cathodic protection - introduction

8/13/2019 Cathodic Protection - Introduction

1/38

P D H en g i n eer .c o m CourseMA-4003

Cathodic Protection

Thisdocumentisthecoursetext.Youmayreviewthismaterialat

yourleisurebeforeorafteryoupurchasethecourse. Ifyouhavenot

alreadypurchasedthecourse,youmaydosonowbyreturningtothe

courseoverviewpagelocatedat:

http://www.pdhengineer.com/pages/MA4003.htm

(Pleasebesuretocapitalizeandusedashasshownabove.)

Oncethecoursehasbeenpurchased,youcaneasilyreturntothe

courseoverview,coursedocumentandquizfromPDHengineersMy

Accountmenu.

Ifyouhaveanyquestionsorconcerns,rememberyoucancontactus

byusingtheLiveSupportChatlinklocatedonanyofourwebpages,

[email protected]

freeat1877PDHengineer.

ThankyouforchoosingPDHengineer.com.

PDHengineer.com,aservicemarkofDecaturProfessionalDevelopment,LLC. MA4003C1


2/38

CATHODIC PROTECTION


3/38


4/38

3

Introduction

Cathodic protection is a technique to control the corrosion of a metal surface by making it workas a cathode of an electrochemical cell. Corrosion is an electrochemical process in which a

current leaves a structure at the anode site, passes through an electrolyte, and re-enters thestructure at the cathode site.

For example, see Figure 1 where one small section of a pipeline isanodic because it is in a soil with low resistivity compared to the rest

of the pipeline. Current leaves the pipeline at that anode site, passesthrough the soil, and re-enters the pipeline at a cathode site. Currentflows because of a potential difference between the anode andcathode. That is, the anode potential is more negative than thecathode potential, and this difference is the driving force for thecorrosion. The total systemanode, cathode, electrolyte, and metallicconnection between anode and cathodeis termed a corrosion cell.

Cathodic protection (CP) is a method to reduce corrosion byminimizing the difference in potential between anode and cathode.This is achieved by applying a current to the structure to be protected

(such as a pipeline) from some outside source. When enough currentis applied, the whole structure will be at one potential; thus, anode andcathode sites will not exist. A cathodic protection system preventscorrosion by converting all of the anodic sites on the metal surface to cathodic sites by supplyingelectrical current from an alternate source. Usually the alternate source is a galvanic anode,which are more active than steel.

Definitions

Anode- electrode at which

oxidation of its surface orsome component of thesolution is occurring.

Cathode- electrode atwhich reduction of itssurface or some componentof the solution is occurring.

Electrolyte- chemicalsubstance or mixture,usually liquid, containingions that migrate in an

electric field such as soil orseawater.


5/38

4

Cathodic protection systems are most commonly used to protect steel, water or fuel pipelines andstorage tanks, steel pier piles, ships, offshore oil platforms and onshore oil well casings. Externalsurfaces of buried metallic structures, surfaces of metal waterfront structures, such as sheetpilings or bearing piles, and the internal surfaces of tanks containing electrolytes, such as water,are applications where cathodic protection is usually technically feasible and cathodic protection

is used in protecting such structures.

When construction of a new buried or submerged system is being planned, the corrosivity of theenvironment should be considered as one of the factors in the design of the system. If experiencewith similar systems in the vicinity of the construction site has shown that the site conditions areaggressive based on leak and failure records, cathodic protection should be considered as ameans of controlling corrosion on the new system. Cathodic protection is one of the few methodsof corrosion control that can be effectively used to control corrosion of existing buried ofsubmerged metal surfaces. Thus, if leak records on an existing system show that corrosion isoccurring, cathodic protection can be applied to stop the corrosion damage from increasing.Cathodic protection can, however, only stop further corrosion from occurring and cannot restore

the material already lost due to corrosion.

In some cases, cathodic protection is required bypolicy or regulation. Regulations by theDepartment of Transportation have establishedstandards for transporting certain liquids andcompressed gas by pipelines in order to establishminimum levels of safety. The regulations requirethat these pipelines be protected by cathodicprotection combined with other means ofcorrosion control, such as protective coatings andelectrical insulation. The regulations provideexcellent guidelines for the application ofcathodic protection to buried and submergedpipelines. In addition to regulations, primarilydue to the safety and environmental consequences of system failure, there are an increasingnumber of federal, state and local governmental regulations regarding the storage andtransportation of certain materials that require corrosion control. Many of these regulations eitherspecify cathodic protection as a primary means of corrosion control or allow its use as analternative method of controlling corrosion.

This course is a brief introduction to cathodic protection. It explains the basics using a couple of

very simple examples. Anyone interested in designing an actual cathodic protection systemshould consult with a corrosion specialist. NACE International is an association of corrosionengineering professionals with design experience in all types of cathodic protection systems.NACE is on the web atwww.NACE.org. NACE also provides industry standards for theapplication of corrosion protection systems.

In the next section we review the basics of how cathodic protection systems operate. Subsequentsections will cover the actual design of a simple galvanic and impressed current system.

Galvanizinggenerally refers to hot-dip

galvanizing which is a way of coating steel with

a layer of metallic zinc. Galvanized coatings are

quite durable in most environments because they

combine the barrier properties of a coating with

some of the benefits of cathodic protection. If the

zinc coating is scratched or otherwise locally

damaged and steel is exposed, the surrounding

areas of zinc coating form a galvanic cell with

the exposed steel and protect it from corrosion.This is a form of localized cathodic protection -

the zinc acts as a sacrificial anode.
http://www.nace.org/http://www.nace.org/http://www.nace.org/http://www.nace.org/


6/38

5

I. Types of Cathodic Protection

There are two main types of cathodic protection systems: galvanic and impressed current. Bothtypes use anodes from which current flows into the electrolyte, a continuous electrolyte from theanode to the protected structure, and an external metallic wire connection. These items are

essential for all cathodic protection systems.

Galvanic system

A galvanic cathodic protection system makes use of the corrosive potentials for different metals.Without cathodic protection, one area of the structure exists at a more negative potential thananother, and corrosion results. If, however, an object with much more negative potential, such asa magnesium anode, is placed adjacent to the structure to be protected, such as a pipeline, and ametallic insulated wire connection is installed between the object and the structure, the objectwill become the anode and the entire structure will become the cathode. Then the new objectcorrodes sacrificially to protect the structure. Thus, the galvanic cathodic protection system is

called asacrificial anode cathodic protection systembecause the anode corrodes sacrificially toprotect the structure. Galvanic anodes are usually made of either magnesium or zinc because ofthese metals higher potentialcompared to steel structures.

