RP0176-2003
NACE International 39
ment). For this exercise, we are given the following inform-
ation:
Structure surface area = 9,300 m
2
(100,000 ft
2
)
Design life = 20 years
ρ = 20 ohm-cm (from Table A1).
Based on this information, we select an anode with the
following characteristics:
Material: aluminum-zinc-mercury alloy
E = 0.25 V driving force between an aluminum or zinc
anode of -1.50 V (Ag/AgCl [sw] reference).
L = 244-cm (96-in.) anode length,
r = 13.7-cm (5.40-in.) anode radius initially for a 22- x 22-
cm (8.5- x 8.5-in.) anode on a 10-cm (4-in.) core,
Wt = 330-kg (725-lb) weight per anode,
CC = 2,750 A-hr/kg (1,250 A-hr/lb) for current capacity for
Al-Zn-Hg alloy anodes from Table B1 or anode
supplier’s specifications,
r
core
= 5.7 cm (2.25 in.), which is one-half the OD of an 11-
cm (4-in.) Sch. 80 pipe.
Therefore, the initial current output per anode is as shown in
Equation (D2):
5.86A
ohm 0.0426
0.25V
1
cm 13.7
cm 2444
ln
244
0.159
20
0.25V
R
E
I ==
−
×
== (D2)
The number of anodes required to protect a structure with 9,300 m
2
(100,000 ft
2
) of exposed surface area is as shown in Equation
(D3):
()
anodes 175=
mA/A 1,000×5.86 anode per output Amps
m 9,300 AreaSurface×mA/m 110Density Current Initial
=N
22
(D3)
The initial current density (110 mA/m2) is obtained from
Table A1.
In order to meet the second current density requirement for
the structure, which determines the number of kg (lb) of
anode material required to protect the structure over the 20-
year design life, use Equation (D4):
()
anodes 99=
mA/A 1,000×kg/anode 330×hr/kgA 2,750
hr/yr 8.760× yr20 Life×m 9,300 AreaSurface×mA/m 55Density Current Mean
= N
22
(D4)
The mean current density (55 mA/m
2
) is obtained from
Table A1.
Lastly, the number of anodes to provide the final current
requirement is calculated in a manner similar to the initial
current requirement, except that the expended dimensions
of the anode are used to represent the anode at the end of
its life (see Equation (D5).
)))
]
((
[
(
6.5cm0.95.713.713.70.9rrrr
coreinitialinitialexpended
=−−=×−−= (D5)
where 0.9 is the anode utilization factor for a standoff
anode. Assume no change in anode length.
The final current output per anode is shown in Equation
(D6):
A4.78=
ohm 0.0523
V0.25
1
cm 6.5
cm 244×4
ln
244
0.159
20
V0.25
=
R
E
=I = (D6)
The number of anodes required to protect a structure with 9,300 m
2
(100,000 ft
2
) of exposed surface area is as shown in
Equation (D7):
()
anodes 146=
mA/A 1,000×4.78 anode per output Amps
m 9,300 AreaSurface×mA/m 75Density Current Final
=N
22
(D7)
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