(14) Select the stub anodes. Because it is desirable to use as small an anode as possible without

exceeding the manufacturers' recommended rate, try using type FC, HSCBCI anode that measures 1-inches

by 9 inches. Use one anode per string as shown in figure D-9. Anode current density is computed as follows:

Output = 1.02/(10 x 0.03) = 0.34 A/sq ft.

Because this exceeds the recommended maximum anode current density shown in table D-1, the type B

anode is the best choice.

0.012P log (D/a)

R'

,

L

where P = 4000 ohm-centimeter, D = 56 feet, L = 5 feet, and a = 16 0.275 = 4.4 feet (factor from fig D-10)

(0.012)(4000 ohm¢imeters) log (56 ft/4.4 ft)

R'

5 ft

48 log 12.73

R'

5

48 1.105

R'

5

R = 10.6 ohms.

L/d = 60/1 = 60 < 100

Using the fringe factor from curve figure D-4, 0.90--

R (adjusted) = 10.6 x 0.90 = 9.54 ohms.

(16) Find the voltage drop in the stub anode circuit.

(a) The electrical conductor to the stub anodes is wire size No.2 AWG, rated at 0.159 ohm/1000

feet, and has an estimated length of 200 feet. Thus,

R = (200 ft/1000 ft) x 0.159 ohm/1000 ft = 0.032 ohm.

(b) To find the voltage drop in the stub anode feeder--

E = IR

where I = 1.02 amperes and R = 0.032 ohm. Thus,

E = 1.02 A x 0.032 ohm

E = 0.033 V.

(c) Find the voltage drop in anode suspension conductors. First, compute the resistance (R) for an

estimated 50-foot-long, No.2 AWG wire rated at 0.159 ohm per 1000 feet:

R = (50/1000) x 0.159 = 0.008 ohm.

E = IR,

where I = 1.02/10 = 0.102 ampere and R = 0.008 ohm. Thus,

E = 1.02 A x 0.008 ohm

E = negligible.