp = 4.0 ft.

outer surface = 5 in. = 0.416 feet, the depth to

center of the pipe Hp = 4.416 ft, and the radius of

pipe = 0.416 feet (see equations, fig 12-5a) and Hp

where

Tw

= water temperature inside pipe

To

= soil temperature at interface of

thawed zone

= 32EF

--The air film is not a factor for a buried pipe of

Rc = Rp + RI + Rs

= 0.1396 + 6.116 + 0.405 (Rp and RI from

previous example)

= 6.66 1 hr~ft~0F/BTU.

--So, the heat loss Q equals

--Depth to center of thawed zone, H2:

H2 = (c)(coth A)

--Radius of thawed zone, r2:

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