DEPARTMENT OF THE ARMYU.S. Army Corps of Engineers

CEMP-ET Washington, DC 20314-1000 ETL 1110-9-10

Technical LetterNo. 1110-9-10 5 January 1991

Engineering and DesignCATHODIC PROTECTION SYSTEM USING CERAMIC ANODES

Distribution Restriction Statement

Approved for public release; distribution is unlimited.

ETL 1110-9-10(FR)

DEPARTMENT OF THE ARMYCEMP-ET U.S. Army Corps of Engineers

Washington, D.C. 20314-1000

Engineer TechnicalLetter 1110-9-10(FR) 5 January 1991

Engineering and DesignCATHODIC PROTECTION SYSTEM USING CERAMIC ANODES

1. Purpose : This letter provides criteria for the design o fimpressed current cathodic protection systems using ceramic anodes.The enclosure to this letter is the revised appendix E of TM 5-811-7: Electrical Design, Cathodic Protection.

2. Applicability : This letter applies to all HQUSACE/OCE elemen tsand all Major Subordinate Commands (MSC) and District Comman ds (DC)having Army military design and construction responsibility.

3. Discussion : Development in ce ramic anodes technology and therecent years of experience with ceramic anodes indicate tha tceramic anodes can be used for general application. A wide varietyof ceramic anodes are manufact ured to satisfy various needs in thefield of cathodic protection. The CEGS 16641 "Cathodic ProtectionSystems for Steel Water Tanks" and CEGS 16642 "Cathodic ProtectionSystems (Impressed Current)" permit the use of precious mixe d metaloxide anodes but no design guidance is provided in TM 5-811-7 forthe us e of conductive ceramic coated titanium or mixed meta lanodes. The enclosed design pr ocedures will assist in design usingthese anodes.

4. Action to be taken : Where the application of an impresse dcathodic protection system is required, the use of ceramic anodesshould be considered. Where ce ramic anode application has been thechosen type of cathodic protection, the enclosed design manua lshall be used in the preparation of final design documents.

5. Implementation : This letter w ill have special application asdefined in paragraph 6c, ER 1110-345-100.

FOR THE DIRECTOR OF MILITARY PROGRAMS:

Encl RICHARD C. ARMSTRONGChief, Engineering DivisionDirectorate of Military Programs

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IMPRESSED CURRENT CATHODIC PROTECTION SYSTEMSUSING CERAMIC ANODES

TABLE OF CONTENTS

Page

SECTION 1 BASIS OF IMPRESSED CURRENT CATHODICPROTECTION DESIGN

1-1 Introduction . . . . . . . . . . . . . . . . . 41-2 Cathodic Protection Design Using

Ceramic Anodes . . . . . . . . . . . . . . . . 4

1-3 References . . . . . . . . . . . . . . . . . . 15

SECTION 2 EXAMPLES OF IMPRESSED CURRENT CATHODICPROTECTION DESIGN

2-1 Purpose . . . . . . . . . . . . . . . . . . . . 18

2-2 Steel Fuel Oil Lines . . . . . . . . . . . . . 18

2-3 Underground Storage Tanks (UST's) . . . . . . . 31

2-4 On-Grade Tank Bottoms . . . . . . . . . . . . . 39

2-5 Gas Distribution System . . . . . . . . . . . . 58

2-6 Elevated Water Tank (Ice Is Expected) . . . . . 70

2-7 Elevated Water Tank (No IcingWill Occur) . . . . . . . . . . . . . . . . . . 78

2-8 On-Grade Water Storage Reservoir (IceIs Expected) . . . . . . . . . . . . . . . . . 104

2-9 Horizontal Anodes (UndergroundApplications) . . . . . . . . . . . . . . . . . 113

2-10 Backfilling Packaged Anodes WithCoke Breeze . . . . . . . . . . . . . . . . . . 116

SECTION 3: Tables . . . . . . . . . . . . . . . . . . . . 117

SECTION 4: Identification of Variables . . . . . . . . . . 128

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List of FiguresPage

2-1A Cathodic Protection System for Fuel Oil LineWith Anode Bed Extended Perpendicular FromPipeline 19

2-1B Cathodic Protection System for Fuel Oil LineWith Anode Bed Installed Parallel to Pipeline 20

2-2 Vertical Groundbed Layout Using PrepackagedCeramic Rod Anode 21

2-3 Prepackaged Ceramic Rod Anode for UndergroundUse 22

2-4 Ceramic Anode Canister Prepackaged Rod 23

2-5 Cathodic Protection for Underground StorageTanks and Piping 33

2-6 On-Grade Tank Bottoms 41

2-7 Typical Deep Anode Groundbed Using Rod Angles 42

2-7A Ceramic Rod Anode for Deep Groundbed Use 43

2-8 Ceramic Anode Tubular Power Rod Used In DeepAnode Bed 44

2-8A Ductile Ceramic Anode Tubular Power Rod 45

2-9A Deep Anode Groundbed Resistance 25 to 50 ft. 48

2-9B Deep Anode Groundbed Resistance 50 to 100 ft. 49

2-9C Deep Anode Groundbed Resistance 100 to 150 ft. 50

2-10 Pole-Mounted Rectifier 57

2-11 Deep Anode Cathodic Protection for GasDistribution System 60

2-12 Typical Deep Anode Groundbed Using TubularAnodes 68

2-13 Ceramic Anode Tubular Power Rod Usedin Deep Anode Bed 69

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2-14 Elevated Pedestal Tank (Icing Conditions)With Hoop Ceramic Anode 76

2-14A A Pressure Entrance Fitting for UnderwaterPower and Reference Cell Wire Penetrationsin Water Storage Tanks 77

2-15 Elevated Water Tank Showing Rectifier andAnode Arrangement 80

2-16 Segmented Elevated Tank for Area Calculations 81

2-17 Equivalent Diameter Factor for Anodes in aCircle in a Water Tank 86

2-18 Fringe Factor for Stub Anodes 87

2-19 Anode Spacing for Elevated Steel Water Tanks 89

2-20 Anode Suspension Arrangement for Elevated SteelWater Tank 90

2-21 Water Tank Hand Role Assembly for Roof-SupportedAnodes and Support Bracket for SubmergedHoop Anodes 103

2-22 Submerged Cathodic Protection System for Ground-Level Water Storage Reservoir 105

2-23 Details of Wire-type Anode System Designed ToWithstand Icing conditions Within Tank 112

2-24 Horizontal Prepackaged Ceramic Anode Groundbed 114

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CATHODIC PROTECTION DESIGN

1-1. Introduction.

Recently, ceramic coated anodes have been incorporated incathodic protection systems. Ceramic or metal-oxide anodes havebeen used for cathodic protection since 1971 in Europe and since1984 in the United States. One of the main advantage of ceramicanodes are that they are not consumed.

Ceramic anodes consist of various shapes of high purity titaniumsubstrates with coatings of precious metal oxides tailored to theenvironment in which they will be used.

Unlike most other metal oxides (or ceramics), these metal oxidesare conductive. Ceramic anodes are dimensionally stable. Theceramic coating is already oxidized (corroded).The current capacity is a function of constituent variables andis rated by the manufacturers. They have design lifeexpectancies of up to 20 years at a rated current output. Thelife can be extended by a reduction in output current density.Their life is limited by time and current density. The end ofthe ceramic anode life is marked by a chemical change in theoxide form and a resultant loss in conductivity. Ceramic anodesare made in a variety of shapes for various applications. Amongthese are wire, rods, tubes, strips, discs, and mesh. Ceramicanodes have excellent ductility, which has eliminated the concernabout mechanical damage during shipment and installation. Ceramic anodes are also a fraction of the size and weight oftraditional anode materials.

Scratches or other minor physical damages to the coating resultin the formation of an inert and nonconductive oxide of thesubstrate (titanium) when operated at less than 60 V in freshwater and underground applications. If they are installed insalt or brackish water, the DC design voltage should be limitedto 12 V. The overall function of the anode is not significantlyimpaired.

1-2. Cathodic Protection Design Using Ceramic Anodes.

The following steps are involved in designing a cathodicprotection system using ceramic anodes:

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a. Collect data.

Design requirements should be established, and certainassumptions will be made.

1) History

Information from occupants in the area can indicatethe severity of corrosion problems. Data on failuresand failure rates of nearby structures can be invalu-able and must be considered.

2) Drawings

Drawings of the structure to be protected and thearea where it is or will be installed are needed toprovide the physical dimensions of the structure fordetermining surface area to be protected, andlocations of other structures in the area that maycause interference, of insulating devices, and ofpower sources. Information on coatings should beobtained.

3) Tests

Current requirement test and potential survey testresults are needed for existing structures that willbe protected. Electrolyte (soil or water) resistivi-ty tests and evaluation of conditions that couldsupport sulfate-reducing bacteria are needed for allcathodic protection designs. This information willindicate the size of the cathodic protection systemthat will be required as well as the probability ofstray current problems. Soil resistivities contrib-ute to both design calculations and location of theanode groundbed.

4) Life

The user must determine the required number of yearsthat the structure needs to be protected, or thedesigner must assume a nominal life span. The struc-ture will begin to deteriorate from corrosion at theend of the cathodic protection system's design lifeunless the system is rejuvenated.

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5) Coatings

Cathodic protection complements the protective coat-ing system. A good coating system substantiallyreduces the amount of cathodic protection currentrequired. The coating efficiency has to beassumed.

6) Short circuits

All short circuits must be eliminated from both newand existing structures for which a cathodicprotection system is being designed.

b. Calculate surface area to be protected.

The overall current requirement of a cathodic protectionsystem is directly proportional to the surface area to beprotected. This includes underground or submerged pipes,buried tanks, and wetted surfaces (up to highwater(level) of watertanks (including risers).

c. Determine current requirement.

For existing structures, a current requirement test willprovide the actual current requirement at the time of thetest. Allowance should be made in the design for futuredegradation of coatings or structure additions that willincrease the current requirement.

For new structures not yet installed, the amount ofcurrent needed to provide protection as defined in Na-tional Association of Corrosion Engineers (NACE) RP-0l-69(reference 22) will be dependent on a number ofvariables. Table 3-1 gives guidelines for currentrequirements in various soil and water conditions.

