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)

106

Integrated Publishing, Inc. |