1.
8: Corrosion/Coatings
Hand-held computer determines concrete coating thickness 204 How can the current output of magnesium rod
National Association of Pipe Coating Applications (NAPCA) used for the cathodic protection of heat exchanger
specifications 208 shells be predicted? 229
How much primer for a mile of pipe? 209 What spacing for test leads to measure current
How much coal-tar enamel for a mile of pipe? 210 on a pipeline? 229
How much wrapping for a mile of pipe? 210 How many magnesium anodes are needed for
Estimating coating and wrapping materials required supplementary protection to a short-circuited bare casing? 230
per mile of pipe 210 Group installation of sacrificial anodes 230
Coefficient of friction for pipe coating materials 211 How can the life of magnesium anodes be predicted? 231
Troubleshooting cathodic protection systems: How to find the voltage rating of a rectifier if
Magnesium anode system 213 it is to deliver a given amount of current through
Cathodic protection for pipelines 214 a given ground bed (graphite or carbon) 231
Estimate the pounds of sacrificial anode material Determining current requirements for coated lines 231
required for offshore pipelines 222 Determining current requirements for coated
Comparison of other reference electrode potentials lines when pipe-to-soil potential values are estimated 231
with that of copper-copper sulfate reference HVDC effects on pipelines 232
electrode at 25° C 224 Troubleshooting cathodic protection systems:
Chart aids in calculating ground bed resistance Rectifier-ground bed , 236
and rectifier power cost 225 How to control corrosion at compressor stations 237
How can output of magnesium anodes be predicted? 226 Project leak growth 238
How to determine the efficiency of a cathodic
protection rectifier 226 Advances in Pipeline Protection 239
How to calculate the voltage drop in ground Methods of locating coating defects 240
bed cable quickly 227 Case histories 243
What is the most economical size for a rectifier cable? 227 Estimate the number of squares of tape for
How to estimate the number of magnesium pipe coating (machine applied) 244
anodes required and their spacing for a bare Estimate the amount of primer required for tape 245
line or for a corrosion quot;hot spotquot; 228 Tape requirements for fittings 245
How can resistivity of fresh water be Induced AC Voltages on Pipelines
determined from chemical analysis? 228
What will be the resistance to earth of a May Present a Serious Hazard 246
single graphite anode? 229 Measuring Unwanted Alternating
How to estimate the monthly power bill for a
cathodic protection rectifier 229 Current in Pipe 248
What will be the resistance to earth of a group of Minimizing shock hazards on pipelines
graphite anodes, in terms of the resistance near HVAC lines 253
of a single anode? 229 Cathodic protection test point installations 254
2.
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, offer Computer program
many challenges to engineers.
Pipe is buoyant, and an empty line may float in water. Wet The following program, written for the Hewlett Packard
silts are like viscous fluids causing inadequately weighted 41CV (Figure 1), calculates the thickness of concrete weight
pipelines to pop up. There are various government entities coating for submarine pipelines expeditiously and with
that have jurisdiction over navigable rivers, bays, marshlands, satisfying results.
and offshore waters. These agencies may stipulate that The prompting feature (Figure 2) is employed to aid
pipelines be buried at certain depths and be stabilized. users' data input. Important calculated values and a recap of
A good way to stabilize a pipeline is to use an adequate the inputs are printed out, and each quantity is identified.
concrete weight coating. Determining the thickness of the Thus, the tape is a complete record. The program is flexible
concrete involves a process of balancing upward forces such in that any inputs can be readily changed for a second run.
as buoyancy of the mud and the downward forces—weights Thus, many quot;what i f questions can be answered quickly.
of pipe, protective coating, and concrete, allowing a factor of Suppose a heavier-weight pipe is used? What if the
60 (negative buoyancy). Such computations with several concrete density is changed? If the negative buoyancy is
variables can become involved and tedious. reduced? Etc.