The efficiency of the coating system, both when new andat the end of design life, is a determining factor in therange of current that will be required over the lifetimeof the system. Total current required is given by thefollowing equation:

I = (A)(I')(l.0 - C ) (eq 1-1)E

where I is the total current requirement, A is the totalsurface area to be protected, I' is the estimated currentdensity, and C is the efficiency of the coating system.EThis procedure should always be followed, even when a

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current requirement test has been performed, as a checkon assumptions made. Current density may be estimatedfrom information given in table 3-1.

d. Select anode type.

Ceramic anodes are made in a variety of shapes, such as,wires rods, tubes, strips, disks, and mesh.,

The 0.062-in. diameter anode wire has a 20-year life at amaximum current rating of 115 mA per linear ft in freshwater, 285 mA in salt water, and 170 mA in brackishwater. Wire anodes are well suited for applications inwater tanks. They are generally not used underground.

Ceramic rod anodes are manufactured bare for aqueousenvironments and prepackaged for installation in soil.Ceramic rod anodes are produced in diameters of 1/8 in.,1/4 in., 3/8 in. and in. and in standard lengths of 4,6, and 8 ft, although almost any length can be customfabricated with self-healing screw connections for fieldassembly to the desired length, or with permanent,factory-molded, cable-to-anode connections. Forunderground applications, rods are frequently packaged in2- or 3-in. diameter steel tubes filled with a highcarbon, low resistivity coke breeze. Their small sizeand high current capacity make rods particularly wellsuited for both underground shallow and deep anodesystems. For marine applications, the rod anode is oftenencased in a perforated PVC package that providesmechanical protection and prevents the possibility of theanode contacting the protected structure. They are usedin a similar manner as high silicon cast iron andgraphite anodes.

For long ceramic anode wires and rods, the voltage dropin the titanium substrate must be considered. Whiletitanium is a relatively good metallic conductor, itsresistance is approximately 33 times that of copper. Themaximum length for solid titanium wire and rod anodeapplications to assure that uniform discharge of currentis achieved in several different environments is providedbelow:

Maximum Anode Length From Connection Point

Solid Titanium Anodes

Anode Diameter 0.062 in. 0.125 in. 0.250 in.

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ENVIRONMENT LENGTH (ft)

Sea Water 3 5 9Coke Breeze 6 10 20Fresh Water 30 50 100

Where a specific design requires longer length anodes thanpro vided for in the above table, the titanium wire or ro dcan be provided with a copper core to reduce the effectiveresistance. This type of construction has been in use forover 15 ye ars and has proved to be very durable. Th etitanium wall thickness should be a minimum of 20 percen t ofthe wire or rod diameter (e.g., for 0.062-in. diameter w ire,the titanium wall thickness should be 0.0124 in., minimum)

The maximum allowable length for copper-core titanium wireand rod anodes is provided in the table below:

Maximum Anode Length From Connection Point

Copper Cored Titanium Anodes

Anode Diameter 0.062 in. 0.125 in. 0.250 in.

ENVIRONMENT LENGTH (ft)

Sea Water 7 12 24Coke Breeze 12 24 54Fresh Water 70 135 300

A strip anode is presently manufactured as a 3- or 7-ft barof ceramic-coated substrate, molded into a multilaye rcomposite of fiberglass-reinforced plastic (FRP) an dpolyuret hane 4 or 8 ft long and 4 in. wide that provide simpact resistance, mechanical support, and electrica linsulation.

A mesh anode is produced using highly expanded titaniu msheet metal and is used where a large area is to b eprotected, where area for anode placement is confined, andwhere future access is not practicable. Its use under astructure's base such as an on-grade tank bottom wit hsecondary membrane containment or select reinforce dconcrete bridges, wharfs, etc., would be appropriate .Because of its size and th e nature of its application, meshis generally restricted to use in new facilities.

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e. Calculate number of anodes (N) or length of bare anode wire(L ).B

1) For required design life

Since ceramic anodes are not consumed during thei rlife, the quantity of ceramic material beyond tha trequire d to form a coating is not relevant. The numberof anod es or length of anode wire required i sdetermined from the total current required and th emanufacturer's published current rating for a give nlife. For rod, strip, tube and disk anodes:

Number of anodes required =

Total current required (eq l-2a)Manufacturers ratedcurrent for specificsize, environment and life.

2) For wire anodes:

Total footage of anode =

Total current requirement (eq l-2b)Manufacturers rated currentcapacity per foot of wirefor a given environment and life.

The number calculated will determine the minimum numberof anodes or anode wire length required.

f. Calculate the total circuit resistance (R ).T

The total circ uit resistance (R ) consists of the anode-to-Tele ctrolyte resistance (R ) plus the interconnecting wir eNresistance (R ) plus the structure-to-elect rolyte resistanceW(R ).C

R = R + R + R (eq 1-3)T N w C

A criterion of 2-ohm maximum groundbed resistance is oftenused to limit the rectifier output voltage and the associ-ated hazards of overprotection. When the total require dcurrent is low, a higher total resistance is often

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accepta ble. As the required current increases, the tota lresistance should be reduced. See table 3-10.

The total anode-to-electrolyte resistance (R ) i sNcalculated in different ways according to the type of anodeins tallation. The anode-to-electrolyte resistance for asingle anode is given by R .A

For a single vertical anode:

R = (0.0052) p [ln (8L/d)-1] (eq 1-4)A (L)

Where R is the anode-to-electrolyte resistance for aA single anode, p is the electrolyte resistivity in ohm-cm,L is the length of the backfill column in feet, and d i sthe diameter of the backfill column in feet.

It should be no ted that the anode dimensions are the over-all length and diameter including backfill, i f the backfillis co ke breeze and is not significantly more than 2 f tlonger than the anode or not significantly more than 20 ftlonger than the anode column in a deep vertical groundbe dconfi guration. For earth backfill, the backfill colum ndimensions should be the dimensions of the manufacturer' sstandard packag ed anode can. Bare ceramic anodes shall notbe installed in ground without coke breeze backfill. Cokebreeze allows venting of gases and effectively reduces R .A

If ve rtical anode dimensions are assumed to be 6 in. i ndiameter and 8 ft in length, the following empirica lrelations may be used:

R = p (eq 1-5)A 398

If the anode di mensions are different, a different empiri-cal relation may be used:

R = p K (eq 1-6)A L

where R is the anode-to-electrolyte resistance, p is th eAelectr olyte resistivity in ohm-cm, L is the length of th ebackfill column in feet, and K is a shape function that isselected from table 3-4.

RA '( 0. 0052) p

L[ 1n 4L

2 % 4L ( 2h) 2 % L 2

2 d h

%2 hL

&( 2 h) 2 % L 2

L& 1]

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(eq 1-7)

Deep anode groundbed resistance graphs are available fo rdeep v ertical ground beds. (See figure 2-9a through 2-9 clocated in section 2-4.)

For a single horizontal anode:

where R is the anode-to-electrolyte resistance in ohms, pAis the electrolyte resistivity in ohm-cm, L is the lengt hof the backfill cylinder in feet, d is the diameter of thebackf ill cylinder in feet, and h is the depth of th ebackfill cylinder in feet.

If the horizont al anode dimensions are assumed to be 6 in.in diameter, 8 ft long, and buried 6 ft below the surface,the following empirical expression may be used.

R = p (eq 1-8)A 441

where R is the anode-to-electro lyte resistance in ohms andAp is the electrolyte resistivity in ohm-cm.

For multiple vertical anodes:

R = (0.0052) p [(ln (8L/d) -1)N N L

+ 2 L ln (.656 N)) (eq 1-9) C c

where R is the anode-to-electrolyte resistance, p is th eNelectro lyte resistivity in ohm-cm, N is the number o fanodes, L is the length of the backfill column in feet, dis t he diameter of the backfill column in feet, and Cc i sthe ce nter-to-center spacing of the anodes in feet. Thi sequation assumes a linear configuration of the groundbe danodes.

If the number o f anodes used does not produce a low enoughanode-to-electr olyte resistance, the number of anodes willhave to be increased accordingly.

RN 'p FADJ398

RN 'RAN

%p PFCc

RN 'p FADJ441

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(eq 1-10)

(eq 1-11)

(eq 1-12)

For optimum res ults, the length of the backfill column (L)should be less than the anode spacing (C ).c

If multiple ano des are assumed to be 6 in. in diameter and8 ft long, the following empirical expression may be used:

where R is the anode-to-electrolyte resistance in ohms, pNis the electrolyte resistivity in ohm-cm and F i sADJsel ected from table 3-9, which compensates both for th enumber and spacing of the anodes, which will be connecte dtogether as one groundbed.

If the anode dimensions are different, another empirica lexpression may be used:

where R is the anode-to-electrolyte resistance, R is th eN Aanode-to-electrolyte resistance for a single anode, p i sthe s oil resistivity in ohm-cm, N is the number of anode sused, P is a paralleling factor selected from table 3-5 ,Fand Cc is the center-to-center spacing of anodes in feet .This e quation assumes a linear configuration of th egroundbed anodes.

For multiple horizontal anodes:

where p is the electrolyte resistivity in ohm-cm and F isADJan adj3sting factor for groups of anodes selected fro mtable 3-9.

For a circle of rod anodes (as in a water storage tank):

where R is the anode-to-electrolyte resistance, p is th eNelectrolyte resistivi ty in ohm-cm, L is the length of eachB

RN '0. 0052 x p x 1n [ D/ 2 AR x DE) ]

LB

RN '0. 0016 p

DR( 1n

8DRDA

% 1n2 DR

H)

RW 'LW RMFT100 ft

Rc 'RSA

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(eq 1-13)

(eq 1-14)

(eq 1-15)

(eq 1-16)

rod anode in feet, D is the tank diameter in feet, A is th eRradius of the anode circle in fee t, and D is an equivalentEdiameter factor from figure 2-17.

For wire anode circle or hoop (in a water storage tanks):

where R is the anode-to-electrolyte resistance, p is th eAelectrolyte resistivity, D is the diameter of the anod eRcircle in feet, D is the diameter of the anode wire i nAfeet, and H is the depth below the high water level i nfeet.

Experience has shown that the diameter of the anode wir ecircle (D) should be typically between 40 and 70 percent ofthe tank diameter.

Wire resistance (R ) is the sum of both the rectifier-to -Wanode lead and the rectifier-to-protected-structure lead.

where L is the length of wire in thousands of feet an dW R is the resistance of the wire in ohms per 1000 ft.MFT

The structure-to-electrolyte resistance (R ) is dependen tCprimarily on the condition of the coating.

