TM 5-811-7
(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.
(15) Find the stub anodes' resistance (from eq D-3):
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.
D-20
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