TM 5-811-7
(c) Find the maximum weight of anode material required (use eq C-1 from appendix C):
YSI
W'
,
E
where Y = 15 years, S = 2.0 pounds per ampere-year, I = 17.2 amperes, and E = 0.50 efficiency. Thus,
(15 yr)(2.0 lb/A&yr)(17.2A)
W'
,
0.50
W = 1032 lb.
c.
Groundbed design.
(1) Anode size is 3-inch by 60-inch (backfilled 10-inch by 84-inch) and weight is 25 pounds per anode
unit.
(2) Find the resistance to earth of a single anode:
PK
Rv '
,
(eq D-1)
L
where P = 1000 ohm-centimeters, L = 7.0 feet (backfilled size), and K = 0.0167, L/d = 8.4 (table 2-6). Thus,
(1000 ohm&cm)(0.0167)
Rv '
,
7.0 feet
Rv = 2.39 ohms
(3) Compute the number of anodes required. The low resistance (2.39 ohms) of a single anode and the
heavy weight of anode material required (1032 pounds) for a 15-year life indicate that the controlling factor
is the amount of anode material, not groundbed resistance. The minimum number of anodes (N) required is
N = 1032/25 = 41.3 or 41 anodes.
These are arranged in a distributed groundbed as shown in figure D-2 based on the following estimates.
(4) Anode distribution:
(a) Conduit area in sections 1 through 6 of figure D-2 are given in table D-3.
Table D-3. Conduit area: heat distribution system
Section
Length (ft)
Surface area (sq ft)
1
1700
3553 + 6239 = 9792
2
500
785 x 1310 = 2095
3
1125
1766 x 3533 = 5299
4
350
550+ 917=1467
5
400
628 + 1048 = 1676
6
275
432 + 721=1153
U.S. Air Force.
(b) The area of conduit protected by one anode is --
A = 21,481/41
A = 524 sq ft/anode.
(c) Anodes will be divided as shown in table I)-4.
D-6
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