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where R is the coating resistance in ohm-square feet an dSA is the total surface area. If the structure surface i sbare, negligible resistance is assumed (R = 0).C

g. Calculate required rectifier voltage and current.

The required rectifier voltage (V and maximum curren tRECrating should include at least an extra 20 pe rcent to allowfor variations in calculations from actual conditions an dfor changes in the system over the system's life.

V = (I) (R ) (1.2) (eq 1-17)REC T

where I is the total current required and R is the tota lTcircuit resistance as calculated above.

I = (I) (1.2) (eq 1-18)REC

where I is the total current required and I is th eRECminimum current rating for a rectifier for this particularapplication.

Select a rectif ier with DC voltage and current capacity ofa sligh tly larger size (as calculated above) from th ecathodic protection rectifier manufacturer's publishe dliterature.

h. Prepare life cycle cost analysis.

The lif e cycle cost analysis should be prepared accordin gto th e guidelines given in TM 5-802-1 (reference 9) .Another source of information on performing l ife cycle costanalyses is NAC E RP-02-72 (reference 21). The choice of aparticula r anode type and configuration for desig ncalculation is somewhat arbitrary. The economics ma ydictate switching to a different design configuration an drepeating the applicable design steps.

i. Prepare plans and specifications.

Prepare plans t hat show the protected structure, locationsof anodes, rectifier, test stations, and powe r source, wirerouting, and details of wire-to-structure connections ,bui lding or structure penetrations, wire color coding ,potential survey test points in paved areas, and othe rpertinent infor mation. Prepare a one-line diagram to showthe entire system including wire sizes, anode type(s) ,power circuit, power circuit protection, and source o f

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power. Prepare specifications to describe req uired featuresof the system components.

1-3. References.

The following documents describe corrosion fundamentals, tradi -tional corrosion control techniques, and particular requirem ents ofU.S. Department of Defense agencies:

1. Bryan, William T., ed., Designing Impressed Current Cathodi cProtection Syst ems With Durco Anodes (The Duriron Company,Dayton, OH, 1970).

2. Burke, N. Dennis, and David H. Kroon, Corrosion and CorrosionControl, Liberty Bell Short Course (Philadelp hia Section ofthe National Association of Corrosion Engineers,Philadelphia).

3. Bushman, James B., Corrosion Control and Cathodic Protectio nTrain ing Manual (Steel Tank Institute, Lake Zurich, IL ,March 1988).

4. Bushman, James B., Cathodic Protection of Underground Storag eTanks and Piping, Paper delivered to the Nationa lAssociation of Corrosion Engineers North Central Regiona lConference (Chicago, 1988).

5. Bushman, James B., and David H. Kroon, "Cathodic Protection ofWater S torage Tanks," Journal of the American Water Work sAssociation (Denver, 1984).

6. Davy, Humphrey, "The Beginnings of Cathodic Protection," ThePhi losophical Transactions of the Royal Society (London,1824).

7. Department of the Air Force, Civil Engineering Corrosio nControl - Cathodic Protection Design, Air Force Manua l88- 45 (Headquarters, U.S. Air Force, Washington D.C. ,1988).

8. Department of the Air Force, Maintenance and Operation ofCathodic Protection Systems, Air Force Manual 85- 5(Headquarters, U.S. Air Force, 1985).

9. Department of the Army , Economic Studies of Military Construc-tio n Design - Application, TM 5-802-1 (Headquarters ,Department of the Army, Washington D.C., 31 D ecember 1988).

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10. Department of the Army, Electrical Design, Cathodic Protection,TM 5-811-7 (Hea dquarters, Department of the Army, 22 April1985).

11. Dwight, H.B., "Calculation of Resistance to Ground," AIE ETransactions, vol. 55 (1939).

12. El ectrical Engineering Cathodic Protection, Military Handboo k1004/10 (31 January 1990).

13. Fontana, Mars G., Corrosion Engineering, 3d ed. (McGraw Hill,Inc., New York, 1986).

14. Hock, V., J. Givens, J. Suarez, and J. Rigsbee, "Structure ,Chemistry and Properties of Mixed Metal Oxide," Paper No .230, Proceedings of the National Association of Corrosio nEngineers (1988).

15. Kumar, A., and M. Armstrong, "Cathodic Protection De signs UsingCeramic Anodes," Materials Performance, vol. 27, No. 1 0(Oct 1988), pp. 19-23.

16. Kumar, A., and M. Armstrong, "New Cathodic Protection DesignsUsing Ceramic A nodes," Materials Performance, vol. 27, No.10 (Oct 1988), pp. 19-23.

17. Kumar, A, and V. Hock, "Survey of Cathodic Protection SystemsUsing Ceramic Coated Anodes," Paper No. 284, Proceedings ofthe National Association of Corrosion Engineers (1989).

18. Morgan, John, Cathodic Protection, 2d ed., National Associationof Corrosion Engineers (Houston, 1987).

19. Myers, James R., Fundamentals and Forms of Corrosion (Air ForceInstitute of Technology, October 1974).

20. Myers, J.R., and M.A. Aimone, Corrosion Control for UndergroundSteel P ipelines: A Treatise on Cathodic Protection (Jul y1980).

21. NACE RP-02-72, "Recommended Practice - Direct Calculation o fEconomic Appraisals of Corrosion Control Measures "(National Association of Corrosion Engineers, Houston ,1972).

22. NACE RP-0l-69, "Recommended Practice - Control of Externa lCorrosion on Underground or Submerged Metallic Pipin gSystems" (National Association of Corrosion Engineers ,Houston, 1969)

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23. NACE RP-02-85, "Recommended Practice - Control of Ex ternal Cor-rosi on on Metallic Buried, Partially Buried or Submerged ,Liquid Storage Systems" (National Association of CorrosionEngineers, Houston, 1985).

24. Paullin, Robert L., Pipeline Corrosion Protection: A Departmentof Transportation Perspective (Materials Transportatio nBureau of the U.S. Department of Transportati on, WashingtonD.C., 1984).

25. Peabody, A.W., Control of Pipeline Corrosion (Nation al Associa-tion of Corrosion Engineers, Houston, 1967).

26. Reding, J.T., Performance of Mixed Metal Oxide Activate dTitanium Anodes (National Association of Corrosio nEngineers, Houston, 1987).

27. Sunde, Earling D., Earth Conduction Effects in Transmissio nSystems (D. VanNostrand Co., New York, 1949).

28. Treseder, R.S., NACE Corrosion Engineer's Reference Boo k(National Association of Corrosion Engineers, Houston ,1980).

29. Turner, Jane M., "Controlling Galvanic Corrosion in Soil WithCathodic Protec tion," Gas Industries Magazine (Park Ridge,IL, 1988).

30. Uhling, Herbert H., The Corrosion Handbook (John Wiley & Sons,New York, 1948).

31. Uhling, Herbert H., and R. Winston Revie, Corrosion an dCorrosion Control: An Introduction to Corrosi on Science andEngineering, 3d ed. (John Wiley & Sons, New York, 1985).

32. West, Lewis H., and Thomas F. Lewicki, Civil Engineerin gCorro sion Control, Vol. 1: Corrosion Control - General ;Vol . II: Cathodic Protection Testing Methods an dIns truments; Vol. III: Cathodic Protection Design (Ai rForce Civil Engineering Center, Tyndal Air Fo rce Base, FL).

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EXAMPLES OF IMPRESSED CURRENTCATHODIC PROTECTION DESIGN

2-1. Purpose.

The following examples show th e use of design procedures explainedin the previous section:

2-2 Steel fuel oil lines2-3 Underground storage tanks2-4 On-grade tank bottoms2-5 Gas distribution systems2-6 Elevated water tanks (ice is expected)2-7 Elevated water tanks (no icing will occur)2-8 On-grade water storage reservoir (ice is expected)

2-2. Steel Fuel Oil Lines.

Impressed current cathodic protection is desired for the 6-in .welded fuel oil line shown in figures 2-1A and 2-lB. Since thi spipeli ne is in existence, current requirement tests have alread ybeen made. There are no other underground structures in the area,so a surface point groundbed is chosen. Figure 2-2 illustrates asurface point groundbed anode system using prepackaged ceramic rodanodes. These prepackaged cer amic rod anodes are further detailedin figures 2-3 and 2-4.

a. Design data.

1) Soil re sistivity in area where groundbed is desired is2000 ohm-cm.

2) Pi pe has 6-in. diameter (outside diameter = 6.62 5in.)

3) Pipe length is 6800 ft.

4) Design life for cathodic protection anodes is 15 yearssince the structure will no longer be needed after thattime.

5) Design current density is 2-mA per sq ft of bare pipe.

6) The pipe is coated with hot-applied coal-tar enamel.

7) 90 percent coating efficiency based on previou sexperience with this type coating.

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8) Effective coating resistance at 15 years is estimate dat 2500 ohm-sq ft.

9) The pipeline is isolated from the pumphouse and tan kwith insulating joints.

10) For this example, we have decided that the cathodi cprotect ion system circuit resistance should not exceed2.5 ohms.

11) Electr ic power is available at 120/240 V single phas eAC from a nearby overhead distribution system.

12) Current requirement testing indicates that 2.8 amp areneeded for adequate cathodic protection.

b. Computations.

1) Calculate the external surface area of the pipe.

Pipe diameter = 6 in.

Pipe length = 6800 ft

Pipe surface area per lin ft = 1. 734 sq ft/lin ft (fro m

table 3-2)

External pipe surface area = 6800 lin ft x 1.734 s q

ft/lin ft = 11,791 sq ft

2) Check the current requirement (I) using equation 1-1:

I = (A)(I')(l.0-C )E

Where: A = 11, 791 sq ft - External pipe surfac e

area from previous calculation.

I' = 2 mA/sq ft - Required current densit yfrom item 5 of paragraph 2-2a.

C = 0.9 0- Coating efficiency expressed i nEdecimal form from item 7 of paragrap h2-2a.

I = 11,791 sq ft x 2 mA/sq ft x (1.0 -0.90)

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I = 2358 mA or approximately 2.4 amp

which agrees well with the current requirement tes tdata provided in item 12 of paragraph 2-2a.