Today, galvanic or sacrificial anodes are made in various shapes using alloys of zinc, magnesiumand aluminum. The electrochemical potential, current capacity and consumption rate of thesealloys are superior for corrosion protection than iron.


7/38

6

Galvanic anodes are designed and selected to have a more "active" voltage (i.e., a more negativeelectrochemical potential) than the metal of the structure. For effective corrosion protection, thepotential of the steel surface ispolarizedor pushed - more negative until the surface has auniform potential. At that stage, the driving force for the corrosionreaction is halted. The galvanic anode continues to corrode;

consuming the anode material until eventually it must be replaced.The polarization is caused by the current flow from the anode to thecathode. The driving force for the corrosion protection current flowis the difference in electrochemical potential between the anode andthe cathode.

In sacrificial anode systems the high energy electrons required for cathodic protection aresupplied by the corrosion of an active metal. Sacrificial anode systems depend on the differencesin corrosion potential that are established by the corrosion reactions that occur on differentmetals or alloys.

For example, the natural corrosion potential of iron is about -0.550 volts in seawater. The naturalcorrosion potential of zinc in seawater is about -1.2 volts. Thus if the two metals are electricallyconnected, the corrosion of the zinc becomes a source of negative charge which preventscorrosion of the iron.

The materials used for sacrificial anodes are either relatively pure active metals, such as zinc ormagnesium, or are magnesium or aluminum alloys that have been specifically developed for useas sacrificial anodes. In applications where the anodes are buried, a special backfill materialsurrounds the anode in order to insure that the anode will produce the desired output.

Sacrificial anodes are normally supplied with either lead wires or cast-in straps to facilitate their

connection to the structure being protected. The lead wires may be attached to the structure bywelding or mechanical connections. These should have a low resistance and should be insulatedto prevent increased resistance or damage due to corrosion. When anodes with cast-in straps areused, the straps can either be welded directly to the structure or the straps can be used aslocations for attachment. A low resistance, mechanically adequate, attachment is required forgood protection and resistance to mechanical damage.

In the process of providing electrons for the cathodic protection of a less active metal the moreactive metal corrodes. The more active metal - the anode - is sacrificed to protect the less activemetalthe cathode. The amount of corrosion depends on the metal being used as an anode but isdirectly proportional to the amount of current supplied.

Because the anodes in a galvanic cathodic protection system are sacrificial they must beinspected periodically and replaced when consumed.

Impressed Current Cathodic Protection Systems

Impressed Current Cathodic Protection systems use the same elements as the galvanic protectionsystem; only the structure is protected by applying a current to it from an anode. The anode and

Polarization- deviation

from the open circuitpotential of an electroderesulting from the passageof current.


8/38

7

the structure are connected by an insulated wire, as for the galvanic system. Current flows fromthe anode through the electrolyte onto the structure, just as in the galvanic system. The maindifference between galvanic and impressed current systems is that the galvanic system relies onthe difference in potential between the anode and structure, whereas the impressed currentsystem uses an external power source to drive the current. See Figure 3.

For larger structures, galvanic anodes cannot economically deliver enough current to providecomplete protection. Impressed current cathodic protection (ICCP) systems use anodesconnected to a DC power source known as a cathodic protection rectifier. Anodes for ICCPsystems are tubular and solid rod shapes or continuous ribbons of various specialized materials.These include high silicon cast iron, graphite, mixed metal oxide, platinum and niobium coatedwire and others.

A typical ICCP system for a pipeline would include an AC powered rectifier with a maximumrated DC output of between 10 and 50 amperes and 50 volts. The positive DC output terminal isconnected via cables to the array of anodes buried in the ground and is known as the anodegroundbed. For many applications the anodes are installed in a 200 foot deep, 10-inch diametervertical hole and backfilled with conductive coke. Conductive cokeis a material that improves

the performance and life of the anodes. A cable rated for the expected current output connectsthe negative terminal of the rectifier to the pipeline. The operating output of the rectifier isadjusted to the optimum level after conducting various tests including measurements ofelectrochemical potential.

As in sacrificial anode systems, impressed current systems depend on a supply of high energyelectrons to stifle anodic reactions on a metal surface. In the case of an impressed current systemthese high energy electrons are supplied by a rectifier.


9/38


10/38

9

II. Cathodic Protection Testing

There are two tests that are important in cathodic protection systems. They are: soil resistivitymeasurements and required current testing.

Soil Resistivity Measurement

Since soil resistivity is a major factor affecting the corrosion rate, the design engineer shouldknow how to measure it. Generally, as soil resistivity decreases, corrosivity increases. Inaddition, as soil moisture content increases, resistivity decreases. It must be stressed that soilresistivity may vary widely within very short distances. Soil resistivity also changes with depthbelow the ground surface. Many samples may be needed for an accurate map of soil resistancein the area.

Buried utilities may produce interference when using the Wenner method in congested areas.Therefore, it is important to position the pins perpendicular to the underground pipeline. Placing

pins parallel to the pipeline would result in measurements lower than actual resistivity.

Temperature greatly affects soil resistivity at temperatures below freezing. Thus, soil resistivityshould not be measured on soil at below-freezing temperatures.

The most common method to determine soil resistivity is to use a ground potential testinstrument. This instrument is commonly called aground megger. The test instrument uses thefall of potentialmethod for determining ground resistance. The unit has both a current sourceand a voltage source. In operation, the instrument causes a current to flow into two outerelectrodes while a potential is measured between the two inner electrodes. The test instrumentoutput is calibrated in ohms.

The soil resistivity is determined from the following equation:

= 191.5 * A * R

Where,

= Soil resistivity, ohms-cm.A = Distance between the electrodes, feet.R = Test instrument resistance reading, ohms.

Note: The above equation yields soil resistivity in ohms-cm, whereas the input is in feet. The

formula takes the conversion into account.

The diagram in Figure 4 below shows the typical setup for measuring soil resistivity.


11/38

10

The distance between the current and potential electrodes (A inthe figure) should be equal to

the proposed electrode length. Readings should also be taken at distances of , 2, and 4 timesthe length of the proposed ground electrode to determine if there are large variances in the soilresistivity. The test electrodes should be driven into the ground to a depth of A/20.

