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2. adjust your wire
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Thermit welding for easy way
Step by step Thermit welding :
1. clean the are
2. adjust your wire
3. please watch this video
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Please contact us : Cathodic Protection System Engineering and Supply
081294392672 or 08565305351
1. clean the are
2. adjust your wire
3. please watch this video
Please like and share
Engineering Service and Material Supply
Please contact us : Cathodic Protection System Engineering and Supply
081294392672 or 08565305351
Dasar cathodic Portection system
Dasar cathodic protection system for presentation, dasar proteksi katodik
Presentation of cathodic protection system, basic about cathodic protection system if any question, please contact me (08565305351 - bajakz@gmail.com)
Presentation of cathodic protection system, basic about cathodic protection system if any question, please contact me (08565305351 - bajakz@gmail.com)
Consideration for Design of Galvanic Anode Cathodic Protection System
by Ernest Klechka, P.E, NACE International CP Instructor
Normally cathodic protection (CP) can be applied by a sacrificial anode or impressed current system. CP can be applied by galvanic or sacrificial system when limited amounts of current are needed, soil resistivity is low (normally less than 5,000 ohm-cm), and electric power is limited or not available. Sacrificial anode system have the added advantage of requiring minimum maintenance.
To design a sacrificial anode system, the following information is needed:
1. Current requirement (I current required)
2. Anode resistance ( R anode), calculated or based on the manufacturer's data
3. Design life
Design Current Requirements
CP current requirements can be determined by calculation based on bare or exposed surface area, assumptions about existing coatings, the coating damage factor, and the current density (CD) required for CO in environment are needed.
To calculate the current required, the total surface area (Atotal) in sq m of the structure is defined. Base on experience, the coating damage factor (fdamage) or percent bare area is determined. The total surface area to be protected by CP (Acp) in sqm is then :
ACP = Atotal fdamage
Based on the environment to which the structure is exposed, the applicable CD for CP (Icd) is estimated in mA/sqm. Using the ACP and Icd, the calculated current required (Icurrent required) is then :
Icurrent required = ACP Icd
Current Requirements Based on Field Tests
Current requirement test are the most reliable way to estimate the current requirements for existing structure and take info account the condition of the coatings and variations in soil resistivity. If the structure is coated and isolated, it is possible to directly determine the current requirements.
A temporary anode (groundbed) is established and a portable power supply (usually a battery, generator, or portable rectifier) is installed. A current is applied (
I) and the change in polarized potential ( V) is determined. The ratio of the current applied dicided by the change in polarized potential (I/V) is used to calculated the current requirement.
For example, if 5A are applied to the structure and a polarized potential shift of 50 mA occur, and a 100 mV shift is needed for CP, then :
Icurrent required = shift needed x (I/V)
= 100 mV x (5A / 50mV)
= 10 A
Determining the Output of a Sacrificial Anode
Often the manufacture of an anode will provide the estimated current output in the form of table or graphs based on the shape and size of their anode and soil resistivity. The average soil (electrolyte) resistivity is needed to use these table or graphs or for calculation.
if this data is not available, the output of an anode can be estimated based on Ohm's Law :
I anode = E driving potential / R anode
Anode Resistance-to-Earth
Typically, since sacrificial anodes are buried vertically. For a single certical anode, such as a package magnesium anode, the resistance-to-earth can be calculated using Dwight's equation for a single vertical rod or pipe :
R = (0.00159
/
L) [ln (8L/d)-1]
where :
R = groundbed resistance (
)
= resistivity ( -m)
d = diameter of anode (m)
The resistivity value used must be representative of the volume resistivity affecting the anode.
Parallel Anode
Sunde's equation can be used to estimate the resistance of distributted parallel anodes:
RN = (0.00159/ L) [ln (8L/d)-1] - 1 + (2L/S) ln(0.656 N)
where :
RN = groundbed resistance ()
= resistivity ( -m)
d = diameter of anode (m)
L = length of anode (m)
S = spacing of anode in the groundbed (m)
If the anode are separate by 6 m or more, the parallel effect of anode is negligible.
Anode Current Output
For a single highh-[ptential (-1.75 Vcse) 7.7 kg (17 lb) magnesium anode, the current output can be calculated or determined based on the manufacturer's data.