3) Select an anode and calculate the number of anode srequire d (N) to meet the design life requirements. Twosizes are selected for trial calculations usin gequation 1-2:

N = I I A

Where: I = 2.8 amp (Curren t requirement from item

12 of paragraph 2-2a)

I = Current rating per anode, varies de -Apending on anode size from table 3-3.

Several different size anodes could be chosen .Experi ence has shown that for this type of groundbed ,60-in. long anode packages are desirable.

For a 3-in. by 60-in. packaged canist ers with a 1/8-in.by 48-i n. ceramic anode rod, I = 1.2 amp/anode (fro mAtable 3-3 for 15-year design life).

N = 2.8 = 2.33; use 3 anodes. 1.2

For a 3 -in. by 60-in. packaged canister with a 1/4-in.by 48-i n. ceramic anode rod, I = 2.4 amp/anode (fro mAtable 3-3 for a 15-year design life).

N = 2.8 = 1.17; use 2 anodes. 2.4

4) Calculate the resistance of a single anode-to-earth (R )Afrom equation 1-6:

R = p K A L

Where: p = 2000 ohm-cm (Soil resistivity in are a

where groundbed is desired from item 1of paragraph 2-2a)

RN 'RAN

%p PFCc

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K = 0.0213 (Shape f unction, from table 3-4[where: L/d = 60 in./3 in. = 20])

L = 5 ft (Effective anode length [canisterlength))

d = 3 in. (Anode ba ckfill diameter [canis-ter diameter])

R = 2000 x 0.0213A5

R = 8.53 ohmsA

5) Calculate the number of anodes requir ed to meet maximumanode groundbed-to-earth resistance requirements fro mequation 1-11:

Where: R = Groundbed-to-earth resistance.N

R = 8.5 3 ohms (Resistance of a singl eAanode-to-earth from the previou scalculation)

N = Assume 5 anodes (discussed below)

p = 2000 ohm-cm (Soil resistivity)

P = 0.00268 (Parall eling factor from tableF3-5 [discussed below])

C = 20 ft (Center-to-center spacing o fCanodes [discussed below])

To determine P , a figure for N must be assumed. ThisFparalle ling factor compensates for mutual interferencebetween anodes and is dependent on spacing. From th elaw of parallel circuits, it appears that five anode smight give the desired circuit resistance of 2.0 ohm smaximum, i.e.:

8.53 ohms/anode = 1.706 ohms 5 anodes

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This is the approximate resistance ba sed on the law forparallel circuits. (When equal resistance values ar ejoined together in a parallel circuit , the total resis-tance v alue of the circuit is approximately equal to asingle resistance value divided by th e number of resis-tance values.)

To calculate the true anode resistanc e for five anodes,we must perform the calculation from equation 1-11.

C = 20 ft (The spacing must be chose nCbased on previous experience to solv ethe equation.) The spacing ca ntyp ically be from 10 to 50 ft. Fo rthis example, we will try 20 ft:

R = 8.53 + 2000 x 0.00268N 5 20

R = 1.706 + 0.268N

R = 1.974 ohmsN

This i s within specification, but very close t oexceeding maximum specified circuit resistance. Tr ysix ano des to allow for variations in soil resistivityand t o allow for wire and pipe-to-earth resistance .Repeat calculation from equation 1-li,

Where: N = 6 anodes

P = 0.00252, parall eling factor from tableF3-5

R - 8.53 + 2000 x 0.00252 N 6 20

R = 1.67 ohmsN

6) Based on the previous selection of six anodes for th enumber of anodes to be used, the total circui tresistance must be determined.

7) Sele ct an area for anode bed placement. Here, th eselecte d area is 100 ft from the pipeline for improvedcurrent distribution. The anode bed location for thistype de sign must be far enough away from the structure

RW 'LW RMFT

1000 ft

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to be protected to assure uniform distribution of th epro tective current to all structure components. Th ehigher the soil or water resistivity, the further awaythe gro und must be located. For this example, previousexperi ence has shown that the nearest anode should b elocated a minimum of 100 ft from the structure t oassure good current distribution. Since there are n oother u nderground utilities in the area, the groundbedmay be installed either perpendicular or parallel t othe pipe (figures 2-IA and 2-lB).

8) Determine the total circuit resistance (R ) fro mTequation 1-3:

R = R + R + RT N W C

Where: R = Groundbed resistance-to-earth (ohms)N R = Header cable and resistance (ohms)W R = Pipe-to-earth resistance (ohms)C

a) Groundbed-to-earth resistance (R ) from step 5.N

R = 1.67 ohmsN

Specify anodes with individual lead wires o fsuff icient length so that each anode wire can b erun directly to the rectifier without splices .(This is virtually always true for anode be ddesigns where the individual lead wire length srequired do not exceed an average of 400 ft.)

b) Anode lead wire resistance from equation 1-15:c

Where: L = Total of each actual lead wir eAVG

lengths/number of anodes= (140 ft + 120 ft + 100 ft + 8 0

ft + 60 ft + 40 ft)/6 anodes = 90 ft per anode

L = L /NW AVG = 90/6 = 15 ft

Rc 'RSA

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= Average anode lead wire length /number of anodes. Thi s is basedon the circuit resistance for 6anode lead wires in parallel.

R = 2. 58 ohms - Resistance for No .MFT14 AWG cable which has bee nselected from table A-6.

R = 15 ft x 2.58 ohmsW 1000 ft

R = 0.039 ohmsW

c) Pipe-to-earth resistance (R ) from equation 1-16:C

Where: R = 2500 ohm-sq ft - Effectiv eS

coating resistance from item 8of paragraph 2-2a.

A = 11,791 sq ft - Externa lpip e surface area calculated i nstep 1 of paragraph 2-2b.

R = 2500 ohm-sq ftC 11,791 sq ft

R = 0.212 ohmC

d) Calc ulate the total circuit resistance (R ) fro mTequation 1-3:

R = R + R + RT N W C

R = 1.67 + 0.039 + 0.212T

R = 1.921 ohmsT

e) Sinc e the design requirements call for a maximu mallowa ble groundbed resistance of 2.0 ohms and R =T1.921 o hms, the design using six (3-in x 5-ft) ceramicanode canisters will work.

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9) Calculate the rectifier voltage (V ) from equationREC1-17:

V = (I) (RT) (120%)REC

Where: I = 2.8 amps (Current requiremen t

from item 12 of paragraph 2-2a.

R = 1.92 ohms (Total circui tTresistance from p reviou scalculation.

120% = Rectifier voltage capacit ydesign safety factor.

V = 2.8 amp x 1.92 ohms x 1.2REC

V = 6.45 VREC

c. Select rectifier.

Based on the design requirement of 2. 8 amperes and 6.45volts, specify a 10-V, 5-amp rectifier, which is thenearest standard capacity available from commercia lcathodic protection rectifier manufacturers.

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2-3. Underground Storage Tanks (USTs).

The service station shown in figure 2-5 has three existin gunderground tanks and associat ed pipe. The quality of the coatingis unknown and it is not feasi ble to install dielectric insulationto isolate the UST system. Because of the anticipated larg ecurren t requirement, an impressed current protection system i schosen. To distribute the current evenly around the tanks an dpiping, and to minimize interference effects on other structures,a distributed anode surface be d using vertical anodes is selected.Vertical anodes can be install ed with relative ease in holes coredthrough the paving around the UST system. Wiring can be installedseveral inches below the paving by cutting and hand excavatin gnarrow slots/trenchs through the paving.

a. Design data.

1) Soil resistivity is 4500 ohm-cm.

2) Pipe is 2 in., nominal size. Total length of al lburied piping is 750 ft.

3) Tanks are 8000 gal, 96 in. diameter by 21 ft 3 in .long.

4) Elect rical continuity of tanks and piping has bee nassured.

5) It is not feasible to install dielectric insulation ;sys tem is therefore not isolated electrically fro mother structures.

6) Design cathodic protection anodes for 20-year life.

7) Coating quality is unknown, assume bare.

8) The cathodic protection system circuit resistanc eshould not exceed 2.5 ohms.

9) Electr ic power is available at 120 V single phase i nthe station building.

10) Current requirement test indicates that 8.2 ampere sare needed for cathodic protection.

b. Computations.

1) Find the external surface area (A) of the storag etanks and piping.

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Area of each tank = 2 B (tank radius) + B (tan k2 diameter) (tank length)

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Area of 3 tanks = 3 [2 x 3.14 x 4 +3.14 x 8 x 21.25]2= 1905 sq ft

Area per lin ft of 2-in. pipe = 0.621 sq ft/ft (fro mtable A-2)

Total pipe area = 750 ft x 0.621 sq ft/ft = 466 sq ft

Total area = 1905 sq ft + 466 sq ft = 2371 sq ft

2) Check the curre nt requirement (I) using equation 1-1:

I = (A)(I')(1.0-C )E

Where: A = 2371 sq ft (External tank and pipin g

surface area from previou scalculation)

I' = 2 mA/sq ft (Required current densit y[assumed])

C = 0.0 0 (Coating efficiency expressed i nEdecimal form from item 2 of paragrap h2-3a)

I = 2371 sq ft (2 mA/sq ft) x (1.0 - 0.0)

I = 4742 mA or 4.7 amp.

The 4.7 amp would be reasonable for the facil ity if itwere insulated. The actual current requirement of 8.2amp occurs because of current loss to othe runderground structures and is also reasonable i nrelation to tha t calculated for an isolated facility.

3) Sel ect an anode and calculate the number of anode srequired (N) to meet the design life requirements.

Calculations can be run on several size anode s, but inthi s case 2-in. by 60-in. packaged rod anodes (ro dsize = 0. 125 in x 4 ft long) are chosen for ease o fconstruction. Using equation 1-2, the number o fanodes required to meet the cathodic protecti on systemdesign life can be calculated:

N = I I A

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Where: I = 8.2 amp (Curren t requirement from item

10 of paragraph 2-3a)

I = 1.0 amps/anode (Current rating pe rAanode from table 3-3)

N = 8.2 = 8.2 anodes; use nine, 2-in.1.0 by 60-in. packaged rod

anodes

4) Cal culate the resistance of a single anode-to-eart h(R ) from equation 1-6:A

R = p K A L

Where: P = 4500 ohm-cm (Soil resistivity fro m

item 1 of paragraph 2-3a)

K = 0.0234 (Shape function from table A- 4[where: L/d = 60 in./2 in. = 30])

L = 5 f t (60 in.) (Effective anod elength).

d = 2 in. (Anode/backfill diameter)

R = 4500 x 0.0234 A 5

R = 21.06 ohmsA

5) Calc ulate the number of anodes required to mee tmaximum anode groundbed resistance requirement. Adis tributed anode array does not lend itself to a nexact calculati on of equation 1-11 because the anodesare positioned at various locations and are no tlocate d in a straight line. Equation 1-11 assumes astraight line configuration. However, to approximat ethe total anode-to-earth resistance, equation 1-li maybe used.

