For magnesium this would be
lb Mg= A yr Z
8760
500
Z 17: 52:
The number of years (YN) for which 1lbofmetal can produce acurrent of
1mAi sd etermined from the following equation:
YN Z
PE
10
K 3
A8760 h = yr
:
For magnesium this would be
500
10
K 3
ð 8760Þ
Z 60 years:
The life expectancy ( L )ofananode of W lb, delivering acurrent of i mA is
calculated as follows:
L Z
YNð W Þ
i
:
For magnesium this would be
L Mg Z
60ð W Þ
i
that is based on a50% anodicefficiency.Because actual efficiencies tend to be
somewhat less, it is advisable to apply asafety factor and multiply the result
by 0.75.
The current requiredtosecure protection of astructure and the available
cell voltagebetweenthe metal structure and sacrificial anodedetermine the
number of anodes required. This can be illustrated by the following
example:
Assume that an undergroundpipeline has an external area of 200 ft.
2
and a
soil resistivity of 600 O cm. Fieldtests indicate that 6mA/ft.
2
is required for
protection. To providep rotection fort he entire pipeline( 6mA/ft.
2
)
(200 ft.
2 ) Z 1200 mA. Magnesium anodes used in this particular soil have a
voltage of K 1.65 V, or ag alvanic cell voltage of
E cell Z E C K E A Z K 0 : 85K ð K 1 : 65Þ Z C 0 : 8V:
The resistance is therefore
R Z
V
I
Z
0 : 8
1 : 2
Z 0 : 67 U :
Cathodic Protection
49
CAT8247—CHAPTER 2—6/10/2006—12:05—SRIDHAR—XML MODEL B–pp. 43–54