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ETL 1110-9-10(FR)
5 Jan 91
Where:
p =
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
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