RN 'RAN

%p PFCc

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Where: R = Groundbed-to-earth resistanceN

R = 21.06 ohms (Resistance of a singl eAanode-to-earth from the previou scalculation)

p = 4500 ohm-cm (Soil resistivity fro mitem 1 of paragraph 2-3a)

C = 10 ft (Estimated approximate spacin gCof anodes)

P = 0.0 0212 (Paralleling factor fro mFtable 3-5; assume 9 anodes; reasonin gis th e same as in step 5, paragrap h2-2b)

N = 9 anodes (Estimated number of anode srequi red; from step 3, paragrap h2-2b)

R = 21.06 + 4500 x 0.00212 N 9 10

R = 3.29 ohms.N

Resistance is t oo high. Additional calculations usingan increasing n umber of anodes (i.e., 11, 12, 13, 14,etc.) have to be made; these calculations show tha tfourteen anodes will yield a groundbed-to-eart hresistance of 2.26 ohms.

6) Selec t the number of anodes to be used. The number sdetermined are:

For life = nine anodes maximum requiredFor resistance = fourteen anodes maximum required

Therefore, the larger number of anodes, fourteen, i sselected.

7) Determine the total circuit resistance (RT) fro mequation 1-3:

RW 'LW RMFT

1000 ft

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R = R + R + RT N W C

Where: R = Groundbed resistance (ohms)N R = Header cable/wire resistance (ohms)W R = Structure-to-earth resistance (ohms).C

a) Groundbed resistance (RN) from step 5.

R = 2.26 ohmsN

b) Header cable/wire resistance (R ) from equation 1 -W15:

Where: L = 150 f t (Effective cable length. The loo pW

circuit makes calculating effective wir eresistance complex. Since current i sdischarged from anodes spaced all along thecable, one-half the total cable length maybe used to approximate the cabl eresis tance. Total cable length = 300 ft .Ef fective cable length = (300 ft) = 15 0ft.)

R = 0.2 54 ohm (This is the resistance pe rMFT1000 li n ft of No. 4 AWG cable, whic hhas been selected for ease o fhandling.)

R = 150 ft x 0.254 ohm W1000 ft

R = 0.038 ohm; use 0.04 ohmW

c) Structure-to-earth resistance.

Since the tanks and piping are essentially bare andar e not electrically isolated, structure-to-eart hresistance may be considered n egligible. ThereforeR = 0.C

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d) Calculate total circuit resistance (RT) fro mequation 1-3:

R = R + R + RT N W C

R = 2.26 + 0.04 + 0 = 2.30 ohmsT

Sinc e the design requirements call for a maximu mgroundbed resistance of 2.5 ohms and RT = 2.3 0ohms, the design using fourteen 2-in. by 60-in .packaged ceramic anodes will work.

8) Cal culate the rectifier voltage (V ) from equatio nREC1-17:

V = (I) (R ) (120%)REC T

Where: I = 8. 2 amp (Current requirement from step 2 ,

paragraph 2-3b)

R = 2.30 ohms (Total circuit resistance fro mTprevious calculation)

120% = Rectifier voltage capacity design safet yfactor.

V = 8.2 amp x 2.30 ohms x 1.2REC

V = 22.6 VREC

c. Select rectifier.

Based on the design requirement of 22.6 V and 8.2 amp, arectifie r can be chosen. A 12-amp, 24-V unit is selecte dbecause this is the nearest standard commercial siz eavailable.

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2-4. On-Grade Tank Bottoms.

Four on-grade fuel oil storage tanks are to be constructed w ith theconfiguration shown on figure 2-6. This design may be prohibitedif secondary containment uses a nonconductive membrane beneath thetanks. The membrane would not allow the cathodic protectio ncurrent to flow from the remotely located deep anode through th enonconductive membrane to the tank bottoms. If this situatio nexists, a distributed anode de sign with the anodes located betweenthe membrane and the tank bottoms would have to be used. Al lpiping will be above grade. T o minimize the extent of undergroundcable, it was decided to use a deep anode groundbed, located justoutside the spill containment dikes. (Note: Some county, state,or federal agencies such as the EPA may have regulations tha taffect the use of deep anode beds because they can provide aconduit for the mixing of water between aquifer levels. In suc hcases, regulations have sometimes required cementing of the annulusbetween the deep anode bed casing and the augered hole to preventthis water migration. The system designer should check with th eapplicable agencies before committing to a deep anode design .Figures 2-7, 2-7A, 2-8, and 2-8A illustrate a typical deep anod egroundbed using ceramic rod an odes. The tank bottoms will be bare.All piping will be above groun d. The tanks will be dielectricallyinsulated from the structures. Field tests were made at the siteand the subsurface geology was determined to be suitable for a deepanode groundbed (reference 25).

a. Design data.

1) Soil resistivity at anode depth is 1500 ohm-cm.

2) Tanks are 75 ft in diameter.

3) Design cathodic protection anodes for a 15-year life.

4) Design current density is 2 A per sq ft of tank bottom.

5) Since the tanks are electrically isolated from eac hother, intertank bonds will be required.

6) The cathodic protection system circuit resistanc eshould not exceed 0.75 ohm.

7) Elec tric power is available at a switch rack in a nunclass ified (nonexplosion proof) area 125 ft from thedesired groundbed location, 230 V AC, single phase.

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8) Since the exterior bottom of these type tanks ar ealways bare, the coating efficiency will equal 0.00.

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b. Computations.

1) Find the surface area to be protected.

Area of each tank bottom = B r :2A = B x 37.5 = 4418 sq ft2

Area of the four bottoms:A = 4 x 4418 = 17,672 sq ft

2) Determine the current requirement (I) from equatio n1-1:

I = (A) (I')(l.0-C )E

Where: A = 17, 672 sq ft (Total surface area fro m

previous calculation)

I' = 2 mA/sq ft (Current density from ite m4 of paragraph 2-4a)

C = 0.0 (bare) (Coating efficiency fro mEitem 8 of paragraph 2-4a)

I = 17,672 sq ft x 2 mA/sq ft x (1.0 -0.0)

I = 35 amp

3) Select an anode and calculate the number of anodes (N)require d to meet the design life. The deep anod egroundbed will consist of a series of rods connected toa conti nuous header cable. For physical strength, the1/8-in. or 1/4-in. diameter rods are usually chosen. Toensure good transmission of current in to the backfillcolumn, long anodes, typically 6 ft or 8 ft are used .For thi s example, 1/4-in. diameter by 72-in. long rodshave been chosen. Using equation 1-2:

N = I I A

Where: I = 35 amp (Current requirement fro m

previous calculation)

I = 6.6 amp/anode (Current rating per heav yAduty coated anode from table 3-3)

N = 35 = 5.3; use 6 anodes 6.6

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4) Calculate the required length of the backfill column.

Spacing between anodes depends primarily on th eresisti vity of both the backfill column (coke breeze).If lo w resistivity calcined fluid petroleum coke i sused, than the spacing between rod anode (1/8-in. an d1/4-in. diameter) should not be greater than twice theindivi dual rod length. Based on previous desig nexperie nce, a spacing between anode rods of 11 ft i sselected.

The minimum length of the backfill column is the ncalculated as follows:

6 anodes at 6 ft per anode = 36 ftSpacing between anodes: 5 at 11 ft = 55 ftSpace above and below anode string* = 20 ft

Total length 111 ft

* Generally, the coke breeze column extends fro m10 ft below the bottom anode t o 10 ft above the topanode as shown in figures 2-7 and 2-8.

5) Calc ulate the backfill column-to-earth resistanc e(R ).A

This can be done from equation 1-4. Because severa lattempts may have to be made to obtain the require dgroundbed resistance, the process is facilitated b yusing t he curves in figures 2-9A through 2-9C. Typicalhole diameters are 6 or 8 in.; a 6-in. diameter hol ehas bee n selected for this groundbed. Figure 2-9C willbe used for this design.

Figu re 2-9C shows that, for a 111-ft long coke breez ebackfil l column, the resistance to earth per 1000 ohm-cm is 0 .31 ohm. In 1500 ohm-cm soil, the resistanceis 0.31 x 1500/1000 = 0.47 ohm (groun dbed resistance in1000 ohm-cm soil, times actual soil r esistivity in ohm-cm per 1000). This is well below the desig nrequirement of 0.75 ohm.

6) Determine total circuit resistance (RT) from equatio n1-3 :

R = R + R + R (Because the cathodi cT A W Cpro tection system utilizes a singl edeep anode groundbed R = R )A N

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50

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Where: R = Backfill column-to-earth resistanc eA

(ohms), calculated in step 5 above

R = Header cable/wire resistance (ohms)W

R = Structure-to-earth resistance (ohms).C

a) Backfill column-to-earth resis tance (RA) from step5:

R = 0.47 ohmA

b) Wire resistance

A deep anode groundbed is defined as one where thetop of the backfill column is at least 50 ft belowthe surface of the earth. Actual depth will vary,depending on subsurface geology and the distanc eover which the current is expected to spread. I nthis e xample, 65 ft was selected as the depth t othe top of the coke breeze column. The anodes canbe sup plied through a single or dual feed. Dua lfeed is preferred to reduce both the resistance ofthe circuit and the chance of a failure due to acable break.

No. 4 AWG cable has been chosen. Cable length shave been calculated based on the followin gdistances, illustrated in the deep groundbe dexample shown in figure 2-7.