A typical 7.7 kg anode is 1,295 mm long by roughly 51 mm square. Because Dwight's equation deals with rods, an equivalent rod with the same circumference as a 51mm square is needed; the rod will be 65 mm in diameter. The equivalent diameter d for square with sides S is given as :
3.14 d = 4 S
d = 4 S /3.14
If the structure is to be polarized to -0.850 Vcse, driving potential is then:
Enet = -1.75 -(-0.85) = -0.90 V
Anode resistance can be calculated using Dwight's equation for 5,000-cm soil :
Ranode = (0.00159 * 5,000 / 3.14*1,295) [ln (8*1,295/65-1]
= 25
Assuming the structure and cable resistance are negligible, the expected current is then :
Ianode = Enet / Ranode
= 0.90 V / 25
= 0.036 A (36mA)
The anticipated current output is then 36mA from single 7.7 kg anode.
Anode supplier literature indicated that in 5,000-cm soil, a high-potential magnesium anode will have an anticipated output of 0.040 A (40 mA) to a structure polarized to -0.85Vcse. The data infers that the structure has negligible resistance to earth and, therefore, no IR drop. The resistance to remote earth of single high-potential (-1.75 Vcse) magnesium anode can be calculated:
Ranode = Enet / Ianode
=(1.75 - 0.85) / 0.04
= 22.5
The two value of current output and resistance are very similar.
Number of Anode Needed Bases on Current Requirements
Once the total current required and the current output from a single anode are determined, the number of anode needed to protect the structure can be calculated. Based on current requirement :
Nanode = Icurrent required / Ianode
After the number of anode is calculated, the number of anode must be rounded up ti the next integer (no partial anode allowed).
Number of Anode Needed Based on Design Life
The number of anode needed can also be calculated based on the current requires an the design life. Total weight need for design life can be based on Faraday's Law :
Wtotal =K I T
where:
K = Consumption (kg/A-h x 24 h/day x 365 days) ( see the table I )
I = current in A (Icurrent required)
T = time in years (design life)
Then the number of anode needed is equal to the total weight needed divided by weight of single anode:
Nanode = W total / W anode
Table I
How Many Anode Are Really Needed?
The number of anode needed, therefore, is the larger of the number needed based on current requirements and the number needed based on design life.
Because the current provided can be greater than the current required, control resistor may be needed to reduce the initial output of anodes. After a period of time, the current required may change due to changes in the condition of coating or the environment. Control resistor may needed to be changed with time.
Galvanic Anode Life
Once the number of anode needed is determined, the life expectancy for the system should be checked. The following equations can be used to determine the anticipated life of the sacrificial anode system.
Magnesium years of life = 0.256 x anode weght in kg x efficiency x utilization factor Current in A
Zinc years of life = 0.0935 x anode weght in kg x efficiency x utilization factor Current in A
Post-Installation Measurements
After the systems installed, the structur-to-soil potentials sould be measured as we;; as the current output from the anode system. IF the current output needed adjustment, more anodes can be added to increase current or control resisters can be added to reduce the current output.
Anode life can be calculated based on the actual measured current of the sacrificial anode system.
Conclusions
The design of a sacrificial anode CP system can be accomplished as long as the current demand and the anode current output can be determined. Anode separated by more tham 6 m can be assumed to have minimum parallel interference.
Sacrificial of galvanic anode CP can be very effective as long as current demands are low and soil resistivities are moderate to low. Added benefits of these system are the reduce maintenance cost, no electrical power is required, and ease of installation. Low-power sacrificialsytem also cause a minimum of interference and can be used to discharge CP interference
Normally cathodic protection (CP) can be applied by a sacrificial anode or impressed current system. CP can be applied by galvanic or sacrificial system when limited amounts of current are needed, soil resistivity is low (normally less than 5,000 ohm-cm), and electric power is limited or not available. Sacrificial anode system have the added advantage of requiring minimum maintenance.
To design a sacrificial anode system, the following information is needed:
1. Current requirement (I current required)
2. Anode resistance ( R anode), calculated or based on the manufacturer's data
3. Design life
Design Current Requirements
CP current requirements can be determined by calculation based on bare or exposed surface area, assumptions about existing coatings, the coating damage factor, and the current density (CD) required for CO in environment are needed.