8ULBL -COHCquot; 37 RCL 15 73 * 188 STO 85 144 STO 18 177 -T=- 213 -SP GR=-
82 ODV 38 RCL 17 74 + 189 RCL 15 145 RCL 21 178 PRCL 85 214 RRCL 14
63 SF 12 39 + 75 Xt2 118 quot;D=? quot; 146 -T INCR=? • 179 PVIEH 215 RVIEU
94 quot; CONCRETE1 40 LPSTX 76 .8855 111 PRCL 15 147 RRCL 21 188 CLR 216 CLP
05 -h COPTIHGquot; 41 * 77 • 112 PROMPT 148 PROMPT 181 quot;HPC=- 217 PDV
96 PRfl 42 .0213 78 STO 12 113 CLP 149 CLR 182 RRCL 86 218 -D=-
97 PDV 43 • 79 RCL 28 114 STO 15 158 STO 21 183 RVIEU 219 RRCL 15
88 CF 12 44 RCL 13 88 • 115 RCL 16 151 RCL 11 184 CLR 228 PVIEU
89 quot;T TRIPL=?quot; 45 • 81 STO 89 116 quot;HP=? quot; 152 quot;NEG BUOY=? - 185 quot;HC=- 221 CLP
18 PROMPT 46 STO 86 82 RCL 08 117 PRCL 16 153 PRCL 11 186 RRCL 87 222 -HP=-
11 STO 95 47 RCL 15 83 X O Y 118 PROHPT 154 PROMPT 187 RVIEH 223 RRCL 16
12 'PIPE DIPH=?quot; 43 RCL 17 84 - 119 CLP 155 CLP 188 CLP 224 PVIEH
13 PROHPT 49 2 85 STO 10 128 STO 16 156 STO 11 189 quot;HT=- 225 CLR
14 STO 15 58 • 86 FS? 81 121 RCL 17 157 GTO 01 198 RRCL 88 226 quot;TH PC=quot;
15 quot;PIPE WT=?quot; 51 + 87 GTO 03 122 -TH PC=? - 191 RVIEU 227 PRCL 17
16 PROHPT 52 RCL 85 88 RCL 11 123 PRCL 17 158*LBL 02 192 CLR 228 RVIEH
17 STO 16 53 + 89 X=8? 124 PROHPT 159 RCL 21 193 quot;HHD=quot; 229 CLP
13 quot;PROT CT TH=?quot; 54 LPSTX 98 GTO 84 125 CLR 168 ST+ 85 194 RRCL 89 238 -DC=quot;
19 PROHPT 55 * 91 RCL 18 126 STO 17 161 GTO 81 195 RVIEU
20 STO 17 56 .8213 92 RCL Ii 127 RCL 19 196 CLR 231 PRCL 19
21 -DEN CONC=?quot; 57 • 93 X<=Y? 128 quot;DC=? quot; 162*LBL 83 197 -HT-HHD=quot; 232 PVIEH
22 PROHPT 58 RCL 19 94 GTO 83 129 PRCL 19 163 RCL 88 198 RRCL 18 233 CLP
23 STO 19 59 • 95 GTO 82 138 PROHPT 164 RCL 12 199 RVIEU 234 -DH=quot;
24 quot;DEN MUD=?quot; 68 STO 87 131 CLR 165 / 288 CLR 235 RRCL 28
25 PROMPT 61 RCL 16 96*LBL 84 132 STO 19 166 STO 13 201 -NEG BUOY=- 236 PVIEU
26 STO 28 62 + 97 RCL 09 133 RCL 28 167 62.4 282 RRCL 11 237 CLP
27 quot;DEN PROT CT=?quot; 63 RCL 86 98 RCL 88 134 quot;DH=? • 168 / 283 RVIEH 238 quot;DPC=quot;
28 PROHPT 64 + 99 X)Y? 135 PRCL 28 169 STO 14 284 CLR 239 RRCL 18
29 STO 18 65 STO 88 188 GTO 83 136 PROHPT 285 quot;VT=- 248 PVIEU
38 -T INCR=?quot; 66 RCL 15 181 GTO 82 137 CLR 178*LBL -DPTP- 286 RRCL 12 241 CLP
31 PROMPT 67 RCL 17 182*LBL 85 138 STO 28 171 PDV 207 RVIEU 242 -T INCR=-
32 STO 21 68 2 183 RCL 85 139 RCL 18 172 CLR 288 CLP 243 RRCL 21
33 -NEG BUOY=?quot; 69 • 184 -T=? quot; 148 quot;DPC=? quot; 173 quot;RESULTS-quot; 289 -BULK DEN=quot; 244 PVIEU
34 PROHPT 70 + 185ftRCL85 141 PRCL 18 174 -I-RECPP.- 218 RRCL 13 245 CLP
35 STO 11 71 RCL 85 186 PROMPT 142 PROMPT 175 RVIEH 211 RVIEU 246 BEEP
36*LBL 81 72 2 187 CLP 143 CLP 176 PDV 212 CLR 247 END
Figure 1. Hewlett Packard-41CV computer program.
3.
Nomenclature and storage registers inputs
RESULTS-RECAP.
T=4.625e
Reg. no. HPC=4.6885
HC=479.3309
05 T = Concrete thickness, in. Assume trial WT=578.6385
value. Final answer is cumulated in MHD=553.4462
Reg. 0.5. HT-HHD=25.1843
HEG BUOY=20.89ee
15 D = Pipe diameter, in. VT=6.1494
BULK DEN=94.8954
16 WP = Weight of pipe, lb/lf. SP GR=1.5979
5=24.8869
17 TH PC = Thickness of protective coating. HP=94.6288
Usually 3/32 in. or 0.0938 in. TH PC=8.8938
DC=165.8808
19 DC = Density of concrete, lb/cu ft. Usually DH=98.0888
DPC=95.6888
140, 165, 190. T IHCR=8.1258
20 DM = Density of mud or fluid,
lb/cu ft. River mud is around 90.
Figure 2. Most common questions asked when checking
18 DPC = Density of protective coating buoyancy.
coating, lb/cu ft. 95 is average.
21 T INCR = Increment of T. Decimal of in. 0.125
is suggested for numerical precision— Formulas
not that a tolerance of this order
can be constantly achieved in the WPC = (D + (THPC))(THPC)0.0218(DPC)
pipe coating world. Increment can be WC = (D + 2(THPC) + T)T 0.0218(DC)
made smaller, if desired.