RWT 'LWT RMFT1000 ft

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From grade to top of backfill column 65 ft

From top of backfill column totop anode 10 ft

From grade to bottom of backfillcolumn 176 ft

From bottom anode to bottom ofbackfill column 10 ft

From anode hole to rectifier 2cables at 125 ft 250 ft

From top of anode assembly tobottom of anode assembly 91 ft

Total cable length from top anodefeed to rectifier 125 ft +65 ft + 10 ft= 200 ft

Total cable length from bottomanode feed to rectifier125 ft + 176 ft - 10 ft = 291 ft

Calc ulate top cable resistance (R ) fro mWTequation 1-15:

Where: L = 200 ft (Cable length from previou sWT

calculation)

R = 0.2 54 ohm (Resistance per 1000 lin f tMFTof No. 4 AWG cable which has bee nsel ected for installation. Thes eresistance values can be found i ntable 3-6).

R = 200 ft x 0.254 ohm = 0.051 ohmWT 1000 ft

Calc ulate bottom cable resistance (R ) fro mWBequation 1-15:

RPOS 'LW RMFT

1000 ftx 1

2'

91 ft ( x 0. 254 ohm/ ft1000 ft

x 12

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Where: L = 291 ft (Cable length from previou sWB

calculation)

R = 0.254 ohm (Cable resistance per 100 0MFTlin ft [same as above])

R = 291 ft x 0.254 ohm = 0.074 ohmWB 1000 ft

These two cables are in parallel, so that their wir eresis tance (R ) is calculated from the law o fT/Bparallel circuits:

1 = 1 + 1 R R RT/B WT WB

1 = 1 + 1 R 0.051 0.074T/B

1 R = 19.6 + 13.5 = 33.1T/B

R = 0.030 ohmT/B

Since current i s dissipating along the portion of thecable to which the anodes are connected, th eresist ance of this cable (R ) is taken as one hal fPOSits total resistance as was done in example 2-2.

*91 ft is the overall anode column length including th einterconnecting wire from the top of the top anode to the botto mof the bottom anode (see item 4 from paragraph 2-4b.)

R = 0.012 ohmPOS

Negative circuit wire resistance must also b ecalculated:

Negativ e cable from rectifier to closest tank = 125 ftof No. 4 AWG

Intertank bonds = 170 ft of No. 4 AWG **

**The two intertank bond circuits are in parallel and of about thesame length. From the law of parallel circuits, total resistanceof two parallel circuits of equal resistance is one half th eresistance of each circuit.

RNEG 'LW RMFT

1000 ft

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Therefore, one half the cable length is used in thi scalculation. The calculation is also conservative since notall the current flows through the complete intertank bon dcircuit.

Total ft of negative circuit wire = 295 ft of No. 4 AWG cable

Negative resistance if calculated from equatio n1-15:

Where: L = 295 ft (Negative cable length)W

R = 0.254 ohm (Negative cable resistanc eMFTper 100 0 un ft of No. 4 AWG cabl e[table A-6])

R = 295 x 0.254 NEG1000

R = 0.074NEG

Total wire resistance therefore is:

R = R + R + R = 0.030 + 0.012 + 0.074W T/B POS NEG

R = 0.116 ohmW

Cable in sulation is also important. Hig hmolecular weight polyethylene insulation ,commonly used for cathodic protection work, tends t oblis ter, become brittle, and then crack in dee pgroundbed use where chlorine gas generation c an occur.This has been m ost prevalent in open holes containingbra ckish water, but may occur in coke breez ebackfilled holes also. Consequently, to minimize th echances of cable failure, one of the two followin gtypes of insulation, which show good resistance t othese oxidizing environments should be used:

Polyvinylidene fluoride (Kynar) 1____________________ Registered trademark of Penwalt Corp.1

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Copolymer of chlor-tri-fluorethylene and ethylen e(Halar) 2

To pr otect the insulation itself and t ofacilitate handling, cables for deep anod egro undbeds also have an outer jacket of high -molecular weight polyethylene extruded over th eKynar or Halar insulation.

c) Structure-to-earth resistance (R ).c

Since the tank bottoms are bare, thei rresistance-to-earth is considered negligible ,therefore, R is taken as zero.C

d) Calculate total circuit resistance (RT) fro mequation 1-3:

R = R + R + RT N W C

R = 0.470 + 0.116 + 0.0T

R = 0.586 ohmT

This is well below the design requirement and ,therefore, this groundbed with 6 - 1/4 in. by 6 ftceramic anode rods with total backfill colum nlength of 111 ft will be used.

7) Calculate rectifier voltage (V ) from equation 1-17 :REC

V = (I) (R ) (120%)REC T

Where: I = 35 amp (Current requirement from ste p

2, paragraph 2-4b)

R = 0.5 86 ohm (Total circuit resistanc eTfrom previous calculation)

120% = Rectifier capacity safety factor

V = 35 amp x 0.586 ohm x 120%REC

V = 24.6 VREC

c. Select rectifier.____________________ Registered trademark of Allied Chemical Corp.2

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Based on the design requirement of 24.6 V and 35 amp, arectifier can be chosen. A 30 V, 42 ampere unit i scommercially av ailable and is selected. This rectifier canbe pole mounted as illustrated in figure 2-10.

d. Installation.

Figures 2-7 and 2-8 show this deep anode design.

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2-5. Gas Distribution System.

It has been decided to install cathodic protection on the ga sdistribution piping in a post housing facility. Figure 2-11 showsa portion of the piping.

The water distribution system has recently been replaced wit hnonmetallic pipe. On this bas is, it has been decided that the gaspiping can be protected with impressed current from a deep anod egroundbed without causing interference to the water pipe. (Note:Some county, state, and federal agencies such as the EPA may haveregulations which prevent the use of deep anode beds because theycan provide a conduit for mixi ng of waters between aquifer layers.In such cases, regulations have required cementing of the annulusbetween the deep anode bed casing and the augered hole to preventthis water migration. The system designer should check with th eapplicable agencies before committing to a deep anode design.)

a. Design data.

1) Experience in the area shows the subsurfac eresistivity at a depth of 50 and 200 ft to be 2000 ohm-cm.

2) Piping consists of:

28,000 ft of 1 1/4-in. diameter pipe1,200 ft of 4-in. diameter pipe3,600 ft of 6-in. diameter pipe

3) Design cathodic protection anode for a 20-year life.

4) Experi ence with similar pipe in the same general soi ltype has shown that a design current density of 2 A persq ft of bare pipe is conservative.

5) Piping is welded steel, poorly coated ; considered to bebare for protection purposes and, therefore, th ecoating efficiency is 0.0 (CE = 0.0).

6) Pi ping is isolated at each house and at the tie-in t othe main supply line.

7) The cathodic protection circuit resistance should no texceed 1 ohm.

8) 120/240 V AC, single phase power is available.

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9) Curre nt requirement tests are not practiced in thi scase, s o design will be based on calculations. (Note:Current requirement tests should be conducted wheneverpracticable.)

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b. Computations.

1) Calcula te the external surface area of the pipe (A) tobe protected.

Unit Area*Pipe di ameter Length (sq ft/lin ft) Area (sq ft)

1 1/4 in. 28,000 ft 0.434 12,152 4 in. 1,200 ft 1.178 1,414 6 in. 3,600 ft 1.734 6,242

*from table A-2 Total 19,808

2) Determine current requirement (I) from equation 1-1:

I = (A)(I')(l.0-C )E

Where: A = 19,808 sq ft (Surface area of pipe t o

be protected from previous calculation)

I' = 2 mA/sq ft (Current density, from item4, paragraph 2-5a)

C = 0.0 (Coating efficiency [bare pipe])E

I = 19,808 sq ft x 2 mA/sq ft x (1.0 - 0.0)

I = 39,616 mA or 39.6 amp; use 40 amp

3) Select an anode and calculate the number of anode sre quired (N) to meet the design life. For thi sdesign, t ubular anodes have been chosen. Th egroundbed will consist of a series of anodes attache dto a continuous header cable. A typical size tubula rcerami c anode used in deep anode beds, 1-in. diamete rby 39.4 in. long, is selected. The number of anode sare determined using equation 1-2:

N = I I A

Where: I = 40 amp (Current requirement fro m

previous calculation)

I = 8.0 amp/anode (Current rat ing per anodeAfrom table 3-3 [in coke breeze])

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N = 40 = 5 anodes 8

To allow a factor of safety, use 6 anodes

4) Calculate the required length of the backfill column.

The maximum allowable spacing between anodes depend sprimarily on the resistivity of the backfill colum n(coke b reeze). For tubular anodes of -in. diameter orgrea ter, maximum spacing between anodes should no texceed four times the anode tube length. For thi sexample, we will try the maximum allowable spacing o f13 ft (39.4 in. x 4/12 = 13.1 ft).

The minimum length of the backfill column i scalculated as:

6 anodes at 39.4 in. (3.3 ft) = 19.8 ftSpacing between anode: 5 at 13 ft = 65.0 ftSpace above and below anode string* 20.0 ft

Total length 104.8 ft(use 105 ft)

* Generally, the coke breeze column extends from 10 ft below th ebottom anode to 10 ft above the top anode.

5) Calc ulate the backfill column-to-earth resistanc e(R ).A

This can be done from equation 1-4. Because severa lattempts may have to be made to obtain the require dre sistance-to-earth, the process is facilitated b yusing t he curve in figure 2-9C. Typical hol edi ameters are 6 or 8 in.; a 6-in. diameter hole ha sbeen selected for this groundbed.

From figure 2-9C for a 105 ft long co ke breeze backfillcolumn, the resistance to earth per 1000 ohm-cm i s0. 325 ohm. In 2000 ohm-cm soil, the resistance i s0. 325 x 2 = 0.65 ohm, which is below the desig nrequirement defined in item 7 of paragraph 2-5a.

6) Determine total circuit resistance (R ) from equatio nT1-3:

RW 'LW RMFT

1000 ft

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R = R + R + R (Because the cathodic protectio nT A W Csystem uses a single deep anode groundbed R = R )A N

Where: R = Backfill column-to-earth resistanc eA

(ohms).

R = Wire resistance (ohms).W

R = Structure-to-earth resistance (ohms).C

a) Backfill column-to-earth resistance (R ) fro mAstep 5.

R = 0.65 ohmA

b) Wire resistance (R ) from general equation 1-15:W

A deep anode groundbed is defined as one wher e the topof th e backfill column is at least 50 ft below th esur face of the earth. Actual depth will vary ,depending on su bsurface geology and the distance overwhich the current is expected to spread. In thi sexample, the depth to the top of the coke breez ebackfil l column was determined to be 100 ft. Th eanodes can be s upplied through a single or dual feed.Dual fe ed is preferred to reduce both the resistanc eof the circuit and the chance of a failure due to acable break.