To calculate the current required, the total surface area (Atotal) in sq m of the structure is defined. Base on experience, the coating damage factor (fdamage) or percent bare area is determined. The total surface area to be protected by CP (Acp) in sqm is then :
ACP = Atotal fdamage
Based on the environment to which the structure is exposed, the applicable CD for CP (Icd) is estimated in mA/sqm. Using the ACP and Icd, the calculated current required (Icurrent required) is then :
Icurrent required = ACP Icd
Current Requirements Based on Field Tests
Current requirement test are the most reliable way to estimate the current requirements for existing structure and take info account the condition of the coatings and variations in soil resistivity. If the structure is coated and isolated, it is possible to directly determine the current requirements.
A temporary anode (groundbed) is established and a portable power supply (usually a battery, generator, or portable rectifier) is installed. A current is applied (
I) and the change in polarized potential ( V) is determined. The ratio of the current applied dicided by the change in polarized potential (I/V) is used to calculated the current requirement.For example, if 5A are applied to the structure and a polarized potential shift of 50 mA occur, and a 100 mV shift is needed for CP, then :
Icurrent required = shift needed x (
I/V)= 100 mV x (5A / 50mV)
= 10 A
Determining the Output of a Sacrificial Anode
Often the manufacture of an anode will provide the estimated current output in the form of table or graphs based on the shape and size of their anode and soil resistivity. The average soil (electrolyte) resistivity is needed to use these table or graphs or for calculation.
if this data is not available, the output of an anode can be estimated based on Ohm's Law :
I anode = E driving potential / R anode
Anode Resistance-to-Earth
Typically, since sacrificial anodes are buried vertically. For a single certical anode, such as a package magnesium anode, the resistance-to-earth can be calculated using Dwight's equation for a single vertical rod or pipe :
R = (0.00159
/
L) [ln (8L/d)-1]where :
R = groundbed resistance (
) = resistivity ( -m)d = diameter of anode (m)
The resistivity value used must be representative of the volume resistivity affecting the anode.
Parallel Anode
Sunde's equation can be used to estimate the resistance of distributted parallel anodes:
RN = (0.00159
/ L) [ln (8L/d)-1] - 1 + (2L/S) ln(0.656 N)where :
RN = groundbed resistance (
) = resistivity ( -m)d = diameter of anode (m)
L = length of anode (m)
S = spacing of anode in the groundbed (m)
If the anode are separate by 6 m or more, the parallel effect of anode is negligible.
Anode Current Output
For a single highh-[ptential (-1.75 Vcse) 7.7 kg (17 lb) magnesium anode, the current output can be calculated or determined based on the manufacturer's data.
A typical 7.7 kg anode is 1,295 mm long by roughly 51 mm square. Because Dwight's equation deals with rods, an equivalent rod with the same circumference as a 51mm square is needed; the rod will be 65 mm in diameter. The equivalent diameter d for square with sides S is given as :
3.14 d = 4 S
d = 4 S /3.14
If the structure is to be polarized to -0.850 Vcse, driving potential is then:
Enet = -1.75 -(-0.85) = -0.90 V
Anode resistance can be calculated using Dwight's equation for 5,000
-cm soil :Ranode = (0.00159 * 5,000 / 3.14*1,295) [ln (8*1,295/65-1]
= 25
Assuming the structure and cable resistance are negligible, the expected current is then :
Ianode = Enet / Ranode
= 0.90 V / 25
= 0.036 A (36mA)
The anticipated current output is then 36mA from single 7.7 kg anode.
Anode supplier literature indicated that in 5,000
-cm soil, a high-potential magnesium anode will have an anticipated output of 0.040 A (40 mA) to a structure polarized to -0.85Vcse. The data infers that the structure has negligible resistance to earth and, therefore, no IR drop. The resistance to remote earth of single high-potential (-1.75 Vcse) magnesium anode can be calculated:Ranode = Enet / Ianode
=(1.75 - 0.85) / 0.04
= 22.5
The two value of current output and resistance are very similar.