11 NEG BUOY = Negative buoyancy. Desired minimum VT = (D + 2(THPC) + 2T)20.0055
value, lb/lf. Can be 0 in special
WMD = (VT)(DM)
circumstances—e.g., to ascertain if a
given pipe will float in water, or will a Note: The user is encouraged to employ the most reliable
certain coated pipe sink in mud? and suitable data available, such as the density of protective
coating, mud, and concrete.
Each value keyed in is printed on the tape as a check of
inputs and the program prompting feature queries user for Examples
value of next named input (Figure 2).
Put program into calculator, XEQ Size 30. The printed
RESULTS-RECAP portions of each of three examples are
Calculated values given. When the user XEQ quot;CONCquot; and inputs the data
from the lower part of the tabulation, he or she will get the
same calculated values above.
Reg. no. Figure 2 includes the most common questions asked
06 WPC = Weight of protective coating, when checking coating buoyancy. Always be sure flag 01 is
clear XEQ quot;CONCquot; and put in new trial T and R/S. Also R/S
lb/lf.
after all other inputs. After all prompted data are keyed
07 WC = Weight of concrete coating, lb/lf. in, the processing starts and results are printed out. Notice
08 WT = Total weight—pipe and coatings, lb/lf. T = 4.6250 in. has an actual negative buoyancy of 25.1843. If
09 WMD = Weight of mud (fluid) displaced, lb/lf. T = 4.5 in., the actual negative buoyancy= 18.4510, which is
less than 20. A precise solution is between 4.5 in. and 4.625 in.
10 WT-WMD = Actual negative buoyancy, lb/lf.
If the user wants to find T, which produces an actual
12 VT = Total volume displaced by pipe and negative buoyancy nearer 20: STO 4.5 in RO 5, STO § in.
coatings, cu ft/If. (0.0625 in.), in R21, XEQ 01. This produces T = 4.5625
13 BULK DEN = Bulk density, wt/vt.lb/cu ft. where actual negative buoyancy is 1.8113 greater than 20.
14 SP GR = Specific gravity, bulk den/den water. This refinement can be found in choosing small T INCR.
Note that the weight of the concrete coating is now
471.8269 lb/lf and we have approached closer to 20.
4.
Calculated values. Weight of protective coating and
RESULTS-RECAP
concrete, total weight and volume of pipe, and coatings per
linear foot will be of value in transporting coated pipe, etc. T=3.2598
HPC=3.9834
Note: Results—RECAP of any run (which is in storage) MO273.9914
can be reprinted as often as desired by XEQ quot;DATA.quot; HT=375.7248
If NEG BUOY is 0 and flag 01 is not set, then the program HHD=352.5529
HT-MHB=23.1728
causes WT to approach WMD and calculates for this NEG BUOY=28.8880
situation. VT=3.9173
BULK DEN=95.9154
If a printer is not available, results can be obtained as SP GR=1.5371
follows:
11=28.8888
1. XEQ 05, put in data and R/S after each item. «P=97.8380
TH PC=6.8938
2. When beep sounds, RCL registers one by one in order DC=165.8900
given above (so as to identify values). BH=90.8«90
DPC=95.9098
Figure 3 illustrates the changing of two input values: T IHCR=B.1250
namely the pipe diameter and weight. The XEQ 05 routine
calls up the contents of the input registers one at a time and Figure 3. Changing of two input valves.
questions them. Always put in a new T value. Then R/S.
When D = ? comes up, key in new value and R/S. Same
with pipe WT = ?. Assume other data remain unchanged, XEQ quot;DflTfiquot;
R/S each time. (Any one value or any combination of values
RESULTS-RECAP. BULK BEN=46.9065
may be changed before rerunning the program.) SP GR=0.7517
Figure 4 determines if a given pipe, 14 in. at 36.71 lb/lf T-e.eeee
UPC-21.3368 6=14.0800
and 1A in. of protective weight coating density of 135 lb/cu ft, HP=36.7180
№=8.0000
will float. T, DC, T INCR, and NEG BUOY are all zero. DM HT=58.8463 TH PC=8.5000
becomes the density of water 62.4. Set flag 01 and XEQ 05. HHD=77.2280 BPC=135.0880
HT-WHB=-19.1733 BC=8.8000
Key in data and R/S. When all data are in place only one NEG BUOY=0.0080 BH=62.4000
computation is allowed, then printed out. WT-WMD tells us VT=I.2375 T INCR=0.0000
the upward forces exceed the downward forces by 19.1733
Figure 4. Result of calculations to determine whether the pipe
lb/lf; also the specific gravity is 0.7517, so the pipe floats.
will float.
Also note that BULK DEN is less than the density of water.
National Association of Pipe Coating Applications (NAPCA) specifications*
RECOMMENDED SPECIFICATION DESIGNATIONS FOR ENAMEL COATINGS
These are recommended enamel coating system designations for
SPECIFICATION coal tar or asphalt coatings, except for specialty requirements.
