corrosion & coatings

8: Corrosion/Coatings

Hand-held computer determines concrete coating thickness 204National Association of Pipe Coating Applications (NAPCA)

specifications 208How much primer for a mile of pipe? 209How much coal-tar enamel for a mile of pipe? 210How much wrapping for a mile of pipe? 210Estimating coating and wrapping materials required

per mile of pipe 210Coefficient of friction for pipe coating materials 211Troubleshooting cathodic protection systems:

Magnesium anode system 213Cathodic protection for pipelines 214Estimate the pounds of sacrificial anode material

required for offshore pipelines 222Comparison of other reference electrode potentials

with that of copper-copper sulfate referenceelectrode at 25° C 224

Chart aids in calculating ground bed resistanceand rectifier power cost 225

How can output of magnesium anodes be predicted? 226How to determine the efficiency of a cathodic

protection rectifier 226How to calculate the voltage drop in ground

bed cable quickly 227What is the most economical size for a rectifier cable? 227How to estimate the number of magnesium

anodes required and their spacing for a bareline or for a corrosion "hot spot" 228

How can resistivity of fresh water bedetermined from chemical analysis? 228

What will be the resistance to earth of asingle graphite anode? 229

How to estimate the monthly power bill for acathodic protection rectifier 229

What will be the resistance to earth of a group ofgraphite anodes, in terms of the resistanceof a single anode? 229

How can the current output of magnesium rodused for the cathodic protection of heat exchangershells be predicted? 229

What spacing for test leads to measure currenton a pipeline? 229

How many magnesium anodes are needed forsupplementary protection to a short-circuited bare casing? 230

Group installation of sacrificial anodes 230How can the life of magnesium anodes be predicted? 231How to find the voltage rating of a rectifier if

it is to deliver a given amount of current througha given ground bed (graphite or carbon) 231

Determining current requirements for coated lines 231Determining current requirements for coated

lines when pipe-to-soil potential values are estimated 231HVDC effects on pipelines 232Troubleshooting cathodic protection systems:

Rectifier-ground bed , 236How to control corrosion at compressor stations 237Project leak growth 238

Advances in Pipeline Protection 239Methods of locating coating defects 240Case histories 243Estimate the number of squares of tape for

pipe coating (machine applied) 244Estimate the amount of primer required for tape 245Tape requirements for fittings 245Induced AC Voltages on Pipelines

May Present a Serious Hazard 246Measuring Unwanted Alternating

Current in Pipe 248Minimizing shock hazards on pipelines

near HVAC lines 253Cathodic protection test point installations 254

Hand-held computer determines concrete coating thickness

Technical data for the engineer's file

Frank E. Hangs, Sovereign Engineering, Inc., Houston

Pipeline crossings, under water or in unstable soil, offermany challenges to engineers.

Pipe is buoyant, and an empty line may float in water. Wetsilts are like viscous fluids causing inadequately weightedpipelines to pop up. There are various government entitiesthat have jurisdiction over navigable rivers, bays, marshlands,and offshore waters. These agencies may stipulate thatpipelines be buried at certain depths and be stabilized.

A good way to stabilize a pipeline is to use an adequateconcrete weight coating. Determining the thickness of theconcrete involves a process of balancing upward forces suchas buoyancy of the mud and the downward forces—weightsof pipe, protective coating, and concrete, allowing a factor of60 (negative buoyancy). Such computations with severalvariables can become involved and tedious.

Computer program

The following program, written for the Hewlett Packard41CV (Figure 1), calculates the thickness of concrete weightcoating for submarine pipelines expeditiously and withsatisfying results.

The prompting feature (Figure 2) is employed to aidusers' data input. Important calculated values and a recap ofthe inputs are printed out, and each quantity is identified.Thus, the tape is a complete record. The program is flexiblein that any inputs can be readily changed for a second run.Thus, many "what i f questions can be answered quickly.Suppose a heavier-weight pipe is used? What if theconcrete density is changed? If the negative buoyancy isreduced? Etc.

8ULBL -COHC"82 ODV63 SF 12

94 " CONCRETE1

05 -h COPTIHG"96 PRfl97 PDV88 CF 12

89 "T TRIPL=?"18 PROMPT11 STO 95

12 'PIPE DIPH=?"13 PROHPT14 STO 15

15 "PIPE WT=?"16 PROHPT17 STO 16

13 "PROT CT TH=?"19 PROHPT20 STO 17

21 -DEN CONC=?"22 PROHPT23 STO 19

24 "DEN MUD=?"25 PROMPT26 STO 28

27 "DEN PROT CT=?"28 PROHPT29 STO 18

38 -T INCR=?"31 PROMPT32 STO 21

33 -NEG BUOY=?"34 PROHPT35 STO 1136*LBL 81

37 RCL 1538 RCL 1739 +40 LPSTX41 *42 .021343 •44 RCL 1345 •46 STO 8647 RCL 1543 RCL 1749 258 •51 +52 RCL 8553 +54 LPSTX55 *56 .821357 •58 RCL 1959 •68 STO 8761 RCL 1662 +63 RCL 8664 +65 STO 8866 RCL 1567 RCL 1768 269 •70 +71 RCL 8572 2

73 *74 +75 Xt276 .885577 •78 STO 1279 RCL 2888 •81 STO 8982 RCL 0883 XOY84 -85 STO 1086 FS? 8187 GTO 0388 RCL 1189 X=8?98 GTO 8491 RCL 1892 RCL Ii93 X<=Y?94 GTO 8395 GTO 82