The as sembly is made on a single length of cable ,beginning at the rectifier, running down through th etubular anodes to the bottom anode and then back u padj acent to the anodes to the rectifier. Resistanc ecalculations are made as though there were thre ecable s as noted below. The conductor provided wit hsome tubular ceramic anodes is designated EPR /HY5O forone-in. diameter anodes and EPR/HY16 for 0.63-in .dia meter anodes. These cables have an ethylene -propylene rubber inner insulation and achlorosulphonated polyethylene outer jacket and ar esuitabl e for deep anode use if the insulation i sprotected with a chlorine-resistant sheath or shield.

RWT 'LWT RMFT1000 ft

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EPR/HY50 cable has been chosen. The cabl elen gths have been calculated based on th efollowing distances, which are illustrated in thedeep groundbed examples sh own in figures 2-12 and2-13:

From grade to top of backfill column 100 ftFrom top of backfill column to top anode 10 ftFrom grade to bottom of backfill column 205 ftFrom bottom anode to bottom of backfill column 10 ftFrom anode hole to rectifier 2 cables at 10 ft 20 ftFrom top of anode assembly to bottom of anode assembly 85 ft

Total cable length from top anode feed to rectifier 10 ft + 100 ft + 10 ft = 120 ft

Total cable length from bottom feed to rectifier 10 ft + 205 ft - 10 ft = 205 ft

Cal culate top cable resistance (R ) fro mWTequation 1-15:

Where:L = 120 ft (Cable length fro mWT

previous calculation)

R = 0.1 183 ohm (Resistance per 100 0MFTlin ft of EPR/HY5O [table 3-6])

R = 120 ft x 0.1183 ohm = 0.014 ohmWT 1000 ft

Calculate bottom cable resistance (R ) fro mWBequation 1-15:

Where:L = 205 ft (Cable length fro mWB

previous calculation)

RPOS 'LW RMFT

1000 ftx 1

2'

85 x 0. 1183 ohm/ ft1000 ft

x 12

RNEG 'LW RMFT

1000 ft

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R = 0.1183 ohm (Cable resi stance perMFT1000 lin ft [same as above])

R = 205 ft x 0.1183 ohm = 0.024 ohmWB 1000 ft

These two cables are in parallel, so that tota lresistance (R ) is calculated from the law o fT/Bparallel circuits:

1 = 1 + 1 R R RT/B WT WB

1 = 1 + 1 R 0.014 0.024T/B

1 R = 71.43 + 41.66 = 113.1T/B

R = 0.009 ohmT/B

Si nce current dissipates along the portion of th ecable to which the anodes are connected, th eresistance of this cable (R ) is taken as one halfPOSits total resistance, as was done in example 2-2.

R = 0.005 ohmPOS

Negative cable resistance: The rectifier i spl aced 25 ft from the connection to the piping ,usin g No. 4 AWG HMWPE insulated cable. Negativ eresistance is calculated from equation 1-15:

Where: L = W

25 ft (Length of cable)

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R = 0.254 ohm (Resistance per 100 0MFTlin ft of No. 4 AWG HMWP Einsulated cable [table A-6])

R = 25 x 0.254 NEG 1000

R = 0.006 ohmNEG

Total wire resistance therefore is:

R =WR + R + R = 0.0 09 + 0.005 +T/B POS NEG0.006

R =W0.020 ohm

c) Structure-to-earth resistance (R ).C

The piping is essentially bare, so structure-to -earth resistance is negligible (R = 0.0).C

d) Calcul ate total circuit resistance (R ) fro mCequation 1-3:

R = R + R + R (Because the cathodi cT A W Cpro tection system uses a single dee panode groundbed R = R )A N

R = 0.65 + 0.020 + 0.0T

R = 0.670 ohmT

7) Calculate rectifier voltage (V ) from equation 1-17:REC

V = (I) (R ) (120%)REC T

Where: I = 40 amp (Current requirement from step

2, paragraph 2-5b)

R = 0.670 ohm (Total circuit resistanc eTfrom previous calculation)

120% = Rectifier voltage capacity desig nsafety factor

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V = 40 amp x 0.670 ohm x 1.2REC

V = 32.2 VREC

c. Select rectifier.

Based on the design requirement of 32.2 V and 40 amp, arecti fier can be chosen. A 36-V, 50-amp unit i scommercially available and is selected.

d. Installation.

Fig ures 2-12 and 2-13 show how the deep anode groundbe dmight look.

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2-6. Elevated Water Tank (Ice Is Expected).

Impressed current cathodic protection is designed for an elevatedwater ta nk (figure 2-14). The tank is already built and curren trequirement tests have been done. Anodes must not be suspende dfrom the tank roof because hea vy ice (up to 2 ft thick) covers thewater surface during winter. The anode cables could not toleratethis weight, so another type of support must be used. A ninternally supported hoop shaped wire anode system is selected.

a. Design data.

1) Tank is a pedestal supported spheroid with a 10-in .riser pipe. Only the bowl will be protected becaus ethe riser pipe is less than 30 inches in diam eter. Forriser pipes see section 2-7.

2) Tank dimensions are:

Capacity = 400,000 galDiameter of bowl = 51 ft 6 in.High water depth = 35 ftHeight of bowl above ground = 100 ft

3) Water resistivity is 2000 ohm-cm.

4) Design cathodic protection anodes for a 15-year life.

5) Wire type ceramic anode will be used.

6) Wetted surfaces are uncoated.

7) Area above high water level is kept well coated.

8) Tank is subject to freezing.

9) The cathodic protection circuit resistance must no texceed 2 ohms.

10) The available electrical power is 120/240 V A C, singlephase.

11) Based on structure current requirement testin grecently perfor med on this tank, the current requiredfor adequate ca thodic protection is 25 amp. This highcur rent requirement indicates that the tank interna lcoating is severely deteriorated.

b. Computations.

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1) Calculate the length of wire in ft (LB) neede d for thecurrent require d from a modification of equation 1-2:

L = I BI A

where:I = 25 amp (Current requirement fro m

adequate cathodic protection fro mitem 11 of paragraph 2-6a)

I = Allow able amp per ft of anode wir eA(va ries depending on desired anod elife and diameter), found in tabl eA-3.

Select 0.0625-in. diameter copper cored anod ewire based on t he current requirement of 25 amp,81 ft of anode wire will be required to provid ean anode life of 15 years.

Length of Anode RingWire Diameter (D )R

For 0.0625 in. L = 25 = 81 ft 25 ft 9 in.B 0.31

2) Calculate the d esired diameter of the anode wire ring(D ). Experience shows that the diameter of the anodeBwire ring should be between 40 and 70 percent of th ebowl diameter. For this example, use a hoop diameterequal to 40 percent of the tank bowl diameter:

D = 51.5 ft x 40% = 20.5 ftR

3) Select anode wire:

Prior to calculating the circuit resistance of th eanode wire ring, it must first be checked to determineif the length i s adequate for the desired anode life.For an anode ring diameter of 20.5 ft, th ecirc umference is 20.5 x B = 64.4 ft. This length i sinadequate for the 0.0625-in. wire (which requires aminimum of 81 ft to meet the desired anode life) .Therefore, we will increase the hoop diameter by 1 0percent to 50 percent of the tank diameter:

D = 51.5 x 50% = 25.75 ftR

RA '0. 0016 p

DR( 1n

8 DRDA

% 1n2 DR

H)

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But this diameter provides an anode length that i sstill s lightly less than that required for a 15-yea rmode life. Therefore, we will use a hoop diamete rthat is 55 percent of the bowl diameter:

D = 51.5 x 55% = 28.3 ftE

4) Cal culate the anode-to-water resistance (R ) for aA0.0625-in. diameter anode wire using equation 1-14:

Where: p = 2000 ohm-cm (Water resistivity from

item 3 of paragraph 2-6a)

D = 28.3 ft (Anode ring diameter fro mRstep 2 of paragraph 2-6b)

D = 0.0 0521 ft (0.0625 in)(Diameter o fAanode determined in step 3 o fparagraph 2-6b)

H = 21 ft (Anode depth below wate rsurface.) (Anode depth determine dfrom the following calculations).

The ano de depth below the high water line i sapproximately 60 percent of the distance between th ehigh water line and the bottom of the tank. Wate rdepth = 35 ft, cf. design data section 2-6a.

Calculate R :A

R = 0.0016 (2000) [ln 8 x 28.3 + ln 2 x 28.3 ]A28.3 0.00521 21

R = 0.113 [ln 43,455 + ln 2.70]A

R = 0.113 (10.68 + 0.99)A

R = 1.32 ohmsA

This is within the design limitation of 2.0 ohms.

RW 'LW RMFT

1000 ft

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5) Determine the total circuit resistance (R ), fro mTequation 1-3:

R = R + R + RT N W C

Where:R = Anode-to-water resistanceNR = Wire resistanceWR = Tank-to-water resistance.C

a) Anode-to-water resistance (R ) = 1.32 ohms fro mNstep 4 ab ove. (R = R since R is equal to on eN A Acontinuous wire anode).

b) Header cable/wire resistance (R ) is solved usingWequation 1-15:

Where: L = 115 ft (Effective wire length. TheW

positive wires from the rectifie rto each end of the anode circl ewill be approximately 115 ft long)

R = 0.51 ohm (Effectiv e wire resistanceMFTper 1000 lin ft. Since there ar epositive wires from the rectifie rto each end of the anode circle ,each wire will car ry about one halfof the current [12.5 amp]. Th ewires selected are No. 10 AWG .Because the two wires are i nparallel, the effective resistanc eis one half the single wir eresistance [1.02 ohms per 1000 li nft/2 = 0.51 ohm])

R = 115 ft x 0.51 ohm W1000 ft

R = 0.06 ohmW

c) Tank-to-water resistance (R ) and negativ eCcircuit resistance.

The negative wire is connected to the tan kstr ucture near the rectifier, so its resistanc e

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is negligible. The tank-to-water resistance i sals o negligible because the coating is ver ydeteriorated.

d) Calculate R :T

R = 1.32 + 0.06 + 0.00T

R = 1.38 ohmsT

This is well below the design requirement.