SYMBOL
FIBERGLASSWRAP
POLYETHYLENE
93.75 MILS MIN.
HEAVY KRAFT
#15 FELT WRAP
WITH KRAFT
ELECTRICAL
SHOT BLAST
INSPECTION
GLASS MAT
SEAL COAT
•PART *1
•• PART #2
ENAMEL
PRIMER
PAPER
*NOTE: These are not all the specifications approved by NAPCA. For other specifications, consult an NAPCA member.
5.
AGSF-
TGM-
AGM-
TGMF-
AGMF-
TGMP-
AGMP-
#
P A R T # 1 SYMBOLS- • # P A R T # 2 SYMBOLS
ENAMEL & WRAPPERS GRADESOF ENAMEL
3— Fully-Plasticized Coal Tar
T-Coal Tar G-Fiberglass
A-Asphalt S-Seal Coat (ordered by penetration
F-#15 Felt GM-Glass Mat dictated by ambient temperature)
P—Polyethylene wrapper 4— High Temperature Surface Coa! Tar
w/kraft paper 7 - Asphalt
O—Outer wrap
To specify Outer Wrap, use EXAMPLE: 93.75 mils (minimum) fully-plasticized coal tar enamel
letter O in place of F.
wrapped with fiberglass, felt and kraft — the symbol would be TGF-3.
Reprinted courtesy of National Association of Pipe Coating Applicators.
STANDARD APPLIED PIPE COATING WEIGHTS
FOR NAPCA COATING SPECIFICATIONS
Refer to NAPCA 1-65-83
Spec.
Symbol
PIPE Weight in Pounds-Per 100 Lineal Feet
SIZE
(continued on next page)
6.
(table continued)
Spec.
Symbol
PIPE Weight in Pounds - Per 100 Lineal Feet
SIZE
Al! weights provide for standard cutbacks.
Weights per 100 lineal feet are figured on the following material weights per square foot of pipe surface:
Weight Weight
Lbs.-Per Lbs.-fer
Sq. Ft. Sq. Ft.
Coal Tar Primer & Enamel—1/32quot; Min. .2461 #15 Pipeline Felt
Coal Tar Primer & Enamel-2/32quot; Min. .4850 (Tar or asphalt saturated) 1271
Coal Tar Primer & Enamel—3/32quot; Min. .7238 Outerwrap 1225
Asphalt Primer & Enamel—1/32quot; Min. .2000 .010 Fiberglass Wrap 0110
Asphalt Primer & Enamel—2/32quot; Min. .3950 Kraft Paper 0200
Asphalt Primer & Enamel—3/32quot; Min. .5912 Glass Mat 0114
Polyethylene 0450
Reprinted courtesy of National Association of Pipe Coating Applicators.
NAPCA SPECIFICATIONS PIPELINE FELTS
PROPERTY #15TAR #15 TAR #•15 ASPH. #15 ASPH.
N.R. R.F. N.R. R.F.
Saturated FaIt - Not Parforatad
Weight (MIn.) lbs. per 100 s.f. 12 12 12 12
(ASTM-D-146-59 Sec. 1-9)
Saturation (Percent) MIN. 18 18 22 22
(ASTM-D-146-59 Sec. 18 MAX. 28 28 40 40
by extraction)
Ash Content (MIN).) (Percent) 70 70 70 70
(ASTM-D-146-59 Sec. 21)
Callper (Min.) (inches) .021 .018 .021 .018
(ASTM-D-645-64T Method C)
Tensile Strength (Min.)
(lbs. per Inch of width)
JASTM-D-146-59 Sec. 12a)
With fiber grain 35 37 35 37
Across fiber grain 12 No. spec. 12 No. spec.
Tear Strength (Min.) (Grams)
(ASTM-D-689-62)
Across fiber grain 400 Not 400 Not
With fiber grain 240 applicable 240 applicable
(continued on next page)
7.
Loss on heating (Max.) (Percent) 10 10 10 10
(AWWA C203-66 Sec. 2.6.6)
Pliability
(AWWA C203-66 Sec. 2.6.4) No cracking No cracking No cracking No cracking
Saturated and Perforated Felt
(1/16quot; perforations on 1 quot;
Staggered Centers)
Tensile Strength (MIn.)
(lbs. per inch of width)
(ASTM-D-146-59 Sec. 12a)
With fiber grain 30 35 30 35
Across fiber grain 10 No spec. 10 No. spec.
Tear Strength (MIN.) (Grams)
(ASTM-D-689-62)
Across fiber grain 348 Not 348 Not
With fiber grain 190 applicable 190 applicable
Umaturated Felt
Asbestos Content (%) (MIN.)
(ASTM-D-1918-67) 85 85 85 85
N.R. —NON REINFORCED
R.F. — REINFORCED LENGTHWISE WITH PARALLEL FIBER GLASS YARNS ON NOMINAL ¥•quot; CENTERS.
Reprinted courtesy of National Association of Pipe Coating Applicators.
Minimum Test Voltages for Various Coating Thicknesses
Coating Thicknesses
32nd Inch Mils Test Volts
Reprinted courtesy of National Association of Pipe Coating Applicators.
How much primer for a mile of pipe?