96*LBL 8497 RCL 0998 RCL 8899 X)Y?188 GTO 83181 GTO 82

182*LBL 85183 RCL 85184 -T=? "185 ftRCL 85186 PROMPT187 CLP

188 STO 85189 RCL 15118 "D=? "111 PRCL 15112 PROMPT113 CLP114 STO 15115 RCL 16116 "HP=? "117 PRCL 16118 PROHPT119 CLP128 STO 16121 RCL 17

122 -TH PC=? -123 PRCL 17124 PROHPT125 CLR126 STO 17127 RCL 19128 "DC=? "129 PRCL 19138 PROHPT131 CLR132 STO 19133 RCL 28134 "DH=? •135 PRCL 28136 PROHPT137 CLR138 STO 28139 RCL 18148 "DPC=? "141 PRCL 18142 PROMPT143 CLP

144 STO 18145 RCL 21

146 -T INCR=? •147 RRCL 21148 PROMPT149 CLR158 STO 21151 RCL 11

152 "NEG BUOY=? -153 PRCL 11154 PROMPT155 CLP156 STO 11157 GTO 01

158*LBL 02159 RCL 21168 ST+ 85161 GTO 81

162*LBL 83163 RCL 88164 RCL 12165 /166 STO 13167 62.4168 /169 STO 14

178*LBL -DPTP-171 PDV172 CLR

173 "RESULTS-"174 -I-RECPP.-175 RVIEH176 PDV

177 -T=-178 PRCL 85179 PVIEH188 CLR181 "HPC=-182 RRCL 86183 RVIEU184 CLR185 "HC=-186 RRCL 87187 RVIEH188 CLP189 "HT=-198 RRCL 88191 RVIEU192 CLR193 "HHD="194 RRCL 89195 RVIEU196 CLR

197 -HT-HHD="198 RRCL 18199 RVIEU288 CLR

201 -NEG BUOY=-282 RRCL 11283 RVIEH284 CLR285 "VT=-286 RRCL 12207 RVIEU288 CLP

289 -BULK DEN="218 RRCL 13211 RVIEU212 CLR

213 -SP GR=-214 RRCL 14215 RVIEU216 CLP217 PDV218 -D=-219 RRCL 15228 PVIEU221 CLP222 -HP=-223 RRCL 16224 PVIEH225 CLR226 "TH PC="227 PRCL 17228 RVIEH229 CLP238 -DC="

231 PRCL 19232 PVIEH233 CLP234 -DH="235 RRCL 28236 PVIEU237 CLP238 "DPC="239 RRCL 18248 PVIEU241 CLP242 -T INCR=-243 RRCL 21244 PVIEU245 CLP246 BEEP247 END

Figure 1. Hewlett Packard-41CV computer program.

Nomenclature and storage registers inputs

Reg. no.05 T = Concrete thickness, in. Assume trial

value. Final answer is cumulated inReg. 0.5.

15 D = Pipe diameter, in.

16 WP = Weight of pipe, lb/lf.

17 TH PC = Thickness of protective coating.Usually 3/32 in. or 0.0938 in.

19 DC = Density of concrete, lb/cu ft. Usually140, 165, 190.

20 DM = Density of mud or fluid,lb/cu ft. River mud is around 90.

18 DPC = Density of protectivecoating, lb/cu ft. 95 is average.

21 T INCR = Increment of T. Decimal of in. 0.125is suggested for numerical precision—not that a tolerance of this ordercan be constantly achieved in thepipe coating world. Increment can bemade smaller, if desired.

11 NEG BUOY = Negative buoyancy. Desired minimumvalue, lb/lf. Can be 0 in specialcircumstances—e.g., to ascertain if agiven pipe will float in water, or will acertain coated pipe sink in mud?

Each value keyed in is printed on the tape as a check ofinputs and the program prompting feature queries user forvalue of next named input (Figure 2).

Calculated values

Reg. no.06 WPC = Weight of protective coating,

lb/lf.

07 WC = Weight of concrete coating, lb/lf.

08 WT = Total weight—pipe and coatings, lb/lf.

09 WMD = Weight of mud (fluid) displaced, lb/lf.

10 WT-WMD = Actual negative buoyancy, lb/lf.

12 VT = Total volume displaced by pipe and

coatings, cu ft/If.

13 BULK DEN = Bulk density, wt/vt.lb/cu ft.

14 SP GR = Specific gravity, bulk den/den water.

RESULTS-RECAP.T=4.625eHPC=4.6885HC=479.3309WT=578.6385MHD=553.4462HT-HHD=25.1843HEG BUOY=20.89eeVT=6.1494BULK DEN=94.8954SP GR=1.59795=24.8869HP=94.6288TH PC=8.8938DC=165.8808DH=98.0888DPC=95.6888T IHCR=8.1258

Figure 2. Most common questions asked when checkingcoating buoyancy.

Formulas

WPC = (D + (THPC))(THPC)0.0218(DPC)

WC = (D + 2(THPC) + T)T 0.0218(DC)

VT = (D + 2(THPC) + 2T)20.0055

WMD = (VT)(DM)

Note: The user is encouraged to employ the most reliableand suitable data available, such as the density of protectivecoating, mud, and concrete.

Examples

Put program into calculator, XEQ Size 30. The printedRESULTS-RECAP portions of each of three examples aregiven. When the user XEQ "CONC" and inputs the datafrom the lower part of the tabulation, he or she will get thesame calculated values above.

