TM 5-852-5/AFR 88-19, Volume 5
r2 = (c)(csch A)
c = (H2p = r2p) = [(f)2 - (0.25)2]1/2 = 399 ft
A = T* arccosh (Hp/rp)
= 22.3 BTU/hr LF of pipe.
--If Hp $ 2rp
--Figures 12-9 and 12-10 can be used to assist in
calculations of this type:
A T*1n (2Hp/rp)
= (0.366)(1n 8/0.25)
= (0.366)(3.466)
= 1.268
= (0.366)(3.48)
H2 = (3.99)(coth 1.268)
= 1.27.
= (3.99)(1.172)
--Then from figure 12-10, with A = 1.27
= 4.68 ft
r2 = (3.99)(csch 1.268)
= (3.99)(0.5215)
= 2.08 ft.
--The thaw zone, under steady state conditions
will be a cylinder of soil enclosing, and parallel to,
= 3.99 ft.
the pipe. The radius of this zone will be about 2 feet
and the axis will be about 5 inches below the bottom
--So:
of the pipe:
H2 = (1.18)(3.992) = 4.7 ft
The axis = H2 - (Hp +rp) = 4.68 - 4.25 = 0.43 ft =
5.2 in. below pipe
r2 = (0.6)(3.992) = 2.39 ft
The heat loss (Q) from this pipe would be
and
If Hp $ 2rp:
--So:
= 22.2 BTU/hr LF of pipe.
This agrees closely with the 22.3 BTU/hr value
determined previously.
12-20
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