Cathodic Protection
301
The number of pounds of metal required to provide a current of 1 A for a
year can be determined from the following equation:
lb metal/A–yr =
8760 h/yr
PE
 
 
For magnesium this would be
lb Mg/A–yr =
8760
500
=17 52
.
The number of years (YN) for which 1 lb of metal can produce a current of 1
mA is determined from the following equation:
YN = 
PE
10 A
h/yr
–3
8760
For magnesium this would be
500
10 (
)
years
–3
8760
60
=
The life expectancy (L) of an anode of W lb, delivering a current of 1 mA is
calculated as follows:
L=
YN(W)
1
For magnesium this would be
L =
60(W)
1
Mg
that is based on a 50% anodic efficiency. Because actual efficiencies tend to be
somewhat less, it is advisable to apply a safety factor and multiply the result
by 0.75.
The current required to secure protection of a structure and the available cell
voltage between the metal structure and sacrificial anode determine the number of anodes required. This can be illustrated by the following example:
Assume that an underground pipeline has an external area of 200 ft 2 and
a soil resistivity of 600 Ω·cm. Field tests indicate that 6 mA/ft 2 is required for
protection. To provide protection for the entire pipeline (6 mA/ft 2 ) (200 ft 2 )