Once the ground resistance measurements are obtained, the soil resistivity may be calculated.For instance, if the test results produce a ground resistance of 5-ohms when the test electrodesare spaced 10 feet apart, then the soil resistivity is:

= 191.5 * 10 * 5.

= 9,575 ohms-cm.

Required Current Testing

A critical element in designing galvanic and impressed current cathodic protection systems is thecurrent required for complete cathodic protection. Complete cathodic protection is achievedwhen the structure potential is -0.85 volts with respect to a copper-copper sulfate referenceelectrode. Note that the electrical term, volt,is always a measurement of the potential differencebetween two points. In cathodic protection systems, the reference point is either a Copper-copper sulfate electrode for structures in contact with soil or fresh water, or a silver chloride

electrode for seawater applications.

Current requirement tests are done by actually applying a current using a temporary test setupand adjusting the current from the power source until a suitable protective potential is obtained.Figure 5 shows a temporary test setup.

Figure 4


12/38

11

In this setup, a DC power supply is used in conjunction with a variable resistor and an ammeter.

At the other end of the pipe, a voltmeter is used to measure the resulting voltage. The variableresistor is adjusted to increase the current until the potential at the location of interest, such as thetest in Figure 5, is at -0.85 volts with respect to a copper-copper sulfate reference cell. Theresulting current supplied is the current required for cathodic protection.


13/38

12

III. Cathodic Protection Design Steps

Before deciding which type, galvanic or impressedcurrent, cathodic protection system will be used andbefore the system is designed, certain preliminary data

must be gathered.

Data Required for Cathodic Protection

Physical dimensions of structure to be protectedOne important element in designing a cathodic protection system is the structure's physicaldimensions (for example, length, width, height, and diameter). These data are used to calculatethe surface area to be protected.

Drawing of structure to be protectedThe installation drawings must include sizes, shapes, material type, and locations of parts of the

structure to be protected.

Electrical isolationIf a structure is to be protected by the cathodic system, it must be electrically connected to theanode. Sometimes parts of a structure or system are electrically isolated from each other byinsulators. For example, in a gas pipeline distribution system, the inlet pipe to each buildingmight contain an electric insulator to isolate in-house piping from the pipeline. Also, anelectrical insulator might be used at a valve along the pipeline to electrically isolate one sectionof the system from another. Since each electrically isolated part of a structure would need itsown cathodic protection, the locations of these insulators must be determined.

Short circuitsAll short circuits must be eliminated from existing and new cathodic protection systems. A shortcircuit can occur when one pipe system contacts another, causing interference with the cathodicprotection system. When updating existing systems, eliminating short circuits would be anecessary first step.

Corrosion history of structures in the areaStudying the corrosion history in the area can prove very helpful when designing a cathodicprotection system. The study should reinforce predictions for corrosivity of a given structure andits environment; in addition, it may reveal abnormal conditions not otherwise suspected.Facilities personnel can be a good source of information for corrosion history.

Electrolyte resistivity surveyA structure's corrosion rate is proportional to the electrolyte resistivity. Without cathodicprotection, as electrolyte resistivity decreases, more current is allowed to flow from the structureinto the electrolyte; thus, the structure corrodes more rapidly. Look at Table 1, as electrolyteresistivity increases, the corrosion rate decreases. Resistivity can be measured either in alaboratory or at the site with the proper instruments. The resistivity data will be used to calculatethe sizes of anodes and rectifier required in designing the cathodic protection system.


14/38

13

Table 1

Probably of Steel Corrosionbased on

Soil ResistivitySoil Resistivity

(ohm-cm)Potential

02,000 Severe

2,00010,000 Moderate

10,00030,000 Mild

30,0000 Not Likely

Electrolyte pH surveyCorrosion is also proportional to electrolyte pH. In general, steel's corrosion rate increases as pHdecreases when soil resistivity remains constant.

Structure versus electrolyte potential surveyFor existing structures, the potential between the structure and the electrolyte will give a directindication of the corrosivity. The potential requirement for cathodic protection is a negativecathodic potential of at least 0.85 volts as measured between the structure and a saturated copper-copper sulfate reference electrode in contact with the electrolyte.

A potential which is less negative than -0.85 volts would probably be corrosive, with corrosivity

increasing as the negative value decreases (becomes more positive).

Current requirementA critical part of design calculations for cathodic protection systems on existing structures is theamount of current required per square foot (called current density) to change the structurespotential to -0.85 volts. The current density required to shift the potential indicates the structure'ssurface condition. A well coated structure, such as pipeline well coated with coal-tar epoxy, willrequire a very low current density of about 0.05 milliampere per square foot. An uncoatedstructure will require a higher current density of about 10 milliamperes per square foot. Theaverage current density required for cathodic protection is 2 milliamperes per square foot of barearea. The amount of current required for complete cathodic protection can be determined three

ways:

1. An actual test on existing structures using a temporary cathodic protection setup.2. A theoretical calculation based on coating efficiency.3. An estimate of current requirements using tables based on field experience.

The second and third methods above can be used on both existing and new structures. Requiredcurrent testing was discussed in the previous section. Current requirements can be calculated


15/38

14

based on coating efficiency and current density desired. The efficiency of the coating as suppliedwill have a direct effect on the total current requirement, as the following equation shows,

I = A * I * (1.00 CE)

Where,I = Total protective current, mA.A = Total structure surface area, square feet.I= Required current density, mA/ft2.CE = Coating efficiency, decimal value.

This equation may be used when a current requirement test is not possible, as on new structures,or as a check of the current requirement test on existing structures. Coating efficiency is directlyaffected by the type of coating used and by quality control during coating application. Consideran example where 5,000 square feet of pipe area is to be protected, the required current density is1.0 mA, and the coating efficiency is 50%. What is the required protective current?

I = 5,000 * 1 * (1.000.50)

I = 2,500 mA.

The importance of coating efficiency is evident in the fact that a bare structure may require100,000 times as much current as would the same structure if it were well coated.

Current density requirements also can be estimated from Table 2. The table gives an estimate ofcurrent density, in milliamperes per square foot, required for complete cathodic protection. Thatvalue, multiplied by the surface area of the structure to be protected gives the total estimated

current required. Caution should be used when estimating, however, as under- or overprotectionmay result.