Figure 2 includes the most common questions askedwhen checking coating buoyancy. Always be sure flag 01 isclear XEQ "CONC" and put in new trial T and R/S. Also R/Safter all other inputs. After all prompted data are keyedin, the processing starts and results are printed out. NoticeT = 4.6250 in. has an actual negative buoyancy of 25.1843. IfT = 4.5 in., the actual negative buoyancy= 18.4510, which isless than 20. A precise solution is between 4.5 in. and 4.625 in.

If the user wants to find T, which produces an actualnegative buoyancy nearer 20: STO 4.5 in RO 5, STO \§ in.(0.0625 in.), in R21, XEQ 01. This produces T = 4.5625where actual negative buoyancy is 1.8113 greater than 20.This refinement can be found in choosing small T INCR.Note that the weight of the concrete coating is now471.8269 lb/lf and we have approached closer to 20.

Calculated values. Weight of protective coating andconcrete, total weight and volume of pipe, and coatings perlinear foot will be of value in transporting coated pipe, etc.

Note: Results—RECAP of any run (which is in storage)can be reprinted as often as desired by XEQ "DATA."

If NEG BUOY is 0 and flag 01 is not set, then the programcauses WT to approach WMD and calculates for thissituation.

If a printer is not available, results can be obtained asfollows:

1. XEQ 05, put in data and R/S after each item.2. When beep sounds, RCL registers one by one in order

given above (so as to identify values).

Figure 3 illustrates the changing of two input values:namely the pipe diameter and weight. The XEQ 05 routinecalls up the contents of the input registers one at a time andquestions them. Always put in a new T value. Then R/S.

When D = ? comes up, key in new value and R/S. Samewith pipe WT = ?. Assume other data remain unchanged,R/S each time. (Any one value or any combination of valuesmay be changed before rerunning the program.)

Figure 4 determines if a given pipe, 14 in. at 36.71 lb/lfand 1A in. of protective weight coating density of 135 lb/cu ft,will float. T, DC, T INCR, and NEG BUOY are all zero. DMbecomes the density of water 62.4. Set flag 01 and XEQ 05.Key in data and R/S. When all data are in place only onecomputation is allowed, then printed out. WT-WMD tells usthe upward forces exceed the downward forces by 19.1733lb/lf; also the specific gravity is 0.7517, so the pipe floats.Also note that BULK DEN is less than the density of water.

RESULTS-RECAP

T=3.2598HPC=3.9834MO273.9914HT=375.7248HHD=352.5529

HT-MHB=23.1728NEG BUOY=28.8880VT=3.9173BULK DEN=95.9154SP GR=1.5371

11=28.8888«P=97.8380TH PC=6.8938DC=165.8900BH=90.8«90DPC=95.9098T IHCR=B.1250

Figure 3. Changing of two input valves.

XEQ "DflTfi"

RESULTS-RECAP.

T-e.eeeeUPC-21.3368№=8.0000HT=58.8463HHD=77.2280HT-WHB=-19.1733NEG BUOY=0.0080VT=I.2375

BULK BEN=46.9065SP GR=0.7517

6=14.0800

HP=36.7180TH PC=8.5000BPC=135.0880BC=8.8000BH=62.4000T INCR=0.0000

Figure 4. Result of calculations to determine whether the pipewill float.

National Association of Pipe Coating Applications (NAPCA) specifications*

RECOMMENDED SPECIFICATION DESIGNATIONS FOR ENAMEL COATINGSThese are recommended enamel coating system designations forcoal tar or asphalt coatings, except for specialty requirements.SPECIFICATION

SYMBOL

•P

AR

T

*1

•• P

AR

T #

2

SH

OT

BL

AS

T

PR

IME

R

93.7

5 M

ILS

MIN

.E

NA

ME

L

FIB

ER

GL

AS

SW

RA

P

#15

FE

LT

WR

AP

SE

AL

CO

AT

GL

AS

S M

AT

PO

LY

ET

HY

LE

NE

WIT

H K

RA

FT

HE

AV

Y K

RA

FT

PA

PE

R

EL

EC

TR

ICA

LIN

SP

EC

TIO

N

*NOTE: These are not all the specifications approved by NAPCA. For other specifications, consult an NAPCA member.

AGSF-TGM-AGM-TGMF-AGMF-TGMP-AGMP-

#PART#1 SYMBOLS-ENAMEL & WRAPPERS

T-Coal Tar G-FiberglassA-Asphalt S-Seal CoatF-#15 Felt GM-Glass MatP—Polyethylene wrapper

w/kraft paperO—Outer wrap

To specify Outer Wrap, useletter O in place of F.

• # PART #2 SYMBOLSGRADESOF ENAMEL

3— Fully-Plasticized Coal Tar(ordered by penetrationdictated by ambient temperature)

4— High Temperature Surface Coa! Tar7 - Asphalt

EXAMPLE: 93.75 mils (minimum) fully-plasticized coal tar enamelwrapped with fiberglass, felt and kraft — the symbol would be TGF-3.

Reprinted courtesy of National Association of Pipe Coating Applicators.

STANDARD APPLIED PIPE COATING WEIGHTSFOR NAPCA COATING SPECIFICATIONS

Refer to NAPCA 1-65-83

Weight in Pounds-Per 100 Lineal Feet

Spec.Symbol

PIPESIZE

(continued on next page)

(table continued)

Spec.Symbol

PIPESIZE Weight in Pounds - Per 100 Lineal Feet

Al! weights provide for standard cutbacks.

