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
244
4
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
Area
Surface
×
mA/m
110
Density
Current
Initial
=
N
2
2
(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/kg
A
2,750
hr/yr
8.760
×
yr
20
Life
×
m
9,300
Area
Surface
×
mA/m
55
Density
Current
Mean
=
N
2
2
(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).
)
) )
]
(
(
[
(
6.5cm
0.9
5.7
13.7
13.7
0.9
r
r
r
r
core
initial
initial
expended
=
−
−
=
×
−
−
=
(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):
A
4.78
=
ohm
0.0523
V
0.25
1
cm
6.5
cm
244
×
4
ln
244
0.159
20
V
0.25
=
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
Area
Surface
×
mA/m
75
Density
Current
Final
=
N
2
2
(D7)
Summary of Contents for CP 1
Page 1: ...CP 1 Cathodic Protection Tester Course Manual February 2005 NACE International 2000 ...
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