Table 2

Current Density Requirements

for

Cathodic Protection of Uncoated Steel

EnvironmentCurrent Density

(mA/ft2)

Neutral Soil 0.41.5

Aerated Neutral Soil 23Wet Soil 16

Acidic Soil 3 -1 5

Soil with Sulfate reducing Bacteria 642

Heated Soil 325

Stationary Freshwater 16

Moving Freshwater 5 -15

Sea Water 3 -10


16/38

15

As you can see from Table 2, a steel pipe in wet soil may only require 1.0 mA/ft2and a steel pipein moving freshwater may require up to 15.0 mA/ft2.

Coating resistance

A coating's resistance decreases greatly with age and directly affects structure-to-electrolyteresistance for design calculations. The coating manufacturers supply coating resistance values.

Protective current requiredBy knowing the physical dimensions of the structure to be protected, the surface area can becalculated. The product of the surface area multiplied by current density obtained previouslygives the total current required.

The need for cathodic protectionFor existing structures, the current requirement survey will verify the need for a cathodicprotection system. For new systems, standard practice is to assume a current density of at least

two milliamperes per square foot of bare area will be needed to protect the structure. However,local corrosion history may demand a different current density. In addition, cathodic protectionis mandatory for underground gas distribution lines and for water storage tanks with a 250,000-gallon capacity or greater. Cathodic protection also is required for underground piping systemslocated within 10 feet of steel reinforced concrete because galvanic corrosion will occur betweenthe steel rebar and the pipeline

Determining type and design of Cathodic Protection System

The first question to ask is: which type - galvanic or impressed current - cathodic protectionsystem is needed? Conditions at the site sometimes dictate the choice. However, when this isnot clear, the criterion used most widely is based on current density required and soil resistivity.If the soil resistivity is low (less than 5,000 ohm-centimeters) and the current densityrequirement is low (less than 1 milliampere per square foot), a galvanic system can be used.However, if the soil resistivity and/or current density requirement exceed the above values, animpressed current system should be used. Each approach is explained in detail below.

Galvanic Cathodic Protection System Design

There are essentially eight steps to designing a galvanic cathodic protection system. Theapproach is,

1.

Review soil resistivity2. Select anode3. Calculate net driving potential for anodes4. Calculate number of anodes needed5. Calculate number of anodes for system's life expectancy6. Select number of anodes to be used7. Select groundbed layout8. Prepare plans and specifications


17/38

16

Each step in designing a galvanic cathodic protection system is explained below. Section IVincludes examples of the application of galvanic cathodic protection designs.

Step 1

Review soil resistivity. The site of lowest resistivity will likely be used for anode location tominimize anode-to-electrolyte resistivity. In addition, if resistivity variations are not significant,the average resistivity will be used for design calculations.

Step 2Select anode. Galvanic anodes are usually either magnesium or zinc. Zinc anodes are used inextremely corrosive soil, such as soils with a resistivity below 2000 ohm- centimeters.Specification data from commercially available anodes include anode weight, anode dimensions,and package dimensions (anode plus backfill). Table 3 shows the specifications for magnesium-alloy anodes. In addition, the anodes drivingpotential must be considered. The choice of anodefrom those available is arbitrary; design calculations will be made for several available anodes,

and the most economical one will be chosen.

Table 3

Dimensions of High-Potential

Magnesium-Alloy AnodesWeight

(lb)

Size

(in.)

Packaged Wt.

(lb)

Packaged Size

(in.)

3 3.75 x 3.75 x 5 12 6x10

5 3.75 x 3.75 x 7.5 17 6x12

9 2.75 x 2.75 x 26 35 6x31

9 3.75 x 3.75 x 13.25 27 6x1712 3.75 x 3.75 x 18 36 6x23

14 2.75 x 2.75 x 41 50 6x46

14 3.75 x 3.75 x 21 42 6.5x26

17 2.75 x 2.75 x 50 60 6x55

17 3.75 x 3.75 x 26 45 6.5x29

20 2.5 x 2.5 x 59.25 70 5x66

24 4.5 x 4.5 x 23 60 7x30

32 5.5 x 5.5 x 21 74 8x28

40 3.75 x 3.75 x 59.25 105 6.5x66

48 5.5 x 5.5 x 30 100 8x38

48 8 x 16 100 12x25

60 4.4x4.4x60 - -Note: Core material is a galvanized 20-gage perforated steel strip.Anodes longer than 24 inches have a 9-gage core. The connecting wire is a 10-foot length ofsolid No. 12 AWG TW insulated copper wire, silver-soldered to the core with joints sealedagainst moisture.


18/38

17

Step 3Calculate net driving potential for anodes. The open-circuit potential of standard alloymagnesium anodes is approximately -1.55 volts to a copper-copper sulfate half-cell. The open-circuit potential of high-manganese magnesium anodes is approximately -1.75 volts to a copper-copper sulfate half-cell.

The potential of iron in contact with soil or water usually ranges around -0.55 volts relative to acopper-copper sulfate cell. When cathodic protection is applied using magnesium anodes, theiron potential assumes some value between -0.55 and - 1.0 volts, depending on the degree ofprotection provided. In highly corrosive soils or waters, the natural potential of iron may be ashigh as -0.82 volts relative to a copper-copper sulfate cell. From this, it is evident that -0.55volts should not be used to calculate the net driving potential available from magnesium anodes.

A more practical approach is to consider iron polarized to -0.85 volts. On this basis, standardalloy magnesium anodes have a driving potential of 0.70 volts (1.55 - 0.85 = 0.70) and highpotential magnesium anodes have a driving potential of 0.90 volts (1.75 - 0.85 = 0.90). For a

cathodic protection design that involves magnesium anodes, these potentials, 0.70 and 0.90 volts,should be used, depending on the alloy selected.

Step 4Calculate number of anodes needed to meet groundbed resistance limitations. The totalresistance of the galvanic circuit is given by,

RT= Ra+ Rw+ Rc

Where,RT= Total resistance, ohms.

RA= Anode-to-electrolyte resistance, ohms.RW= Anode lead wire resistance, ohms.RC= Structure-to-electrolyte resistance, ohms.

As an alternative, the total resistance also can be found by using,

RT =

Where,RT= Total resistance, ohms.

E = The anodes driving potential, volts.I = Total protective current required to achieve cathodic protection, amps.