Weights per 100 lineal feet are figured on the following material weights per square foot of pipe surface:

WeightLbs.-PerSq. Ft.

Coal Tar Primer & Enamel—1/32" Min. .2461Coal Tar Primer & Enamel-2/32" Min. .4850Coal Tar Primer & Enamel—3/32" Min. .7238Asphalt Primer & Enamel—1/32" Min. .2000Asphalt Primer & Enamel—2/32" Min. .3950Asphalt Primer & Enamel—3/32" Min. .5912

WeightLbs.-ferSq. Ft.

#15 Pipeline Felt(Tar or asphalt saturated) 1271

Outerwrap 1225.010 Fiberglass Wrap 0110Kraft Paper 0200Glass Mat 0114Polyethylene 0450


NAPCA SPECIFICATIONS PIPELINE FELTS

PROPERTY

Saturated FaIt - Not Parforatad

Weight (MIn.) lbs. per 100 s.f.(ASTM-D-146-59 Sec. 1-9)

Saturation (Percent) M IN .(ASTM-D-146-59 Sec. 18 MAX.by extraction)

Ash Content (MIN).) (Percent)(ASTM-D-146-59 Sec. 21)

Callper (Min.) (inches)(ASTM-D-645-64T Method C)

Tensile Strength (Min.)(lbs. per Inch of width)JASTM-D-146-59 Sec. 12a)With fiber grainAcross fiber grain

Tear Strength (Min.) (Grams)(ASTM-D-689-62)

Across fiber grainWith fiber grain

#15TARN.R.

12

1828

70

.021

3512

400240

#15 TARR.F.

12

1828

70

.018

37No. spec.

Notapplicable

#•15 ASPH.N.R.

12

2240

70

.021

3512

400240

#15 ASPH.R.F.

12

2240

70

.018

37No. spec.

Notapplicable

(continued on next page)

Loss on heating (Max.) (Percent)(AWWA C203-66 Sec. 2.6.6)Pliability(AWWA C203-66 Sec. 2.6.4)

Saturated and Perforated Felt

(1/16" perforations on 1"Staggered Centers)

Tensile Strength (MIn.)(lbs. per inch of width)(ASTM-D-146-59 Sec. 12a)With fiber grainAcross fiber grain

Tear Strength (MIN.) (Grams)(ASTM-D-689-62)Across fiber grainWith fiber grain

Umaturated Felt

Asbestos Content (%) (MIN.)(ASTM-D-1918-67)

10

No cracking

3010

348190

85

10

No cracking

35No spec.

Notapplicable

85

10

No cracking

3010

348190

85

10

No cracking

35No. spec.

Notapplicable

85

N.R. —NON REINFORCEDR.F. — REINFORCED LENGTHWISE WITH PARALLEL FIBER GLASS YARNS ON NOMINAL ¥•" CENTERS.


Minimum Test Voltages for Various Coating Thicknesses

32nd InchCoating Thicknesses

Mils Test Volts


How much primer for a mile of pipe?

Multiply pipe O.D. by 2% to get gallons of primer permile.

Examples.Pipe 24 in. in diameter will require 24 x 2% or 66 gallons

of primer per mile;Pipe 16 in. in diameter will require 16 x 2% or 44 gallons

of primer per mile; and

Pipe 41Z2 in. in diameter will take 41Z2 x 2% or 123/8 gallonsof primer per mile (12 gallons in round figures).

The given multiplier is applicable only where the pipe isnew and in good condition. For rough pipe the multiplyingfactor should be upped from 2% to 3.45. In other words,rough 20-in. pipe will require 20 x 3.45 or 69 gallons ofprimer per mile.

It is hard to estimate the amount of spillage and waste.Temperature and relative humidity affect the way the primergoes on the pipe.

How much coal-tar enamel for a mile of pipe?

For an average cover of %2 in., multiply outsidediameter of pipe by 0.6. The answer is in tons of enamelper mile. (Add 15% to take care of gate valves, drips, bends,etc.)

Examples. How many tons of coal-tar enamel will berequired to apply a %2-in. coat to a 26-in. pipeline?

26 x 0.6 = 15.6 tons per mile

Add 15% to take care of gate valves, bends etc.

15.6 + 2.3 = 17.9 or 18 tons per mile

How many tons of coal-tar enamel will be required toapply a %2~in- coat to a 8.625-in. pipeline?

How much wrapping for a mile of pipe?

For 14-in. to 48-in. pipe multiply the diameter in inchesby 15 to get the number of squares (100 ft ). For pipe sizesof 12% in. and smaller, add 1 to the nominal size of pipe andmultiply by 15.

Examples. How many squares of asbestos felt would berequired to wrap 1 mile of 30-in. pipeline?

30 x 15 = 450 squares of felt

How many squares of felt would be required to wrap 1mile of 4-in. pipe?

8.625 x 0.6 =5.2 tons per mile5.2x15% =0.8

= 6 tons per mile

These factors strongly affect the amount of coating usedper mile on pipelines: 1. Temperature of the pipe andambient temperatures. (It is assumed that the temperatureof the enamel can be controlled.) 2. The kind of dopemachine. 3. Humidity. 4. The number of bends per mile. 5.The number of valves, crossovers, etc. 6. Condition of thepipe.

This latter point has a great effect on the amount ofenamel required for a mile of pipeline. In fact, wherereconditioned pipe is being coated, it is recommended thatthe amount of enamel be increased by about 20%.