Number of Anode Needed Bases on Current Requirements
Once the total current required and the current output from a single anode are determined, the number of anode needed to protect the structure can be calculated. Based on current requirement :
Nanode = Icurrent required / Ianode
After the number of anode is calculated, the number of anode must be rounded up ti the next integer (no partial anode allowed).
Number of Anode Needed Based on Design Life
The number of anode needed can also be calculated based on the current requires an the design life. Total weight need for design life can be based on Faraday's Law :
Wtotal =K I T
where:
K = Consumption (kg/A-h x 24 h/day x 365 days) ( see the table I )
I = current in A (Icurrent required)
T = time in years (design life)
Then the number of anode needed is equal to the total weight needed divided by weight of single anode:
Nanode = W total / W anode
Table I
How Many Anode Are Really Needed?
The number of anode needed, therefore, is the larger of the number needed based on current requirements and the number needed based on design life.
Because the current provided can be greater than the current required, control resistor may be needed to reduce the initial output of anodes. After a period of time, the current required may change due to changes in the condition of coating or the environment. Control resistor may needed to be changed with time.
Galvanic Anode Life
Once the number of anode needed is determined, the life expectancy for the system should be checked. The following equations can be used to determine the anticipated life of the sacrificial anode system.
Magnesium years of life = 0.256 x anode weght in kg x efficiency x utilization factor Current in A
Zinc years of life = 0.0935 x anode weght in kg x efficiency x utilization factor Current in A
Post-Installation Measurements
After the systems installed, the structur-to-soil potentials sould be measured as we;; as the current output from the anode system. IF the current output needed adjustment, more anodes can be added to increase current or control resisters can be added to reduce the current output.
Anode life can be calculated based on the actual measured current of the sacrificial anode system.
Conclusions
The design of a sacrificial anode CP system can be accomplished as long as the current demand and the anode current output can be determined. Anode separated by more tham 6 m can be assumed to have minimum parallel interference.
Sacrificial of galvanic anode CP can be very effective as long as current demands are low and soil resistivities are moderate to low. Added benefits of these system are the reduce maintenance cost, no electrical power is required, and ease of installation. Low-power sacrificialsytem also cause a minimum of interference and can be used to discharge CP interference
Examples of Design for Cathodic Protection Systems
From Estimated Exposed Surface Area
Estimating current requirements from expected exposed surface is always subject to
error. There are many factors, which affect the results.
Consider:
• Total surface area in contact with soil or other electrolyte.
• Dielectric properties of any protective coating.
• Factors which may damage a protective coating during installation.
• Expected protective coating life under service conditions.
• Expected percentage coverage by protective coating.
• Past experience with coating applicators and construction contractors.
• Current density required for cathodic protection of the metal(s) in the
environment.
In the end, the expected current requirement depends on calculating the area of
exposed metal in contact with the electrolyte and multiplying it by the “best estimate”
of current density for the conditions present.
There is an alternate approach for coated electrically isolated structures (pipes, under-
ground storage tanks, etc.) where there is data available on existing cathodic protection
systems.
The approach requires reliable local data on:
• Expected leakage conductance (Siemens/unit area) in 1000 ohm cm. soil for a
class of coating (epoxy, polyethylene tape, etc.) and type of service
(transmission pipeline, gas distribution, fuel tank).
• Soil resistivity in the service area.
• Structure to soil potential shift required to produce polarization needed to meet
cathodic protection criteria. This is the immediate change in potential of an
isolated structure measured to a point at “remote earth” when cathodic
protection is applied.
The value is not a criteria for protection. However, under a given set of operating and exposure conditions, a potential shift will provide a good estimate of current needed to meet accepted criteria.
The approach is best understood by using an example.
Example 5.1
A gas utility is planning to install 3049 meters (10,000 feet) of 5.1 cm (2 inch) coated
steel distribution mains in a new development. The average soil resistivity in the area
is 5,000 ohm cm. The corrosion engineer wishes to estimate the approximate current
required to cathodically protect the pipes.
Experience in the utility has developed the following data on cathodic protection
current requirements:
Average leakage conductance G for distribution type service is 2.14 × 10−3S/m2in
1000 ohm cm soil.
Average potential shift measured to “remote earth” to achieve protection is −0.250
volt.