For example, using a high-potential magnesium anode with a driving potential of 0.9 volts, and arequired current of 2,500 mA, what is the total resistance of the galvanic system?

RT =0.90

2.5


19/38

18

RT = 0.36 ohms.

The structure-to-electrolyte resistance, RC, is calculated by using the following formula,

RC =

Where,RC= Structure-to-electrolyte resistance, ohms.R = Average coating resistance, ohms/ft2.A = Structures surface area, square feet.

If the average coating resistance is 1,000-ohms/ft2, and the surface area of the pipe is 5,000square feet, then the structure-to-electrolyte resistance is,

RC =

1,000

5,000

RC = 0.20 ohms.

If we assume that the average wire coating resistance, Rw, is negligible the anode-to-electrolyteresistance (i.e. the groundbed resistance) is then found by rearranging the total resistanceformula,

Ra= RT- Rc

Where,RA= Anode-to-electrolyte resistance, ohms.RT= Total resistance, ohms.RC= Structure-to-electrolyte resistance.

This gives the maximum allowable groundbed resistance; this will dictate the minimum numberof anodes required. For instance if the total resistance is 0.36 ohms and the structure-to-electrolyte resistance is 0.20 ohms, then the anode-to-electrolyte resistance is,

Ra= 0.360.20

Ra= 0.16 ohms.

Note: As number of anodes decreases, groundbed resistance increases.

The following formula is used to calculate the number of anodes required,

N =. ((

))


20/38

19

Where,N = Number of anodes,

= Soil resistivity, ohms-cm.Ra= Maximum allowable groundbed resistance, ohms.

L = Length of the backfill column, feet (See Table 3).d = Diameter of the backfill column, feet (See Table 3).

For example, if the soil resistivity is 2,000 ohms-cm, and Rais 0.16-ohms, how many 12-poundanodes are required?

From Table 3, we see that for a 12-pound anode, the package length is 23 inches and thediameter is 6 inches. Converting to feet, we have L=1.92 and d=0.5. The number of anodes isthen,

N =0.0052 2,0000.16

1.92

((8 1.920.5

) 1)

N = 82 anodes.

Step 5Calculate number of anodes for the system's life expectancy. Each cathodic protection systemwill be designed to protect a structure for a given number of years. To meet this lifetimerequirement, the number of anodes must be calculated using,

N =

.

Where,N = Number of anodes required.n = Expected lifetime, years.W = Weight of one anode, pounds.I = Total protective current required to achieve cathodic protection, mA.

If the life expectancy is 25 years and the current required to protect the system is 2,500 mA, thenwith a 12-pound anode the number of anodes required is,

N =25 2,50049.3 12

N = 106 anodes.

Step 6The actual number of anodes required is the greater of the answer from Step 4 and Step 5 above.In our examples, we found the system need 82 anodes for protection. But, in Step 5 we saw thatto achieve the life expectancy, 106 anodes are required. Therefore 106 anodes should be used.


21/38

20

Step 7Select groundbed layout. When the required number of anodes has been calculated, the area to beprotected by each anode is calculated by,

A =

Where,A = Area to be protected by one anode, square feet/anode.AT= Total surface area to be protected, square feet.N = Total number of anodes to be used.

For example, if the total area to be protected is 5,000 ft2and the number of anodes required is106 anodes, then the area protected by one anode is,

A =

5,000

106

A = 47 ft2/anode.

For galvanic cathodic protection systems, the anodes should be spaced equally along thestructure to be protected.

Step 8Calculate life-cycle cost for proposed design. The design process should be done for severaldifferent anode sizes to find the one with minimal life-cycle cost.

Step 9Prepare plans and specifications. When the design procedure has been done for several differentanodes and the final anode has been chosen, plans and specifications can be completed.

Impressed Current Cathodic Protection System Design

A similar process is used for an impressed current cathodic protection system design. In thiscase, thirteen steps are required when designing impressed current cathodic protection systems.The steps are,

1. Review soil resistivity2. Review current requirement test3. Select anode4. Calculate number of anodes needed to satisfy current density limitations5. Calculate number of anodes needed to meet design life requirement6. Calculate number of anodes needed to meet groundbed resistance requirements7. Select number of anodes to be used8. Select area for placement of anode bed9. Determine total circuit resistance


22/38

21

10.Calculate rectifier voltage11.Select a rectifier12.Calculate system cost13.Prepare plans and specifications

The steps are explained below.

Step 1Review soil resistivity. As with galvanic systems, this information will contribute to both designcalculations and location of anode groundbed.

Step 2Review current requirement test. The required current will be used throughout the designcalculations. The calculated current required to protect one square foot of bare pipe should becompared with the values in Table 2 as a reasonableness test.

Step 3Select anode. As with the galvanic system, the choice of anode is arbitrary at this time; economywill determine which anode is best. Table 4 gives common anode sizes and specifications.

Table 4

Dimensions of

Circular High-Silicon Chromium-Bearing Cast Iron

AnodesAnode

Weight

(lb)

Anode

Dimension

(in.)

Anode

Surface Size

(ft2)

Package

Area

(ft2)

12 1x60 1.4 10x8444 2x60 2.6 10x84

60 2x60 2.8 10x84

110 3x60 4.9 10x84

The most common anodes are made of high silicon chromium-bearing cast-iron (HSCBCI).When impressed current-type cathodic protection systems are used to mitigate corrosion on anunderground steel structure, the auxiliary anodes often are surrounded by a carbonaceousbackfill. Backfill materials commonly used include coal coke breeze, calcined petroleum cokebreeze, and natural graphite particles. The backfill serves three basic functions it:

Decreases the anode-to-earth resistance by increasing the anode's effective size. Extends the system's operational life by providing additional anode material. Provides a uniform environment around the anode, minimizing deleterious localized

attack.

The carbonaceous backfill, however, cannot be expected to increase the groundbed lifeexpectancy unless it is well compacted around the anodes. In addition to HSCBCI anodes, theceramic anode should be considered as a possible alternative for long-term cathodic protection of


23/38

22

water storage tanks and underground pipes in soils with resistivities less than 5000 ohm-centimeters. The ceramic anode consumption rate is 0.0035 ounce per ampere-year compared toone pound per ampere-year for HSCBCI anodes.

Step 4

Calculate number of anodes needed to satisfy manufacturer's current density limitations.Impressed current anodes are supplied with a recommended maximum current density. Highercurrent densities will reduce anode life. To determine the number of anodes needed to meet thecurrent density limitations, use the following equation,

N =

Where,N = Number of anodes required.I = Total protection current, milliamperes.