(4 + 1) x 15 = 75 squares of felt

These figures are conservative and do not allow forwrapping drips, for patching holidays, and for short ends thatare usually discarded. They allow for machine wrappingusing /^-in. lap on diameters up to 10/4 m- a n d using 1-in. lapon larger diameters.

For smaller diameters of pipe, this figure should beincreased about 5% to take care of additional wastage andpatching.

Estimating coating and wrapping materials required per mile of pipe

Primer and coal-tar enamel

Gallons of primer and tons of coal-tar enamel are based oncovering new pipe in good condition. For rough pipe,increase primer quantities by 25% and the coal-tar quantitiesby 20%. No allowance is included for spillage or waste.

Wrappings

Squares are based on machine wrapping, allowing /4-in.laps on pipe sizes through 10 in. and 1-in. laps on pipe sizesgreater than 10 in. No allowance is included for wrappingdrips, patching holidays, or short ends, which are usuallydiscarded.

NominalPipeSize

PrimerGallons

Coal-tarEnamelTons

GlassWrap

Squares

KraftPaper

Squares

15-PoundFelt

Squares

Coefficient of friction for pipe coating materials

Tests indicate that previous values appear to be valid for thinfilm epoxies and conservative for coal tars

J. B. Ligon, Assistant Professor, Mechanical Engineering-Engineering Mechanics Dept., Michigan TechnologicalUniversity, Houghton, Mich., and G. R. Mayer, Project Engineering Manager, Bechtel, Inc., San Francisco

A major factor in the stress analysis of buried pipelines isthe movement that pipe undergoes in the presence of tem-perature and pressure differentials during its life. This move-ment is highly dependent upon friction resistance of the soil.

Although ample information is available on the staticcoefficient of friction for many materials, there is a lack ofdata on friction between soils and various coatings used inthe pipeline industry. In the past, friction coefficientinformation was extrapolated from data in the literaturethat were believed to have a similarity to the external pipecoating to soil interface.

However, with the development of thinfilm epoxy resincoating systems and the increasing use of these systems inthe pipeline industry, a change from a conventional coal tarfelt coating to a thinfilm epoxy coating would indicate asignificant change in the friction coefficient design criteriadue to the extreme contrast in the surface texture of the twomaterials.

To evaluate the effect of the difference in surface textureon a pipeline system, test procedures were developed todetermine the coefficient of friction for both coal tar felt andthinfilm epoxy to various soils and to obtain more reliableinformation for future pipeline designs using these coatingmaterials.

Static friction tests were conducted to find the coefficientof friction between coal tar felt and thinfilm epoxy pipecoating and eight representative backfill soil samples fromtypical locations along a pipeline right of way. The resultsindicate that the friction coefficients are significantly largerthan those previously extrapolated from literature and thatcoal tar has a higher friction resistance in respect toanchorage of a pipeline.

For the coal tar felt coating, the coefficient varies from0.59 to 0.91 depending on the soil and moisture content. Thethinfilm epoxy coating varies from 0.51 to 0.71 under thesame conditions.

Definitions

Static coefficient of friction. The theoretical long-itudinal soil force acting on the pipe surface can becalculated from the relation:

where: F = Longitudinal soil friction force (Ib)/ji = Coefficient of friction (dimensionless)p = Normal soil pressure acting on the pipe

surface (psi)dA = Soil to pipe differential contact area (in.2)

fA pdA = Total normal soil force on pipe surface (Ib)

The above relation is independent of the pressure per unitarea of the mating surfaces as long as very high contactpressures are not encountered, which is the case for a buriedpipeline.

In terms of a buried pipeline supporting a soil burden upto three pipe diameters of depth, the soil force relationbecomes:

where: Wp = Weight of pipe and contents (lb/in.)D = Pipe diameter (in.)H = Depth of the pipe centerline (in.)Y = Specific weight of the soil (lb/in.3)

The sensitivity of the soil friction forces to the coefficientof friction is readily apparent from the above relation. Sincethis force is inversely proportional to the active length that apipeline moves as a result of temperature and pressureexpansion, the coefficient of friction becomes a major factorin pipeline stress design.

Test system. The theoretical soil friction force acting onthe surface of a coated plate can be calculated from therelation:

F = ^N

where: N = Normal force acting on the plate surfaces (Ib)W = Total weight of plate and external load (Ib)

COATED PLATE

Since the coefficient of friction parameter is the same inall of the previous expressions, the plate test system can beused to simulate the soil-to-pipeline interface.

Coal tar felt. The coal tar coating, an enamel and feltcombination, was prepared to simulate an actual coatedpipeline.

The surface of a steel plate was coal tar coated and thenwrapped in a 15-lb felt. Therefore, the soil surface contactwas predominantly with the felt material.

Thinfilm epoxy. The epoxy coating simulated a thinfilmepoxy resin pipeline coating applied by the fusion method.

Plates used in the tests were coated by the fluidized bedmethod. However, the surface texture was the same as thatexpected by coating plant production equipment.

Results

Coal tar to soils. Figure 1 illustrates that the frictioncoefficient varies from 0.59 to 0.91 for the various soils testedat field moisture content.

Figure 1. Coefficient of friction for coal tar coatingto soil.

Figure 2. Coefficient of friction for thin film epoxyto soil.