Multiply pipe O.D. by 2% to get gallons of primer per Pipe 41Z2 in. in diameter will take 41Z2 x 2% or 123/8 gallons
mile. of primer per mile (12 gallons in round figures).
The given multiplier is applicable only where the pipe is
new and in good condition. For rough pipe the multiplying
factor should be upped from 2% to 3.45. In other words,
Examples. rough 20-in. pipe will require 20 x 3.45 or 69 gallons of
Pipe 24 in. in diameter will require 24 x 2% or 66 gallons primer per mile.
of primer per mile; It is hard to estimate the amount of spillage and waste.
Pipe 16 in. in diameter will require 16 x 2% or 44 gallons Temperature and relative humidity affect the way the primer
of primer per mile; and goes on the pipe.
8.
How much coal-tar enamel for a mile of pipe?
For an average cover of %2 in., multiply outside 8.625 x 0.6 =5.2 tons per mile
diameter of pipe by 0.6. The answer is in tons of enamel 5.2x15% =0.8
per mile. (Add 15% to take care of gate valves, drips, bends, = 6 tons per mile
etc.)
These factors strongly affect the amount of coating used
per mile on pipelines: 1. Temperature of the pipe and
Examples. How many tons of coal-tar enamel will be
ambient temperatures. (It is assumed that the temperature
required to apply a %2-in. coat to a 26-in. pipeline?
of the enamel can be controlled.) 2. The kind of dope
26 x 0.6 = 15.6 tons per mile machine. 3. Humidity. 4. The number of bends per mile. 5.
The number of valves, crossovers, etc. 6. Condition of the
Add 15% to take care of gate valves, bends etc. pipe.
This latter point has a great effect on the amount of
15.6 + 2.3 = 17.9 or 18 tons per mile
enamel required for a mile of pipeline. In fact, where
How many tons of coal-tar enamel will be required to reconditioned pipe is being coated, it is recommended that
apply a %2~in- coat to a 8.625-in. pipeline? the amount of enamel be increased by about 20%.
How much wrapping for a mile of pipe?
For 14-in. to 48-in. pipe multiply the diameter in inches (4 + 1) x 15 = 75 squares of felt
by 15 to get the number of squares (100 ft ). For pipe sizes
These figures are conservative and do not allow for
of 12% in. and smaller, add 1 to the nominal size of pipe and
wrapping drips, for patching holidays, and for short ends that
multiply by 15.
are usually discarded. They allow for machine wrapping
Examples. How many squares of asbestos felt would be
using /^-in. lap on diameters up to 10/4 m - a n d using 1-in. lap
required to wrap 1 mile of 30-in. pipeline?
on larger diameters.
30 x 15 = 450 squares of felt For smaller diameters of pipe, this figure should be
increased about 5% to take care of additional wastage and
How many squares of felt would be required to wrap 1
patching.
mile of 4-in. pipe?
Estimating coating and wrapping materials required per mile of pipe
Nominal Glass Kraft 15-Pound
Primer and coal-tar enamel Pipe Primer Coal-tar Wrap Paper Felt
Size Gallons Enamel Squares Squares Squares
Tons
Gallons of primer and tons of coal-tar enamel are based on
covering new pipe in good condition. For rough pipe,
increase primer quantities by 25% and the coal-tar quantities
by 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 sizes
greater than 10 in. No allowance is included for wrapping
drips, patching holidays, or short ends, which are usually
discarded.
9.
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 Technological
University, Houghton, Mich., and G. R. Mayer, Project Engineering Manager, Bechtel, Inc., San Francisco
A major factor in the stress analysis of buried pipelines is where: F = Longitudinal soil friction force (Ib)
the movement that pipe undergoes in the presence of tem- /ji = Coefficient of friction (dimensionless)
perature and pressure differentials during its life. This move- p = Normal soil pressure acting on the pipe
ment is highly dependent upon friction resistance of the soil. surface (psi)
Although ample information is available on the static dA = Soil to pipe differential contact area (in.2)
coefficient of friction for many materials, there is a lack of fA pdA = Total normal soil force on pipe surface (Ib)
data on friction between soils and various coatings used in
The above relation is independent of the pressure per unit
the pipeline industry. In the past, friction coefficient
area of the mating surfaces as long as very high contact
information was extrapolated from data in the literature
pressures are not encountered, which is the case for a buried
that were believed to have a similarity to the external pipe
pipeline.
coating to soil interface.
In terms of a buried pipeline supporting a soil burden up
However, with the development of thinfilm epoxy resin
to three pipe diameters of depth, the soil force relation
coating systems and the increasing use of these systems in
becomes:
the pipeline industry, a change from a conventional coal tar
felt coating to a thinfilm epoxy coating would indicate a
significant change in the friction coefficient design criteria
due to the extreme contrast in the surface texture of the two
materials.