A1= Anode surface area, square feet per anode.I1= Recommended maximum current density output, milliamperes.

Note: The recommended maximum current density output for high-silicon chromium-bearingcast-iron anodes 1,000 mA/ft2.

As an example, a pipeline is to be protected and the total protection current required is 2,500mA, and the recommended maximum current density is 1,000 mA/ft2. How many anodes areneeded?

Select an anode from Table 4. The choice is arbitrary, but assume the choice is a 12 lb anode.

From Table 4, the surface area of the anode is 1.4 square feet. Therefore the number of anodesrequired is,

N =2,500

1.4 1,000

N = 1.8 anodes.

Step 5Calculate number of anodes needed to meet design life requirement. The following equation isused to find the number of anodes:

N =

,

Where,N = Number of anodes.I = Total protection current, mA.n = Life in years.


24/38

23

W = Weight of one anode, pounds.

Continuing with our example, if the life is 25-years, then the number of anodes required is,

N =25 2,500

1,000 12

N = 5.2 anodes

Step 6Calculate number of anodes needed to meet maximum anode groundbed resistance requirements.The formula for anode resistance is,

Ra=

+

This equation can be rearranged to find the number of anodes if the anode resistance is known.In this case, the number of anodes required is found by,

N =

Where,N = Number of anodes.Ra= Anode resistance, ohms.

= Soil resistivity, ohm-centimeters.K = Anode shape factor, Table 5.

L = Length of the anode backfill column in feet.P = Paralleling factor from Table 6.S = Center-to-center spacing between anode backfill columns, in feet.

The design parameters may give us a maximum anode resistance and anode spacing. With thisinformation we can determine the number of anodes required.

For an example, assume the maximum anode resistance is 2.0-ohms, and the spacing is 25 feet.If we are using a 12-lb anode, the package dimensions are 10x84, which means the L/d ratio is84/10 or 8.4. From Table 5, an L/d of 8.4 (choose 8) is 0.0165. Remember the anode length isin feet, so for 84 inches the factor, L, is 7.0. For the first pass at calculating the number of

anodes, assume the paralleling factor is zero. Therefore, the solution looks like this,

N =2,000 0.0165

72.02,000 025

N = 2.36 anodes.


25/38

24

In this case, with N=2.36 anodes we would likely choose to use three anodes. Therefore, thecalculation should be redone with a paralleling factor from Table 6 for three anodes. The resultis,

N =2,000 0.0165

72.02,000 0.00289

25

N = 2.67 anodes.

Adding the paralleling factor did not change the recommendation for three anodes. If thecalculation had yielded a different number of anodes, then we would need to re-calculate with anew paralleling factor.

Table 5

Shape Functions for

Impressed CurrentCathodic Protection AnodesL/d K L/d K

5 0.014 20 0.0213

6 0.015 25 0.0224

7 0.0158 30 0.0234

8 0.0165 35 0.0242

9 0.0171 40 0.0249

10 0.0177 45 0.0255

12 0.0186 50 0.0261

14 0.0194 55 0.026616 0.0201 60 0.0270

18 0.0207Notes: L = Effective anode lengthD = anode/backfill diameter

Step 7Select number of anodes to be used. The highest number calculated by the equations in eitherStep 4, Step 5, or Step 6 will be the number of anodes used. In the previous examples, thegreatest number of anodes came from Step 5, based on the life expectancy of the anodes and is

5.2 anodes. We should select 6 anodes.

Step 8Select an area for placement of the anode bed. The area with the lowest soil resistivity will bechosen to minimize anode-to-electrolyte resistance.

Table 6

Anode Paralleling Factors

Number

of

Anodes

(N)

Factor

(P)

Number

of

Anodes

(N)

Factor

(P)

2 0.00261 14 0.00168

3 0.00289 16 0.00155

4 0.00283 18 0.00145

5 0.00268 20 0.00135

6 0.00252 22 0.00128

7 0.00237 24 0.001218 0.00224 26 0.00114

9 0.00212 28 0.00109

10 0.00201 30 0.00104

12 0.00182


26/38

25

Step 9Determine total circuit resistance. The total circuit resistance will be used to calculate therectifier size needed. But, first we need to calculate the anode groundbed resistance. The anodegroundbed resistance is found using the equation in step 6.

Ra=

+

Where,Ra= Anode resistance, ohms.N = Number of anodes.

= Soil resistivity, ohm-centimeters.K = Anode shape factor, Table 5.L = Length of the anode backfill column in feet.P = Paralleling factor from Table 6.S = Center-to-center spacing between anode backfill columns, in feet.

Continuing with the previous examples, the number of anodes is now 6, which changes theparalleling factor to 0.00252. Rais,

Ra=2,000 0.0165

6 7 +2,000 0.00252

25

Ra= 0.99 ohms.

Rais well below the design criteria specified of 2.0 ohms.

Calculate groundbed header cable resistance. The cable is typically supplied with a specifiedresistance in ohms per 1,000 feet. The wire resistance then is calculated from,

RW= Rx* X

Where,Rw= Total wire resistance, ohms.Rx= Resistance of the wire, ohms/1000 ft.X = Structure's length in feet, per 1000 ft.

If we assume the anodes bed will be placed 150 feet from the pipeline and the wire resistance is0.16 ohms/1,000 ft, the term RWis,

RW= 150/1000 * 0.16

RW= 0.024 ohms.

Calculate structure-to-electrolyte resistance,


27/38

26

RC=

Where,RC= Structure-to-electrolyte resistance, ohms.

R = Coating resistance, ohms per square feet.A = Coated pipe area, square feet.

If the coating resistance is 2,500 ohms/ft2and the coated pipe to be protected consists of 3,185linear feet of 6 outer diameter pipe. What is the structure-to-electrolyte resistance?

First, we need to calculate the area of the coated pipe. The formula is,

A = Len * 3.14 * dia

Where,A = Area of coated pipe, square feet.Len = Linear footage of pipe, feet.Dia = Outer diameter of the pipe, feet.

A = 3,185 * 3.14 * 6/12

A = 5,000 ft2.

RC=2,500

5,000

RC= 0.5 ohms.