UMlFiED SGlL CLASSIFICATION SYMBOL

LOCATION SOIL DESCRIPTION MOISTURE

SANOY SIiT (ML)SiLTY SAND {SM>SItT (MtISiLTY SAND (SM)SIlTY SANO (SM)SAND - SILT (SM)SANDY SILT (ML)SlLTY • SAND (SM)

LOCATION SOiI OESCWPTiON MOISTURE

SANDY SiIT (ML)1

SiLTY SANO (SM)SfLT |ML)SIlTY SAND (SM)SlLTY SANO CSM)SAND-SO (SM)SANDY SILT (ML)SILTY SAND (SM)SRAVa-SAKOSILT (GM)

UNFIEO SOIL CtASSJFiCATiON SYMBOL

The mechanism by which the friction coefficient changesamong the various soils is not apparent from the results ofthese tests, since the soil samples used did not varyconsiderably in description. However, the range overwhich the factor changes is considerably greater than thatpreviously extrapolated from the literature, as follows:

Since such information on this coating was not available inthe literature, no comparison to similar tests could be made.

Temperatures in the 1200F range should have little or noeffect on the coefficient of friction of epoxy to soil.

Conclusions

Although it is virtually impossible to precisely simulate thesurface contact situation of a pipeline in a back-filled trench,the test procedure and apparatus reported here are a meansof approximating it. The results indicate that coal tar coatingshave a higher friction resistance than epoxy coatings as far asanchoring capabilities of the soil are concerned.

The selection of a coating, based on its soil frictionresistance, could be of economic value in reducing thosesituations where extreme expansions call for reinforcedfittings or elaborate culverts to overcome excessive stresslevels in a pipeline system.

The test results show that previous values commonly usedfor the coefficient of friction were conservative for similarsoils, and it is suggested that some conservatism still beincorporated in future analyses.

Since the tests indicated that the friction coefficient of theepoxies was similar to those previously used for coal tarcoatings, their continued use for epoxy-coated pipelinesshould be valid. A more conservative approach to thesevalues is recommended in soils where the presence ofexcessive moisture could change the friction factor substan-tially when interfaced with a smooth epoxy coating.

Troubleshooting cathodic protection systems: Magnesium anode system

Soil description

Silt

Sand

Gravel

Coefficient of frictioncommonly used

0.3

0.4

0.5

The tests also indicate that the moisture content alters thefriction factor to some extent, as would be expected.

In order to investigate the influence of temperature, testswere conducted with coal tar felt wrapping heated to 1200F.Only a slight softening beneath the coating surface wasobserved, and it is believed that temperatures up to thisrange will not significantly affect the magnitude of thefriction coefficient.

Thinfilm epoxy to soils. As expected, results shown inFigure 2 indicate that the friction factor range of 0.51 to 0.71for the epoxy is somewhat lower than that of the coal tar.

Again, the mechanism is not clear, but the results are fairlyconsistent with the friction coefficient increasing in mostinstances in the same order of magnitude as for coal tar.

The basic technique is the same as that outlined earlier;more measurements are required because of the multiplicityof drain points. First, the current output of stations nearestthe point of low potential should be checked; if these aresatisfactory, a similar check should be extended in bothdirections until it is clear that the trouble must be on theline. When a given anode group shows a marked drop incurrent output, the cause may be drying out, shrinkage ofbackfill, or severed or broken lead wires. If the current iszero, the pipe-to-soil potential of the lead wire will showwhether it is still connected to the pipe or the anode, andthus indicate the direction to the failure. If the current islow, there may be a loss of one or more anodes by asevered wire; a pipe-to-soil potential survey over the anodeswill show which are active, just as in the case of rectifieranodes.

Trouble indicated on the line, rather than at the anodestations, is tracked down in the same manner as that used for

<• Broken Wire

Figure 1. Locating idle anodes by surface potentials. The solidline shows the potentials found along a line of anodes when allare delivering current; the dotted line exhibits the change whenthere is a break in the anode lead at the point indicated. Singledisconnected anodes may also be located by this method. Adriven ground rod, a pipe lead, or even a rectifier terminal maybe used for the reference ground; all readings should bereferred to the same reference.

line protected by rectifiers—first, investigating locationswhere something that has been done might be responsible;

Cathodic protection for pipelines

Estimate the rectifier size required for an infinite line.Refer to Figure 1. If coating conductance tests have not beenperformed, pick a value from Table 1.

Table 1Typical Values of Coating Conductance

Micromhos/sq ft Coating Condition

1-10 Excellent coating—high resistivity soil

10-50 Good coating—high resistivity soil

50-100 Excellent coating—low resistivity soil

100-250 Good coating—low resistivity soil

250-500 Average coating—low resistivity soil

500-1,000 Poor coating—low resistivity soil

next, checking potentials; and finally, making a line currentsurvey.

Use a value of—0.3 volts for AEx. This is usually enoughto raise the potential of coated steel to about—0.85 volts.

Use a value of 1.5 volts for AE at the drain point. Highervalues may be used in some circumstances; however, theremay be a risk of some coating disbondment at highervoltages.

Calculate I at the drain point.

AIA = amp./in. x AEx /0.3 x D

where D = pipe OD, inches.

Example. 30-in. OD line with coating conductivity= 100micromhos/sq ft. What is the current change at the drain

Based on 0.25-inch wall pipe and steel resistivity of 18.8 microhm-cm

_* x

— ,

am

pere

s pe

r in

chD

A

Ex

Thousands of Feet, x

Figure 1. Drainage current versus distance for a coated infinite line.

Based on 0.25-inch wall pipe and steel resistivity of 18.8 microhm-cm

IA 0

3

. u

— x

— ,

am

pere

s pe

r in

chD

A

EJ

Thousands of Feet, L

Figure 2. Drainage current versus distance for a coated finite line.

point, and how far will this current protect an infinitely longpipeline?

