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

Integrated Publishing, Inc. |