Now, we can calculate total circuit resistance. Remember that the equation for the totalresistance is,

RT= RA+ RW + RC

Where,RT= Total resistance, ohms.RA= Anode-to-electrolyte resistance, ohms.RW= Anode lead wire resistance, ohms.R

C= Structure-to-electrolyte resistance, ohms.

From our examples, where Rais 0.99 ohms, Rwis 0.024 ohms, and Rcis 0.5 ohms. TotalResistance is,

RT= 0.99 + 0.024 + 0.5

RT=1.514 ohms.


28/38

27

Step 10Calculate rectifier voltage. The following equation is used to determine voltage output (VREC) ofthe rectifier:

VREC= 1.5 * I * RT

Where,VREC= Rectifier voltage, volts.I = Total protection current, amperes.RT= Total circuit resistance, ohms.

Note: The 150 percent is a factor is to allow for aging of the rectifier stacks.

From our examples, the total protection current is 2,500 mA and the total resistance is 1.514ohms. The rectifier voltage is,

VREC= 1.5 * 2.5 * 1.514.

VREC= 5.7 volts.

Step 11Select a rectifier. A rectifier must be chosen based on the results from Step 10. Many rectifiersare available commercially and it is just a matter of selecting one that satisfies the minimumcurrent and voltage requirements from the calculations. In our examples, a rectifier that iscapable of generating 5.7 volts and 2,500 mA should be selected.

Besides the more common rectifiers being marketed, a solar cathodic protection power supply(for DC power) may be considered for remote sites with no electrical power. Three factors thatshould be considered when specifying a solar cathodic protection power supply are:

1. The cost of the solar cathodic protection power supply in dollars per watt of continuouspower.

2. The solar cathodic protection power supply's much higher initial cost compared toselenium rectifiers operated by AC power.

3. The additional maintenance required for a solar cathodic protection power supply, mainlyto keep the solar panels free of dirt deposits.

Step 12Calculate system cost. As with the galvanic cathodic protection system, the choice of anode fordesign calculation is arbitrary. When several anodes have been used in the design calculations,an economic analysis should be done. An economic analysis is beyond the scope of this course,but the following example will show how the economics of how the insulated wire is analyzed.


29/38

28

Economics are probably the controlling factor in selecting a cable for the anode connection. Todetermine the total annual cable cost, Kelvin's Economic Law can be used as shown below,

T =.

+ 0.15 * S * X

Where,T = Total annual cost, dollars per year.I = Total protection current, amperes.R = Cable resistance, ohms per 1,000 feet.X = Cable length, feet.P = Cost of electrical energy, dollars per kilowatt-hour.E = Rectifier efficiency, decimal.S = Cable's initial cost, dollars per foot.

For instance, if the total protection current is 2,500 mA, the cable resistance is 0.16 ohms/1,000

ft, the cable length is 150 feet, the energy costs is $0.12 per kWh, the rectifier efficiency is 75%,and the cables initial cost is $0.35 per foot. What is the total annual cost of the cable in dollarsper year?

T =0.0876 2.52 0.16 150 0.12

0.75 + 0.15 * 0.35 * 150

T = $9.98 year

A similar approach can be used for an economic analysis for other components of the protectionsystem.

Step 13Prepare plans and specifications.


30/38

29

IV. Design Example

In this section we will review two complete design examples, one using a galvanic protectionsystem and the other using an impressed current system. The calculations are the same as foundin the previous section.

Galvanic Protection Example

Design a galvanic protection system for a linear steel pipeline with the following conditions.

Design Conditions:

Galvanized steel pipe:

Section 13,000 ft, 14 O.D.

Section 21,000 ft, 8. O.D.

Section 31,000 ft., 2 O.D.Coating efficiency: 85%

Average coating resistance, 1,500 ohms/ft2.

Soil resistivity: 3,000 ohms-cm.

Design life: 25 years.

Required current density: 2.0 mA.

Anode: High potential magnesium alloy

14# (50# package anode), per Table 3.

(Driving potential = 0.90 volts.)

Assume the connecting wire resistance is negligible.

Solution:

First we need to determine the total area to protect. Using

A = Len * 3.14 * dia

The area is,

Section 1 3,000 14 10,990Section 2 1,000 8 2,093Section 3 990 2 527

Total 13,600

The total surface area of all of the piping is 13,600 feet. Next we calculate the total protectivecurrent required. We know the coating efficiency is 85% and the required current density is 2.0mA, so,


31/38

30

I = 13,600 * 2 * (1.000.85)

I = 4,080 mA

The required current is 4,080 mA or 4.08 amps. Now, we just follow the design steps as

presented in the previous section.

Step 1. Review soil resistivity.

The soil resistivity is given in the specifications as 3,000 ohms-cm.

Step 2. Select anode.

The design specifications call for a 14# anode. From Table 3 the length of the anode is 46 andthe diameter is 6. Therefore L = 3.83 feet and d = 0.5 feet.

Step 3. Calculate net driving potential for anodes.

The design specifications define the driving potential as 0.90 volts.

Step 4. Calculate number of anodes needed to meet groundbed resistance limitations.

With a design specification driving potential of 0.90 volts and a total protective current of 4.08amps, the total resistance of the galvanic circuit is given by,

RT =

Which results in,

RT =0.90

4.08

RT = 0.22 ohms.

The structure-to-electrolyte resistance is given by,

RC =

With an average coating resistance of 1,500 ohms/ft2and 13,600 ft2of pipe area, Rcis,

RC =1,500

13,600

RC = 0.11 ohms.


32/38

31

The groundbed resistance (also called the anode-to-electrolyte resistance is given by,

Ra= RT- Rc

Therefore the groundbed resistance is,

Ra= 0.220.11

Ra= 0.11 ohms.

Now the number of anodes can be found from,

N =0.0052 ((

8 ) 1)

The soil resistivity was given as 3,000 ohms-cm, the anodes length and diameter are 3.83 and0.5 respectively, so the number of anodes is,

N =0.0052 3,0000.11 3.83 ((

8 3.830.5

) 1)

N = 12.69 anodes.

Step 5. Calculate number of anodes for system's life expectancy.

We found we needed 12.69 anodes in the previous step, but since the anodes are sacrificial we

need to determine how many anodes will be needed to last 25-years based on the current required

and the size of the anode (pounds.)

N =

49.3

The total protective current is 4,080 mA and a 14# anode has been selected. The number ofanodes is then,

N =25 4,08049.3 14

N = 147.78 anodes.