AEA = 1.5

AEx = 0.3

AEA/AEx = 5.0

Refer to Figure 1 and read

L = 30,000 ft

AIA/D x 0.3/AEx = 0.7 (from Figure 1)

AIA = 0.7x 30 = 21 amp.

21 amps will protect the line for 30,000 ft in either directionfrom the drain point.

Estimate rectifier size for a finite line. Refer to Figure 2.A 31Z2 -in. OD line 20,000 ft long is protected by insulating

flanges at both ends and has a coating conductivity of 500

micromhos/sq ft. What is AEA/ET and what AIA is required?Assume that

E x = 0.3

From Figure 2 read

AEA/EX = 5.8

AEA = 1.74

AIA/D x 0.3/Ex= 1.8 amp./in.

AIA =1.8 x 3.5 = 6.3 amp.

Estimate ground bed resistance for a rectifierinstallation.

Example. A ground bed for a rectifier is to be installedin 1,000 ohm-cm soil. Seven 3-in. x 60-in. vertical graphiteanodes with backfill and a spacing of loft will be used. Whatis the resistance of the ground bed?

Refer to Figure 3. The resistance is 0.56 ohms.

Ano

de B

ed R

esis

tanc

e, R

A (o

hms)

Ano

de R

esis

tanc

e, R

A (o

hms)

AnodeSpacing

Figure 3. Anode bed resistance versus numberof anodes. 3-in. x 60-in. vertical graphite anodesin backfill.

Number of Anodes

Figure 4. Anode bed resistance versus number ofanodes. 2-in. x 60-in. vertical bare anodes.

AnodeSpacing

Figure 3 is based on a soil resistivity of 1,000 ohms-cm. Ifthe soil resistivity is different, use a ratio of the actual soilresistivity divided by 1,000 and multiply this by the readingobtained from Figure 3.

A rectifier ground bed is to be composed of 10 — 2 x60-in. bare "Duriron" silicon anodes spaced 20 ft apart. The

Number of anodes

soil resistivity is 3,000 ohms-cm. What is the ground bedresistance? Refer to Figure 4. The resistivity from the chartis 0.55 ohms. Since the chart is based on soil resistivity of1,000 ohms-cm, the ground bed resistivity is 3,000/1,000 x0.55 or 1.65 ohms.

The resistance of multiple anodes installed vertically andconnected in parallel may be calculated with the followingequation:

R = 0.00521P/NL x (2.3 Log 8L/d - 1 + 2L/S Log 0.656N)

(1)

where:R = Ground bed resistance, ohmsP = Soil resistivity, ohm-cmN = Number of anodesd = Diameter of anode, feetL = Length of anode, feetS = Anode spacing, feet

If the anode is installed with backfill such as coke breeze,use the diameter and length of the hole in which the anode isinstalled. If the anode is installed bare, use the actualdimensions of the anode.

Figure 5 is based on Equation 1 and does not include theinternal resistivity of the anode. The resistivity of a singlevertical anode may be calculated with Equation 2.

R = 0.00521P/L x (2.3 Log 8L/d - l) (2)

If the anode is installed with backfill, calculate theresistivity using the length and diameter of the hole inwhich the anode is installed. Calculate the resistivity usingthe actual anode dimensions. The difference between thesetwo values is the internal resistance of the anode. Use thevalue of P, typically about 50 ohm-cm, for the backfillmedium.

Figure 5 is based on 1,000 ohm-cm soil and a 7-ft x 8-in.hole with a 2-in. x 60-in. anode.

Example. Determine the resistivity of 20 anodes installedvertically in 1,500 ohm-cm soil with a spacing of 20 ft.

Read the ground bed resistivity from Figure 5.

An

od

e B

ed

Re

s.

Ohm

s

Spacing

Number of AnodesFigure 5. Anode bed resistance.

R = 0.202 ohm

Since the anodes are to be installed in 1,500 ohm-cm soil andFigure 5 is based on 1,000 ohm-cm soil, multiply R by theratio of the actual soil resistivity to 1,000 ohm-cm.

R = 0.202 x 1,500/1,000

R = 0.303 ohm

The internal resistivity for a single 2-in. x 60-in. verticalanode installed in 50 ohm-cm backfill (7ftx8-in. hole) is0.106 ohm.

Since 20 anodes will be installed in parallel, divide theresistivity for one anode by the number of anodes to obtainthe internal resistivity of the anode bank.

0.106/20 = 0.005 ohm

The total resistivity of the 20 anodes installed vertically willtherefore be 0.308 ohm (0.303 + 0.005).

Galvanic Anodes

Zinc and magnesium are the most commonly usedmaterials for galvanic anodes. Magnesium is available

either in standard alloy or high purity alloy. Galvanicanodes are usually pre-packaged with backfill to facilitatetheir installation. They may also be ordered bare if desired.Galvanic anodes offer the advantage of more uniformlydistributing the cathodic protection current along thepipeline, and it may be possible to protect the pipelinewith a smaller amount of current than would be requiredwith an impressed current system but not necessarily at alower cost. Another advantage is that interference with otherstructures is minimized when galvanic anodes are used.

Galvanic anodes are not an economical source of cathodicprotection current in areas of high soil resistivity. Their use isgenerally limited to soils of 3,000 ohm-cm except wheresmall amounts of current are needed.

