MIL-HDBK-1004/10
where
I
=
14.2
A
R
=
3.03
ohms
E
=
14.2
x 3.03
E
=
43.0
V
(4)
Total voltage drop in main anode circuit:
E
= 0.45 + 43.0
T
E
= 43.45 or 45 V
T
Use a multiplying factor of 1.5, or 67.5 V.
(5) The nearest commercially available rectifier meeting the
above requirement is a single-phase, 80-V unit.
n) Selection of 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 measuring 1-1/2-inch by 9 inches. Use
one anode per string as shown in Figure 99. Anode current density is computed
as follows:
Output = 1.02/(10 x 0.03) = 0.34 A/ft
Because this exceeds the recommended maximum anode current density (refer to
Table 27), the type B anode is the best choice.
o)
Resistance of stub anodes:
R
=
0.012P log D/a/L
where
P=
4,000 ohm-cm
D=
56 feet
L=
5 feet
a=
16 x 0.275 = 4.4 feet (factor from Figure 100)
R=
(0.012 x 4,000 log 56/4.4)/5
R=
48 log 12.73/5
R=
48 x 1.105/5
R=
10.6 ohms
L/d
= 60/1 = 60<100
Fringe factor from curve Figure 101, 0.90 R (adjusted) = 10.6 x
0.90 = 9.54 ohms
p)
Voltage drop in stub anode circuit:
(1) Electrical conductor to stub anodes. Wire size No. 2 AWG,
0.159 ohms/1,000 feet (refer to Table 10), estimated length 200 feet:
R = (200/1,000) x 0.159 = 0.032 ohm
166
Previous Page
