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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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