Magnesium is the most-used material for galvanic anodesfor pipeline protection. Magnesium offers a higher solutionpotential than zinc and may therefore be used in areas ofhigher soil resistivity. A smaller amount of magnesium willgenerally be required for a comparable amount of current.Refer to Figure 6 for typical magnesium anode performancedata. These curves are based on driving potentials of —0.70volts for H-I alloy and —0.90 volts for Galvomag workingagainst a structure potential of —0.85 volts referenced tocopper sulfate.

The driving potential with respect to steel for zinc is lessthan for magnesium. The efficiency of zinc at low currentlevels does not decrease as rapidly as the efficiency formagnesium. The solution potential for zinc referenced to a

Soil Resistivity

- Ohms-Cm.

Anode Current - Mi l l iamperes

Figure 6. Magnesium anode current.

Soil

Resistivity

Ohms-Cm

Soil

Resistivity

Ohms-Cm

1 Anode2 Anodes3 Anodes4 Anodes8 AnodesSix Anodes

Spacing 15 ft.

1.4 In.xl.4 In.By 5 Ft. Long

Current Output Mi l l iamperes

Figure 7a. Current output zinc anodes.

Spacing 15 f t .

1.4-In.xi.4-In.By 5- f t . Long


Figure 7b. Current output zinc anodes.

Water Resistivity Ohms-Cm

Ano

de B

ed R

esis

tanc

e, R

A (o

hms)

S p a c i n g 15 f t .

l.4-In.xl.4-In.Jy 5-ft. Long


Figure 8a. Current output zinc anodes.

AnodeSpacing

Number of Anodes

Figure 8b. Current output zinc anodes.

copper sulfate cell is —1.1 volts; standard magnesium has asolution potential of —1.55 volts; and high purity magnesiumhas a solution potential of —1.8 volts.

If, for example, a pipeline is protected with zinc anodes ata polarization potential of —0.9 volts, the driving potentialwill be —1.1 — (—0.9) or —0.2 volts. If standard magnesium isused, the driving potential will be —1.55 —(—0.9) or —0.65volts. The circuit resistance for magnesium will be approxi-mately three times as great as for zinc. This would behandled by using fewer magnesium anodes, smaller anodes,or using series resistors.

If the current demands for the system are increased due tocoating deterioration, contact with foreign structures, or byoxygen reaching the pipe and causing depolarization, thepotential drop will be less for zinc than for magnesiumanodes. With zinc anodes, the current needs could increaseby as much as 50% and the pipe polarization potential wouldstill be about 0.8 volts. The polarization potential woulddrop to about 0.8 volts with only a 15% increase in currentneeds if magnesium were used.

The current efficiency for zinc is 90%, and this value holdsover a wide range of current densities. Magnesium anodes

have an efficiency of 50% at normal current densities.Magnesium anodes may be consumed by self-corrosion ifoperated at very low current densities. Refer to Figures 7a,7b, 8a, and 8b for zinc anode performance data. The data inFigures 7a and 7b are based on the anodes being installed ina gypsum-clay backfill and having a driving potential of —0.2volts. Figures 8a and 8b are based on the anodes beinginstalled in water and having a driving potential of —0.2volts.1

Example. Estimate the number of packaged anodesrequired to protect a pipeline.

What is the anode resistance of a packaged magnesiumanode installation consisting of nine 32-lb. anodes spaced 7 ftapart in 2,000 ohm-cm soil?

Refer to Figure 9. This chart is based on 17# packagedanodes in 1,000 ohm-cm soil. For nine 32-lb. anodes, theresistivity will be:

1 x 2,000/1,000 x 0.9 = 1.8 ohm

See Figure 10 for a table of multiplying factors for othersize anodes.

Ano

de B

ed R

esis

tanc

e, R

A (o

hms)

AnodeSpacing

Number of Anodes

Figure 9. Anode bed resistance versus number of anodes; 17# packaged magnesium anodes.

Chart based on 17-Ib. magnesium anodes installed in 1,000 ohm-cm soil in groups of10 spaced on 10-ft centers.

For other conditions multiply number of anodes by the following multiplying factors:

For soil resistivity: MF = —£— For 9-lb. anodes: MF = 1.25

For conventional magnesium: MF = 1.3 For 32-Ib. anodes: MF = 0.9

Num

ber

of

Ano

des

per

1,00

0 ft

Num

ber

of A

node

s pe

r M

ile

Coating Conductivity (micromhos/sq.ft)

Figure 10. Number of anodes required for coated line protection.

Example. A coated pipeline has a coating conductivityof 100 micromhos/sq. ft and is 10,000 ft long, and thediameter is 10/4 in. How many 17-lb magnesium anodes willbe required to protect 1,000 ft? Refer to Figure 7 and read 2anodes per 1,000 ft. A total of twenty 171b. anodes will berequired for the entire line.

Source

Pipeline Corrosion and Cathodic Protection, Gulf PublishingCompany, Houston, Texas.

Reference

1. From data prepared for the American Zinc Institute.

Estimate the pounds of sacrificial anode material required for offshore pipelines

This rule of thumb is based on the following assumptions:

• 5% of pipeline considered bare• 3 milliamperes/sq. ft required for protection• 30-year design life

Basis for rule: sq. ft of surface area of pipeline/linearft x % of bare pipe x 3.0 m.a./sq. ft x lb./amp.-year x design

life in years = the Ib. of galvanic anode material that will berequired.

Multiplier = OD x lb./amp.-yr x (12 PI/144)

x 30 x 0.05 x 0.003 x 1,000

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