ETL 1110-9-10(FR)

5 Jan 91

Where:

4000 ohm-cm (Water resistivity from

item 3 of paragraph 2-7a)

D

=

56 ft (Tank diameter from item 2 of

paragraph 2-7a)

AR =

7 ft (Radius of stub anode circle

from step lib of paragraph 2-7b)

DE =

0.07 (Equivalent

diameter

factor

from figure 2-17)

LB =

4 ft (Length of each stub anode)

RN =

0.0052 x 4000 x ln [56/(14 x 0.07)]

4

RN =

21.03 ohms.

Because the stub anodes are

short,

their

anode-to-water resistance may have to be adjusted

by the fringe factor. The fringe factor depends on

the ratio of length-to-diameter (L/d). If L/d is

less than 100, obtain the fringe factor from figure

2-18. Multiply the calculated stub anode-to-water

resistance by the fringe factor (F) to obtain the

RADJ = RN x F

(eq 2-7)

Where:

RN =

Stub anode-to-water resistance from

step 12.

F

=

Fringe factor from figure 2-18.

In this example, L/d for the stub anodes is:

L = 4 ft

94

Integrated Publishing, Inc. |