Calculations:
Total surface area of the proposed pipe.
As=πd L = (5.1 × 3.1416/100) × 3049 = 488 sq. meters
Estimated leakage conductance of new pipe in 1000 ohm cm soil.
g = G × A = 2.14 × 10−3×488 = 1.04 Siemens
Since resistance = 1/conductance
Resistance to remote earth = 1/1.04 = 0.96 ohm
Estimated resistance to remote earth in 5000 ohm cm soil. (Resistance is directly pro-
portional to resistivity).
0.96 × 5 = 4.8 ohms
Estimated current to shift pipe potential to remote earth −0.250 volt. From Ohm’s
Law (I = E/R)
0.250/4.8 = 0.052 A.
for detail you can buy PEABODY or cantact me for getting ebook
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Estimating current requirements from expected exposed surface is always subject to
error. There are many factors, which affect the results.
Consider:
• Total surface area in contact with soil or other electrolyte.
• Dielectric properties of any protective coating.
• Factors which may damage a protective coating during installation.
• Expected protective coating life under service conditions.
• Expected percentage coverage by protective coating.
• Past experience with coating applicators and construction contractors.
• Current density required for cathodic protection of the metal(s) in the
environment.
In the end, the expected current requirement depends on calculating the area of
exposed metal in contact with the electrolyte and multiplying it by the “best estimate”
of current density for the conditions present.
There is an alternate approach for coated electrically isolated structures (pipes, under-
ground storage tanks, etc.) where there is data available on existing cathodic protection
systems.
The approach requires reliable local data on:
• Expected leakage conductance (Siemens/unit area) in 1000 ohm cm. soil for a
class of coating (epoxy, polyethylene tape, etc.) and type of service
(transmission pipeline, gas distribution, fuel tank).
• Soil resistivity in the service area.
• Structure to soil potential shift required to produce polarization needed to meet
cathodic protection criteria. This is the immediate change in potential of an
isolated structure measured to a point at “remote earth” when cathodic
protection is applied.
The value is not a criteria for protection. However, under a given set of operating and exposure conditions, a potential shift will provide a good estimate of current needed to meet accepted criteria.
The approach is best understood by using an example.
Example 5.1
A gas utility is planning to install 3049 meters (10,000 feet) of 5.1 cm (2 inch) coated
steel distribution mains in a new development. The average soil resistivity in the area
is 5,000 ohm cm. The corrosion engineer wishes to estimate the approximate current
required to cathodically protect the pipes.
Experience in the utility has developed the following data on cathodic protection
current requirements:
Average leakage conductance G for distribution type service is 2.14 × 10−3S/m2in
1000 ohm cm soil.
Average potential shift measured to “remote earth” to achieve protection is −0.250
volt.
Calculations:
Total surface area of the proposed pipe.
As=πd L = (5.1 × 3.1416/100) × 3049 = 488 sq. meters
Estimated leakage conductance of new pipe in 1000 ohm cm soil.
g = G × A = 2.14 × 10−3×488 = 1.04 Siemens
Since resistance = 1/conductance
Resistance to remote earth = 1/1.04 = 0.96 ohm
Estimated resistance to remote earth in 5000 ohm cm soil. (Resistance is directly pro-
portional to resistivity).
0.96 × 5 = 4.8 ohms
Estimated current to shift pipe potential to remote earth −0.250 volt. From Ohm’s
Law (I = E/R)
0.250/4.8 = 0.052 A.
for detail you can buy PEABODY or cantact me for getting ebookthks for all
bajakz
INTERNATIONAL STANDARD ISO 15589-2
Foreword
ISO 15589-2:2004(E)
ISO (the International Organization for Standardization) is a worldwide federation of national standards bodies
(ISO member bodies). The work of preparing International Standards is normally carried out through ISO
technical committees. Each member body interested in a subject for which a technical committee has been
established has the right to be represented on that committee. International organizations, governmental and
non-governmental, in liaison with ISO, also take part in the work. ISO collaborates closely with the
International Electrotechnical Commission (IEC) on all matters of electrotechnical standardization.
International Standards are drafted in accordance with the rules given in the ISO/IEC Directives, Part 2.
