(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.

(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.