To evaluate the effect of the difference in surface texture where: Wp = Weight of pipe and contents (lb/in.)
on a pipeline system, test procedures were developed to D = Pipe diameter (in.)
determine the coefficient of friction for both coal tar felt and H = Depth of the pipe centerline (in.)
thinfilm epoxy to various soils and to obtain more reliable Y = Specific weight of the soil (lb/in.3)
information for future pipeline designs using these coating The sensitivity of the soil friction forces to the coefficient
materials. of friction is readily apparent from the above relation. Since
Static friction tests were conducted to find the coefficient this force is inversely proportional to the active length that a
of friction between coal tar felt and thinfilm epoxy pipe pipeline moves as a result of temperature and pressure
coating and eight representative backfill soil samples from expansion, the coefficient of friction becomes a major factor
typical locations along a pipeline right of way. The results in pipeline stress design.
indicate that the friction coefficients are significantly larger
than those previously extrapolated from literature and that Test system. The theoretical soil friction force acting on
coal tar has a higher friction resistance in respect to the surface of a coated plate can be calculated from the
anchorage of a pipeline. relation:
For the coal tar felt coating, the coefficient varies from
0.59 to 0.91 depending on the soil and moisture content. The F = ^N
thinfilm epoxy coating varies from 0.51 to 0.71 under the
same conditions. where: N = Normal force acting on the plate surfaces (Ib)
W = Total weight of plate and external load (Ib)
Definitions
Static coefficient of friction. The theoretical long-
itudinal soil force acting on the pipe surface can be
COATED PLATE
calculated from the relation:
10.
Since the coefficient of friction parameter is the same in Thinfilm epoxy. The epoxy coating simulated a thinfilm
all of the previous expressions, the plate test system can be epoxy resin pipeline coating applied by the fusion method.
used to simulate the soil-to-pipeline interface. Plates used in the tests were coated by the fluidized bed
method. However, the surface texture was the same as that
expected by coating plant production equipment.
Coal tar felt. The coal tar coating, an enamel and felt
combination, was prepared to simulate an actual coated
Results
pipeline.
The surface of a steel plate was coal tar coated and then Coal tar to soils. Figure 1 illustrates that the friction
wrapped in a 15-lb felt. Therefore, the soil surface contact coefficient varies from 0.59 to 0.91 for the various soils tested
was predominantly with the felt material. at field moisture content.
LOCATION SOIL DESCRIPTION MOISTURE
SANOY SIiT (ML)
SiLTY SAND {SM>
SItT ( M t I
SiLTY SAND (SM)
SIlTY SANO (SM)
SAND - SILT (SM)
SANDY SILT (ML)
SlLTY • SAND (SM)
UMlFiED SGlL CLASSIFICATION SYMBOL
Figure 1. Coefficient of friction for coal tar coating
to soil.
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-SAKO
SILT (GM)
UNFIEO SOIL CtASSJFiCATiON SYMBOL
Figure 2. Coefficient of friction for thin film epoxy
to soil.
11.
The mechanism by which the friction coefficient changes Since such information on this coating was not available in
among the various soils is not apparent from the results of the literature, no comparison to similar tests could be made.
these tests, since the soil samples used did not vary Temperatures in the 1200F range should have little or no
considerably in description. However, the range over effect on the coefficient of friction of epoxy to soil.
which the factor changes is considerably greater than that
previously extrapolated from the literature, as follows:
Conclusions
Coefficient of friction
Soil description commonly used Although it is virtually impossible to precisely simulate the
surface contact situation of a pipeline in a back-filled trench,
Silt 0.3
the test procedure and apparatus reported here are a means
Sand 0.4 of approximating it. The results indicate that coal tar coatings
Gravel 0.5 have a higher friction resistance than epoxy coatings as far as
anchoring capabilities of the soil are concerned.
The tests also indicate that the moisture content alters the The selection of a coating, based on its soil friction
friction factor to some extent, as would be expected. resistance, could be of economic value in reducing those
In order to investigate the influence of temperature, tests situations where extreme expansions call for reinforced
were conducted with coal tar felt wrapping heated to 1200F. fittings or elaborate culverts to overcome excessive stress
Only a slight softening beneath the coating surface was levels in a pipeline system.
observed, and it is believed that temperatures up to this The test results show that previous values commonly used
range will not significantly affect the magnitude of the for the coefficient of friction were conservative for similar
friction coefficient. soils, and it is suggested that some conservatism still be
incorporated in future analyses.
Since the tests indicated that the friction coefficient of the
Thinfilm epoxy to soils. As expected, results shown in epoxies was similar to those previously used for coal tar
Figure 2 indicate that the friction factor range of 0.51 to 0.71 coatings, their continued use for epoxy-coated pipelines
for the epoxy is somewhat lower than that of the coal tar. should be valid. A more conservative approach to these
Again, the mechanism is not clear, but the results are fairly values is recommended in soils where the presence of
consistent with the friction coefficient increasing in most excessive moisture could change the friction factor substan-
instances in the same order of magnitude as for coal tar. tially when interfaced with a smooth epoxy coating.