The main task of technical committees is to prepare International Standards. Draft International Standards
adopted by the technical committees are circulated to the member bodies for voting. Publication as an
International Standard requires approval by at least 75 % of the member bodies casting a vote.
Attention is drawn to the possibility that some of the elements of this document may be the subject of patent
rights. ISO shall not be held responsible for identifying any or all such patent rights.
ISO 15589-2 was prepared by Technical Committee ISO/TC 67, Materials, equipment and offshore structures
for petroleum, petrochemical and natural gas industries, Subcommittee SC 2, Pipeline transportation systems.
ISO 15589 consists of the following parts, under the general title Petroleum and natural gas industries —
Cathodic protection of pipeline transportation systems:
— Part 1: On-land pipelines
— Part 2: Offshore pipelines
Introduction
Pipeline cathodic protection is achieved by the supply of sufficient direct current to the external pipe surface,
so that the steel-to-electrolyte potential is lowered to values at which external corrosion is reduced to an
insignificant rate.
Cathodic protection is normally used in combination with a suitable protective coating system to protect the
external surfaces of steel pipelines from corrosion.
External corrosion control in general is covered by ISO 13623.
Users of this part of ISO 15589 should be aware that further or differing requirements may be needed for
individual applications. This part of ISO 15589 is not intended to inhibit alternative equipment or engineering
solutions to be used for the individual application. This may be particularly applicable where there is innovative
or developing technology. Where an alternative is offered, any variations from this part of ISO 15589 should
be identified.
Deviations from this part of ISO 15589 may be warranted in specific situations, provided it is demonstrated
that the objectives expressed in this part of ISO 15589 have been achieved.
More... you can download the file
http://www.4shared.com/file/142724789/8ca67b26/ISO15589-2-2004forOR.html
thks for all
bajakz
ISO 15589-2:2004(E)
ISO (the International Organization for Standardization) is a worldwide federation of national standards bodies
(ISO member bodies). The work of preparing International Standards is normally carried out through ISO
technical committees. Each member body interested in a subject for which a technical committee has been
established has the right to be represented on that committee. International organizations, governmental and
non-governmental, in liaison with ISO, also take part in the work. ISO collaborates closely with the
International Electrotechnical Commission (IEC) on all matters of electrotechnical standardization.
International Standards are drafted in accordance with the rules given in the ISO/IEC Directives, Part 2.
The main task of technical committees is to prepare International Standards. Draft International Standards
adopted by the technical committees are circulated to the member bodies for voting. Publication as an
International Standard requires approval by at least 75 % of the member bodies casting a vote.
Attention is drawn to the possibility that some of the elements of this document may be the subject of patent
rights. ISO shall not be held responsible for identifying any or all such patent rights.
ISO 15589-2 was prepared by Technical Committee ISO/TC 67, Materials, equipment and offshore structures
for petroleum, petrochemical and natural gas industries, Subcommittee SC 2, Pipeline transportation systems.
ISO 15589 consists of the following parts, under the general title Petroleum and natural gas industries —
Cathodic protection of pipeline transportation systems:
— Part 1: On-land pipelines
— Part 2: Offshore pipelines
Introduction
Pipeline cathodic protection is achieved by the supply of sufficient direct current to the external pipe surface,
so that the steel-to-electrolyte potential is lowered to values at which external corrosion is reduced to an
insignificant rate.
Cathodic protection is normally used in combination with a suitable protective coating system to protect the
external surfaces of steel pipelines from corrosion.
External corrosion control in general is covered by ISO 13623.
Users of this part of ISO 15589 should be aware that further or differing requirements may be needed for
individual applications. This part of ISO 15589 is not intended to inhibit alternative equipment or engineering
solutions to be used for the individual application. This may be particularly applicable where there is innovative
or developing technology. Where an alternative is offered, any variations from this part of ISO 15589 should
be identified.
Deviations from this part of ISO 15589 may be warranted in specific situations, provided it is demonstrated
that the objectives expressed in this part of ISO 15589 have been achieved.
