length along the tank bottom equal to 1 times the spacing of the anode from the bottom.

(b) The anode suspension arrangement for the tank being considered is shown in figure D-9. It can

be seen that stub anodes are required for this design. Ten stub anodes are spaced equally on a circumference

with a radius of 8 feet in a way shown in figure D-8. For smaller diameter tanks, stub anodes may not be

required.

(12) Find the current division between main and stub anodes.

(a) The area of tank bottom protected by stub anodes is found by equation D-10 (fig D-9):

As = B (r22 - r12),

(eq D-10)

where r2 = 13 feet (protected segment radius) and r1 = 2.5 feet (riser radius). Thus,

As = 3.1416 (132 sq ft - 2.52 sq ft)

As = 3.1416 x 162.75

As = 511.3 sq ft.

(b) The maximum current for stub anodes is therefore--

Is = 2.0 x 511.3

Is = 1022.6 milliamperes or 1.02 amperes.

(c) The maximum current for the tank bowl is 15.2 amperes.

(d) The maximum current for main anodes is--

Im = 15.2A - 1.02A

Im = 14.2A.

(13) Find the rectifier voltage rating.

(a) The electrical conductor to the main anode is wire size No.2 AWG, rated at 0.159 ohm per 1000

feet, and has an estimated length of 200 feet. Thus, the resistance of the wire, R, is:

200 ft

R = ------ x 0.159 ohm = 0.032 ohm.

1000 ft

(b) For the voltage drop in the main anode feeder--

E = IR,

where I = 14.2 amperes and R = 0.032 ohm. Thus,

E = 14.2 A x 0.032 ohm

E = 0.45 V.

(c) For the voltage drop through the main anodes--

E = IR,

where I = 14.2 amperes and R = 3.03 ohms. Thus,

E = 14.2 A x 3.03 ohms.

E = 43.0 V.

(d) The total voltage drop in main anode circuit is thus--

ET = 0.45 + 43.0

ET = 43.45 or 45 V.

Use a multiplying factor (safety factor) of 1.5 to get 67.5 volts.