(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.

(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.

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.