Troubleshooting cathodic protection systems: Magnesium anode system
The basic technique is the same as that outlined earlier;
more measurements are required because of the multiplicity
of drain points. First, the current output of stations nearest
the point of low potential should be checked; if these are
satisfactory, a similar check should be extended in both
directions until it is clear that the trouble must be on the
line. When a given anode group shows a marked drop in
current output, the cause may be drying out, shrinkage of
backfill, or severed or broken lead wires. If the current is
zero, the pipe-to-soil potential of the lead wire will show
whether it is still connected to the pipe or the anode, and < Broken Wire
•
thus indicate the direction to the failure. If the current is Figure 1. Locating idle anodes by surface potentials. The solid
low, there may be a loss of one or more anodes by a line shows the potentials found along a line of anodes when all
severed wire; a pipe-to-soil potential survey over the anodes are delivering current; the dotted line exhibits the change when
will show which are active, just as in the case of rectifier there is a break in the anode lead at the point indicated. Single
disconnected anodes may also be located by this method. A
anodes. driven ground rod, a pipe lead, or even a rectifier terminal may
Trouble indicated on the line, rather than at the anode be used for the reference ground; all readings should be
stations, is tracked down in the same manner as that used for referred to the same reference.
12.
line protected by rectifiers—first, investigating locations next, checking potentials; and finally, making a line current
where something that has been done might be responsible; survey.
Cathodic protection for pipelines
Estimate the rectifier size required for an infinite line. Use a value of—0.3 volts for AE x . This is usually enough
Refer to Figure 1. If coating conductance tests have not been to raise the potential of coated steel to about—0.85 volts.
performed, pick a value from Table 1. Use a value of 1.5 volts for AE at the drain point. Higher
values may be used in some circumstances; however, there
may be a risk of some coating disbondment at higher
Table 1 voltages.
Typical Values of Coating Conductance Calculate I at the drain point.
Micromhos/sq ft Coating Condition
1-10 Excellent coating—high resistivity soil AI A = amp./in. x AE x /0.3 x D
10-50 Good coating—high resistivity soil
50-100 Excellent coating—low resistivity soil
100-250 Good coating—low resistivity soil
where D = pipe OD, inches.
250-500 Average coating—low resistivity soil
Example. 30-in. OD line with coating conductivity= 100
500-1,000 Poor coating—low resistivity soil micromhos/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 — , amperes per inch
D AEx
Thousands of Feet, x
Figure 1. Drainage current versus distance for a coated infinite line.
13.
Based on 0.25-inch wall pipe and steel resistivity of 18.8 microhm-cm
— x — , amperes per inch
. u
D AEJ
IA 0 3
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 long micromhos/sq ft. What is AE A /E T and what AI A is required?
pipeline? Assume that
AE A = 1.5 E x = 0.3
AE x = 0.3 From Figure 2 read
AE A /AE x = 5.0 AE A /E X = 5.8
AE A = 1.74
Refer to Figure 1 and read
AI A /D x 0.3/E x = 1.8 amp./in.
L = 30,000 ft
AI A =1.8 x 3.5 = 6.3 amp.
AIA/D x 0.3/AEx = 0.7 (from Figure 1)
Estimate ground bed resistance for a rectifier
AIA = 0.7x 30 = 21 amp. installation.
21 amps will protect the line for 30,000 ft in either direction Example. A ground bed for a rectifier is to be installed
from the drain point. in 1,000 ohm-cm soil. Seven 3-in. x 60-in. vertical graphite
Estimate rectifier size for a finite line. Refer to Figure 2. anodes with backfill and a spacing of loft will be used. What
A 31Z2 -in. OD line 20,000 ft long is protected by insulating is the resistance of the ground bed?
flanges at both ends and has a coating conductivity of 500 Refer to Figure 3. The resistance is 0.56 ohms.
14.
Anode Bed Resistance, R A (ohms)
Anode
Spacing
Figure 3. Anode bed resistance versus number
of anodes. 3-in. x 60-in. vertical graphite anodes
in backfill.
Number of Anodes
Anode Resistance, RA (ohms)
Figure 4. Anode bed resistance versus number of Anode
anodes. 2-in. x 60-in. vertical bare anodes. Spacing
Number of anodes
Figure 3 is based on a soil resistivity of 1,000 ohms-cm. If soil resistivity is 3,000 ohms-cm. What is the ground bed
the soil resistivity is different, use a ratio of the actual soil resistance? Refer to Figure 4. The resistivity from the chart
resistivity divided by 1,000 and multiply this by the reading is 0.55 ohms. Since the chart is based on soil resistivity of
obtained from Figure 3. 1,000 ohms-cm, the ground bed resistivity is 3,000/1,000 x
A rectifier ground bed is to be composed of 10 — 2 x 0.55 or 1.65 ohms.
60-in. bare quot;Durironquot; silicon anodes spaced 20 ft apart. The
15.
The resistance of multiple anodes installed vertically and Figure 5 is based on Equation 1 and does not include the
connected in parallel may be calculated with the following internal resistivity of the anode. The resistivity of a single
equation: vertical anode may be calculated with Equation 2.
R = 0.00521P/NL x (2.3 Log 8L/d - 1 + 2L/S Log 0.656N) R = 0.00521P/L x (2.3 Log 8L/d - l) (2)
(1)
If the anode is installed with backfill, calculate the
where: resistivity using the length and diameter of the hole in
R = Ground bed resistance, ohms which the anode is installed. Calculate the resistivity using
P = Soil resistivity, ohm-cm the actual anode dimensions. The difference between these
N = Number of anodes two values is the internal resistance of the anode. Use the
d = Diameter of anode, feet value of P, typically about 50 ohm-cm, for the backfill
L = Length of anode, feet medium.