More... you can download the file
http://www.4shared.com/file/142724789/8ca67b26/ISO15589-2-2004forOR.html
thks for all
bajakz
Basic Calculation For Sacrificial System
A. Current requirement ( I ) = SA x CD
Where :
SA = Surface area (taken from Norton Corrosion calculation)
CD = Current density (taken from NACE standard RP 0176-83)
(Standard Recommended Practice Corrosion Control of Steel, fixed Offshore Platform
Associated with Petroleum Production)
Sea water = 55 mA.sq.m
Seabed = 11 mA/sq.m
B. Anode Weight = N
N = (I x Con x L) : U
Where,
I = Current requirement (Amp)
Con = Consumption rate of aluminium anode = 3.35 Kg/AY
Consumption rate of magnesium anode = 7.7 Kg/AY
L = Life time = 10 years
U = Utilization factor = 0.9
C. Example
Data:
Atmospheric Storage Tank:
- Length (L) : 5.54 m
- Diameter (D) : 9.058 m
- Water level Assume : 1 m
Current Density (CD) : 55 mA/m2
Consumption Rate of Aluminum Anode : 3.5 kg/Amp Year
Years design : 10 tahun
Utility Factor : 0.9
Calculation:
1 Surface Area (SA)
- Surface Area of Ground = pi x D2 = 3.14 x 82.047 = 257.62 m2
- Surface Area of wall = pi x D x L = 3.14 x 9.058 x 5.54 = 157.56 m2
Total Surface Area = 415.18 m2
2 Current Requirement (Ip)
Ip = SA x CD
= 415.18 m2 x 55 mA/m2
= 22,834.9 mA = 22.835 Amp
3 Weight of Anode (W)
W.Al = Ip x Y x 3.5 kg/Amp Year
U = 22.835 Amp x 10 years x 3.5 kg/Amp Year : 0.9
= 799.225 kgs
4 Number of Aluminium Anode (N)
Proposed Aluminium Anode type 50 kg - Net weight = 50 kg/ea
N.Al = W.Al : 50 kg
= 799.225 kgs : 50 kg
= 15.985 pcs = 16 pcs
Total = 16 pcs Aluminum Anode type 50 kg
regards
bajakz
email : bajakz@gmail.com
note : if you need more of calculation for cathodic protection, please contact me
Where :
SA = Surface area (taken from Norton Corrosion calculation)
CD = Current density (taken from NACE standard RP 0176-83)
(Standard Recommended Practice Corrosion Control of Steel, fixed Offshore Platform
Associated with Petroleum Production)
Sea water = 55 mA.sq.m
Seabed = 11 mA/sq.m
B. Anode Weight = N
N = (I x Con x L) : U
Where,
I = Current requirement (Amp)
Con = Consumption rate of aluminium anode = 3.35 Kg/AY
Consumption rate of magnesium anode = 7.7 Kg/AY
L = Life time = 10 years
U = Utilization factor = 0.9
C. Example
Data:
Atmospheric Storage Tank:
- Length (L) : 5.54 m
- Diameter (D) : 9.058 m
- Water level Assume : 1 m
Current Density (CD) : 55 mA/m2
Consumption Rate of Aluminum Anode : 3.5 kg/Amp Year
Years design : 10 tahun
Utility Factor : 0.9
Calculation:
1 Surface Area (SA)
- Surface Area of Ground = pi x D2 = 3.14 x 82.047 = 257.62 m2
- Surface Area of wall = pi x D x L = 3.14 x 9.058 x 5.54 = 157.56 m2
Total Surface Area = 415.18 m2
2 Current Requirement (Ip)
Ip = SA x CD
= 415.18 m2 x 55 mA/m2
= 22,834.9 mA = 22.835 Amp
3 Weight of Anode (W)
W.Al = Ip x Y x 3.5 kg/Amp Year
U = 22.835 Amp x 10 years x 3.5 kg/Amp Year : 0.9
= 799.225 kgs
4 Number of Aluminium Anode (N)
Proposed Aluminium Anode type 50 kg - Net weight = 50 kg/ea
N.Al = W.Al : 50 kg
= 799.225 kgs : 50 kg
= 15.985 pcs = 16 pcs
Total = 16 pcs Aluminum Anode type 50 kg
regards
bajakz
email : bajakz@gmail.com
note : if you need more of calculation for cathodic protection, please contact me
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