TM 5-811-7
(6) Design for a maximum current density of 5 milliamperes per square foot.
(7) Design for a 5-year life.
(8) Use HSCBCI anodes.
(9) Electrical current is available at 115 volts a.c., single phase.
b. Computations.
(1) Find the tank's interior area using equation D-2:
r2 +
AT = 2
dL,
where r=1.92 feet, d=3.83 feet, and L = 12 feet. Thus,
AT = 2 x 3.1416 x (1.92)2 + 3.1416 x 3.38 x 12
AT = 167.5 sq ft.
(2) Find the maximum protective current required:
I = 167.5 x 5
I = 838 mA or 0.84 A.
(3) Find the minimum weight of anode material needed for a 5-year life (eq C-I from appendix C):
YSI
W'
,
E
where Y = 5 years, S = 1.0 pound per ampere-year, I = 0.84 ampere, and E = 0.50. Thus,
(5 yr)(1.0 lb/A&yr)(0.84 A)
W'
0.50
W = 8.4 lb.
(4) Compute the number of anodes required. An anode 1 inches in diameter by 9 inches long weighing
4 pounds is chosen as the most suitable size. For proper current distribution, three anodes are required.
(5) Find the resistance of a single anode using equation D-3:
0.012P log (d/D)
R'
(eq D-3)
L
where P =8600 ohm-centimeters, D = 3.83 feet (tank diameter), d = 1 inches or 0.125 foot (anode
diameter), L = 9 inches or 0.75 foot (anode length). Thus,
0.012 (8600 ohm&cm) log (3.83 ft/0.125 ft)
R'
0.75 ft
103.2 log 30.64
R'
0.75
R = 204.5 ohms
This resistance must be corrected by the fringe factor because the anodes are short. The fringe factor is 0.48
from the curve in figure D-4 for an L/d = 9/1.5 = 6:
R = 204.5 x 0.48
R = 98.2 ohms.
D-8
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