Step 6. Determine the number of anodes required as the greater of either Step 4 or Step 5.

The system design dictated the need for about 13 anodes. However, for the system to last 25-years 148 anodes are required. Therefore, the design will be based on 148 anodes.

Step 7. Select groundbed layout.


33/38

32

The groundbed layout is based on how much area each anode can cover. With 13,600 squarefeet of surface area and 148 anodes each anode can cover,

A =

A =13,600

148

A = 91.89 ft2.

The groundbed layout should be designed with the assumption that each anode is coveringapproximately 92 square feet of surface area.

Lets now look at an impressed current protection example.

Impressed Current Protection Example

Design Conditions:

Galvanized steel pipe:

Section 13,000 ft, 14 O.D.

Section 21,000 ft, 8. O.D.

Section 31,000 ft., 2 O.D.

Coating efficiency: 85%

Average coating resistance: 3,250 ohms/ft2.

Soil resistivity: 2,600 ohms-cm.

Design life: 25 years.

Anode spacing: 40 feet.

Maximum groundbed resistance: 2.0 ohms.

Required current density: 1 mA.

Anode: High potential magnesium alloy, 44#, per Table 4.

Total protection current: 2.3 amps.Recommended maximum current density output: 1000 mA/Ft2.Wire resistance: 0.28 ohms/1,000 ft.Wire length: 200 ft.

Solution:

Just as with the galvanic protection system, we first start by determining the area to be protected.

The total area to protect is found using,

A = Len * 3.14 * dia


34/38

33

The area is,

Section 1 3,000 14 10,990Section 2 1,000 8 2,093

Section 3 990 2 527Total 13,600

The required protective current is given in the design specifications, but lets compare the

specification current to the current found from the required current formula. The coating

efficiency is 85% and the current density is 1.0 mA, so from,

I = A * I * (1.00 CE)

I = 13,600 * 1.0 * (1.000.85)

I = 2,040 mA.

The required current from the formula is 2.04 amps, which is close to the total protective current

specified of 2.3 amps.

Step 1. Review soil resistivity.

The soil resistivity is given in the specifications as 2,600 ohms-cm.

Step 2. Review current requirement test.

The current test yielded a total protective current requirement of 2.3 amps, which is reasonablewhen compared to the formula approach.

Step 3. Select anode.

The design specifications call for a 44# anode. From Table 4 the length of the anode is 84 (7feet), the diameter is 10 (0.83 feet), and the anode surface area is 2.6 ft2. Therefore L = 7.0 feetand d = 0.83 feet.

Step 4. Calculate number of anodes needed to satisfy manufacturer's current density limitations.

The number of anodes required is based on the total protective current, the area of a single

anode, and the maximum current density output. The area of a single anode was given as 2.6 ft2

and the current density is 1000 mA/ft2. Therefore,


35/38

34

N =

1 1

N =2,300

2.6 1,000

N = 0.88 anodes.

Step 5. Calculate number of anodes needed to meet design life requirement.

Just as with a galvanic system, the number of anodes required may be different than thecalculations yield because some of the anode is lost over the life of the system. It is not a majorissue with impressed current protection systems, but it still must be considered. With a 25-yearlife and a 2.3 amp current, the number of 44# anodes required is,

N =

1,000

N =25 2,3001,000 44

N = 1.31 anodes

Step 6. Calculate number of anodes to meet anode groundbed resistance requirements.

We also need to look at the number of anodes required to meet the groundbed resistancerequirements. The soil resistivity was given as 2,600 ohms-cm, the maximum anode resistance

was given as 2.0 ohms. From Table 5 we find that an anode with an L/d ratio of 7/0.83 (8.4) willhave a K factor of about 0.0165. With a spacing factor of 40 the calculation is,

N =

N =2,600 0.0165

72.02,600 040

N = 3.06 anodes.

Since we know the number of anodes is around three, we need to re-calculate the number ofanodes required including a paralleling factor from Table 6. For three anodes the parallelingfactor is 0.00289. The number of anodes is now,

N =2,600 0.0165

72.02,600 0.0028940


36/38

35

N = 3.38 anodes.

Step 7. Select number of anodes to be used.

From the previous steps we found the number of anodes required to be 0.88, 1.31, or 3.38.

Therefore, select 4 anodes for the system.

Step 8. Select area for placement of anode bed.

The anode bed is generally placed off to one side of the pipeline in an area of good soilresistivity as shown is the figure below.

Step 9. Determine total circuit resistance.

Once we know the number of anodes the total circuit resistance can be calculated.

The actual groundbed resistance is,

Ra= +

Note that the paralleling factor is now 0.00283 since we are using 4 anodes.

Ra=2,600 0.0165

4 7 +2,600 0.00283

40

Ra= 1.72 ohms.

The wire resistance is based on 200 feet of wire at 0.28 ohms/1000 ft.


37/38

36

RW= Rx* X

RW= 200/1000 * 0.28

RW= 0.06 ohms.

The pipeline to electrolyte resistance is based on the coating resistance of the pipeline and thesurface area of the pipeline. The coating resistance was given as 3,250 ohms/ft2and the area is13,600 square feet.

RC=

RC=3,250

13,600

RC= 0.24 ohms.

Therefore the total circuit resistance is,

RT= RA+ RW + RC

RT= 1.72 + 0.06 + 0.24

RT=2.02 ohms.

Step 10. Calculate rectifier voltage.

With the total circuit resistance known, we can now calculate the rectifier voltage requirement.

Using 2.3 amps and 2.02 ohms the rectifier voltage is,

VREC= 1.5 * I * RT

VREC= 1.5 * 2.3 * 2.02

VREC= 6.97 volts.

Step 11. Select a rectifier.

A rectifier should be selected that is capable of delivering 2.3 amps at 6.97 volts.

At this point, all that is left to do is to perform an economic evaluation using several different

anode sizes and then prepare the plans and specifications for the system.


38/38

Conclusion

This course has been a brief introduction to cathodic protection systems. The theory of corrosionprotection was covered as well as the testing methods used in the design of cathodic protectionsystems. Both galvanic cathodic protection systems and impressed current cathodic protectionsystems were reviewed in the context of a simple straight-line pipeline system. The basiccalculations involved in selecting the type and number of anodes required and two designexamples were explored.

Copyright 2009 Lee Layton. All Rights Reserved.

+++

cathodic protection - introduction

Documents