0108im.jpg)
ETL 1110-9-10(FR)
5 Jan 91
Where:
r=
23 ft (Tank radius
from
item
3
of
paragraph 2-8a)
h=
20 ft (High water depth from item 3 of
paragraph 2-8a)
A=
2 it x 23 x 20 + it x 232
A=
2890 + 1662
A=
4552 sq ft
2)
Compute the current requirement (I) using equation
1-11:
I = (A)(I')(l.0 - CE)
Where:
A=
4552 sq ft
(Area of tank to be
protected from previous calculation)
I' =
2.5 mA/sq ft (Required current
density
from item 6 paragraph 2-8b)
CE =
0.0 (Coating efficiency, assuming tank
will eventually be essentially bare)
Current required:
I=
2.5 IDA/sq ft x 4552 sq ft
I=
11,380 IDA; use 12 amp
Since the computed 12 amp is larger than the tested
requirement of 9 amp, use the 12 amp as the required
current.
3)
Calculate the length of anode wire in ft (LB) needed
for the current required, using a modification of
equation 1-2:
LB =
I
IA
Where:
I=
12
amp
(Current
requirement
from
previous calculation)
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