ETL 1110-9-10(FR)

5 Jan 91

Area of 3 tanks = 3 [2 x 3.14 x 42 +3.14 x 8 x 21.25]

= 1905 sq ft

Area per lin ft of 2-in. pipe = 0.621 sq ft/ft (from

table A-2)

Total pipe area = 750 ft x 0.621 sq ft/ft = 466 sq ft

Total area = 1905 sq ft + 466 sq ft = 2371 sq ft

2)

Check the current requirement (I) using equation 1-1:

I = (A)(I')(1.0-CE)

Where:

A =

2371 sq ft (External tank and piping

surface

area

from

previous

calculation)

I' =

2 mA/sq ft (Required current density

[assumed])

CE

=

0.00 (Coating efficiency expressed in

decimal form from item 2 of paragraph

2-3a)

I

=

2371 sq ft (2 mA/sq ft) x (1.0 - 0.0)

I

=

4742 mA or 4.7 amp.

The 4.7 amp would be reasonable for the facility if it

were insulated. The actual current requirement of 8.2

amp occurs because of current loss to other

underground structures and is also reasonable in

relation to that calculated for an isolated facility.

3)

Select an anode and calculate the number of anodes

required (N) to meet the design life requirements.

Calculations can be run on several size anodes, but in

this case 2-in. by 60-in. packaged rod anodes (rod

size = 0.125 in x 4 ft long) are chosen for ease of

construction.

Using equation 1-2, the number of

anodes required to meet the cathodic protection system

design life can be calculated:

N=

I

IA

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