6) Calculate rectifier voltage (V ) from equation 1-17:REC

V = (I) (RT) (120%)REC

Where: I = 25 amp (Current requirement fro m

step 11, paragraph 2-6a)

R = 1.38 ohms (Total circui tTresistance from previou scalculation)

120% = Rectifier volta ge capacity designsafety factor

V = 25 amp x 1.38 ohms x 1.2REC

V = 41.4 VREC

c. Select rectifier.

Based on the design requirements of 41.4 V and 25 amp, acommercially available 48-V, 28-amp unit is selected. Toprevent over- or under-protection as the water leve lvaries, automatic potential control is specified. Th eta nk-to-water potential is maintained by the controlle rthrough two permanent copper-copper sulfate referenc eel ectrodes suspended beneath the anode wire circle. Th ereferenc e electrodes should have a life of at least 5years. The tank-to-water potential measured by th econtroller should be free of IR drop error.

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d. Installation.

Figure 2-14 shows a typical installation while figure 2-14 Aprovides a typical detail for a pressure entrance fittingfor underwater power and referenc e cell wire penetrations.

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2-7. Elevated Water Tank (No Icing Will Occur).

This impressed current design is for a tank (figure 2-15) that hasnot been built; thus, it is not possible to measure curren trequirements and other factors. Calculated estimates are used.

a. Design data.

1) Tank will be ellipsoidal on both top and bottom.

2) Tank dimensions will be:

Capacity = 500,000 galDiameter = 56 ftTank height (from ground to bottom of bowl) = 115 ftOverall tank depth = 39 ftVertical shell height = 11 ftHigh water level = 35 ftRiser pipe diameter = 5 ft

3) Water resistivity is 4000 ohm-cm.

4) Design for a 20-year life.

5) The tank water will not be subjected to freezing.

6) Segmented rod anodes will be used; 4-ft long by 0.138-in. diameter. Note: Segmented rods have an advantag ein that they can be field assembled using fac tory-madeconnections. On the other hand, continuous wir e(.0625-in. diameter) is also available in lon g lengths(typically 500 ft). These can be fabricated wit hfactory-made wire connections but their overa ll lengthmust be specifi ed. Continuous wire will almost alwaysbe less expensive.

7) All wetted inner surfaces will be uncoated. Are aabove water will be coated.

8) Electric power available will be 120/240 V AC, singlephase.

9) Desig n for a current requirement of 2.5 A per sq f tfo r the bowl and 8.0 A per sq ft for the riser. Du eto the velocity of the water in the riser, th e riser'scurrent require ment is typically much higher than thebowl.

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b. Computations.

1) Find the area of wetted surface or tank bowl (A) shownin figure 2-16 from equation 2-1:

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AB ' 2 A r h2 % r 2

AB ' 2 3. 1416 28 ( 14)2 % ( 28) 2

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(eq 2-3)

A = A + A + A (eq 2-1)T C B

Where:

A = Wetted area of the top sectionTA = Area of the center sectionCA = Area of the bottom section.B

a) Fi nd the appropriate wetted area of the to psection (A ) using equation 2-2:T

A = 2 B rh (eq 2-2)T

Where:r = 28 ft (Tank radius)h = 10 ft (Water height)

A = 2 x 3.1416 x 28 ft x 10 ftT

A = 1759 sq ft (approximate).

b) Fin d the wetted area of the center section (A )Cusing equation 2-2:

A = 2 B rh (eq 2-2)C

r = 28 ft (Tank radiush = 11 ft (Water height)

A = 2 x 3.1416 x 28 ft x 11 ftC

A = 1935 sq ft

c) Fin d the wetted area of the bottom section (A )Bfrom equation 2-3:

Where:r = 28 ft (Tank radius)h = 14 ft (Water height)

A = 3894 sq ftB

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d) Calculate (A):

A = 1759 sq ft + 1935 sq ft + 3894 sq ftA = 7588 sq ft

2) Find the riser pipe area (A ) using equation 2-2:R

A = 2 B r h (eq 2-2)R R R

Where:r = 2.5 ft (Riser radius)Rh = 115 ft (Height of riser)R

A = 2 x 3.1416 x 2.5 ft x 115 ftR

A = 1806 sq ftR

3) Find the maximum design current f or the tank bowl (I )Tusing equation 1-1:

I = (A) (I) (1.0 - C )T E

Where:A = 7588 sq ft (Total wetted area o f

tankbowl from step 1 of paragrap h2-7b)

I = 2/ 5 mA/sq ft (Required curren tdensity from item 9 of paragrap h2-7a)

C = 0.0 (Coating efficiency, wette dEin ner surfaces will be uncoated ,from item 7 of paragraph 2-7a)

I = 7588 sq ft x 2.5 mA/sq ft x (1.0 -T0.0).

I = 18,970 mA; use 19.0 ampT

4) Find the maximum design current for the rise r(I ),using equation 1-1:)R

I = (A ) (I) (1.0 - C )R R E

Where:A = 1806 sq ft (Total surface area o fR

riser pipe from step 2 of paragraph2-7b)

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I' = 8. 0 mA/sq ft (Required curren tdensity for riser pipe from item 9of paragraph 2-7a)

C = 0.0 (Coating efficiency. Inne rEsurfaces will be uncoated, fro m item7 of paragraph 2-7a)

I = 1806 sq ft x 8.0 mA/sq ft x (1.0 -R0.0).

I = 14,448 mA; use 14.5 amp.R

5) Select the numb er of anodes required for the bowl, tomeet the anode system's 20-year design life usin gequation 1-2:

N = I I A

Where:I = 19 amp (Current requiremen t

determined in step 3)

I = 1 amp/anode (Current rating pe rAanode from table 3-3 for 4-ft lon gby 0.138-in. diameter ceramic rods)

N = 19 1

N = 19 rod segments (4-ft long segment)

6) Select the numb er of anodes required for the riser tomeet the anode system's 20 year design life usin gequation 1-2:

N = I I A

Where:I = 14.5 amp (Current requiremen t

determined in step 4)

I = 1 amp/anode (Current rating pe rAanode from table 3-3 for 4 ft lon gby 0.138-in. diameter ceramic rods)

AR 'D x N

2 ( A % N)

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(eq 2-4)

N = 14.5 1

N = 14.5; use 15 rod segments (4 ft .long)

7) Calcu late the radius of the main anode circle (AR) ,using equation 2-4:

Where:D = 56 ft (Tank diameter, from ite m 2 of

paragraph 2-7a)

N = 10 anode strings (assumed: it i snecessary to assume a number o fanode strings since there are tw ounknowns in this equation.

A = 56 ft x 10 R2 (3.1416 + 10)

A = 560 ft R26.28

A = 21.3 ft; use 22 ft for the mai nRanode circle radius

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CC '2 A AR

N

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(eq 2-5)

8) Determine the center-to-center spacing (Cc) for th emain anode strings.

a) To find circumference spacing (C ), use equationC2-5:

Where:A = 22 ft (Radius of main anode circl eR

from previous calculation.

N = 10 anode strings ([Assumed] Used inprevious calculation)

C = 2 x 3.1416 x 22 ft C10

C = 13.8 ft; use 14 ft for center-to -Ccenter spacing between anod estrings.

b) The c ord spacing is approximately the same a scircumferential spacing; 14 ft will be use d(figure 2-19).

9) Select main anode system

Number of 4-ft long rod segments needed for th ecur rent requirement (from step 5) = 19. Number o fanode strings u tilized (from step 7) = 10. Using tworod segments per string will provide twenty anodes ,which is sufficient for the current requirement .However, for rod type anode designs, the main anod erods should extend from a distance of 4 ft above th etan k bottom to within 4 ft of the HWL. Due to th ecurvature of the tank bottom, the total water depth atthe l ocation of the main anodes is approximately 3 0ft. Therefore, the minimum rod length should be 22 ft(30 ft - 4 ft - 4 ft = 22 ft). Since these rods comein 4 ft lengths, we will use six segments per anod ewith a total length of 24 ft.

10) Calcula te the resistance of the main anodes to wate r(R ) using equation 1-13:N

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RN '0. 0052 p 1n [ D/ ( 2 AR x DE) ]

LB

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Where:p = 4000 ohm-cm (Water resistivity from

item 3 of paragraph 2-7a)

D = 56 ft (Tank diameter from item 2 ofparagraph 2-7a)

A = 22 ft (Radius of main anode circl eRfrom step 7 of paragraph 2-7b)

D = 0.275 (Equivalent diameter facto rEfrom figure 2-17)

L = 24 ft (Length of each anode fro mBstep 9 of paragraph 2-7b)

R = 0.0052 x 4000 x ln(56/(44 x 0.275))N 24

R = 1.33 ohms.N

If th e anode rod length-to-diameter ratio (L/d) i sless than 100, the anode-to-water resistance needs tobe adjusted by the fringe factor. (See step 12 for adiscussion of fringe factor.)

In this c ase, L = 24 ft and d = 0.0115 ft. L/d ,there fore, is (24/0.0115) = 2086. No fringe facto rcorrection is required.

11) Calculate the stub anode requirement (N ):S

a) The mai n anode radius has been calculated to b e22 ft. The main anodes are spaced to provid eapproximately the same distance from the side sand the bottom of the tank. The main anode willprote ct a length inward along the tank botto mequal to approx imately the same spacing that theanode is spaced away from the tank wall.

b) The anode suspension arrangement for the tan kbeing considered is shown in figure 2-20. It can

ETL 1110-9-10(FR)5 Jan 91

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be se en that stub anodes are required for thi sdesig n. For a two-ring anode design (main an done-stub anode ring), which is usually suffic ientfor tan ks up to 1 million gal storage capacity ,the 4-ft long stub anodes are located on a ra diusone-f ourth of the bowl radius, or 7 ft (28 ft x0.25 = 7 ft). T ypically, there will be about onehal f as many stub anodes (two ring design) a sthere are main anodes so we will plan for fiv e 4-ft lo ng stub anodes on a 7-ft radius. Th eoutside radius of the area to be protected by thestub anodes is approximately 13 ft and the in sideradius is 2.5 f t (riser radius). The stub anodesare thus locate d on an 7-ft radius to place themin the center of the area to be protected. (N ote:For smaller diameter tanks, stub anodes may no tbe required.)

c) Find the current division between main and stu banodes.

(1) The area of tank bottom protected by stu banodes (As) is found by equation 2-6 (se efigure 2-20):

A = B