S = Anode spacing, feet 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.
If the anode is installed with backfill such as coke breeze,
use the diameter and length of the hole in which the anode is Example. Determine the resistivity of 20 anodes installed
installed. If the anode is installed bare, use the actual vertically in 1,500 ohm-cm soil with a spacing of 20 ft.
dimensions of the anode. Read the ground bed resistivity from Figure 5.
Anode Bed R e s . Ohms
Spacing
Number of Anodes
Figure 5. Anode bed resistance.
16.
R = 0.202 ohm either in standard alloy or high purity alloy. Galvanic
anodes are usually pre-packaged with backfill to facilitate
Since the anodes are to be installed in 1,500 ohm-cm soil and
their installation. They may also be ordered bare if desired.
Figure 5 is based on 1,000 ohm-cm soil, multiply R by the
Galvanic anodes offer the advantage of more uniformly
ratio of the actual soil resistivity to 1,000 ohm-cm.
distributing the cathodic protection current along the
pipeline, and it may be possible to protect the pipeline
R = 0.202 x 1,500/1,000 with a smaller amount of current than would be required
with an impressed current system but not necessarily at a
R = 0.303 ohm lower cost. Another advantage is that interference with other
structures is minimized when galvanic anodes are used.
The internal resistivity for a single 2-in. x 60-in. vertical Galvanic anodes are not an economical source of cathodic
anode installed in 50 ohm-cm backfill (7ftx8-in. hole) is protection current in areas of high soil resistivity. Their use is
0.106 ohm. generally limited to soils of 3,000 ohm-cm except where
Since 20 anodes will be installed in parallel, divide the small amounts of current are needed.
resistivity for one anode by the number of anodes to obtain Magnesium is the most-used material for galvanic anodes
the internal resistivity of the anode bank. for pipeline protection. Magnesium offers a higher solution
potential than zinc and may therefore be used in areas of
0.106/20 = 0.005 ohm higher soil resistivity. A smaller amount of magnesium will
generally be required for a comparable amount of current.
Refer to Figure 6 for typical magnesium anode performance
The total resistivity of the 20 anodes installed vertically will data. These curves are based on driving potentials of —0.70
therefore be 0.308 ohm (0.303 + 0.005). volts for H-I alloy and —0.90 volts for Galvomag working
against a structure potential of —0.85 volts referenced to
copper sulfate.
Galvanic Anodes The driving potential with respect to steel for zinc is less
than for magnesium. The efficiency of zinc at low current
Zinc and magnesium are the most commonly used levels does not decrease as rapidly as the efficiency for
materials for galvanic anodes. Magnesium is available magnesium. The solution potential for zinc referenced to a
Soil Resistivity - Ohms-Cm.
Anode Current - Milliamperes
Figure 6. Magnesium anode current.
17.
1 Anode
2 Anodes
3 Anodes
4 Anodes
8 Anodes
Six Anodes
Soil Resistivity Ohms-Cm
Spacing 15 ft.
1.4 In.xl.4 In.
By 5 Ft. Long
Current Output M i l l i a m p e r e s
Figure 7a. Current output zinc anodes.
Soil Resistivity Ohms-Cm
Spacing 15 f t .
1.4-In.xi.4-In.
B 5 - f t . Long
y
Current Output M i l l i a m p e r e s
Figure 7b. Current output zinc anodes.
18.
Water Resistivity Ohms-Cm
Spacing 15 ft.
l.4-In.xl.4-In.
Jy 5-ft. Long
Current Output M i l l i a m p e r e s
Figure 8a. Current output zinc anodes.
Anode Bed Resistance, RA (ohms)
Anode
Spacing
Number of Anodes
Figure 8b. Current output zinc anodes.
19.
copper sulfate cell is —1.1 volts; standard magnesium has a have an efficiency of 50% at normal current densities.
solution potential of —1.55 volts; and high purity magnesium Magnesium anodes may be consumed by self-corrosion if
has a solution potential of —1.8 volts. operated at very low current densities. Refer to Figures 7a,
If, for example, a pipeline is protected with zinc anodes at 7b, 8a, and 8b for zinc anode performance data. The data in
a polarization potential of —0.9 volts, the driving potential Figures 7a and 7b are based on the anodes being installed in
will be —1.1 — (—0.9) or —0.2 volts. If standard magnesium is a gypsum-clay backfill and having a driving potential of —0.2
used, the driving potential will be —1.55 —(—0.9) or —0.65 volts. Figures 8a and 8b are based on the anodes being
volts. The circuit resistance for magnesium will be approxi- installed in water and having a driving potential of —0.2
mately three times as great as for zinc. This would be volts.1
handled by using fewer magnesium anodes, smaller anodes, Example. Estimate the number of packaged anodes
or using series resistors. required to protect a pipeline.
If the current demands for the system are increased due to What is the anode resistance of a packaged magnesium
coating deterioration, contact with foreign structures, or by anode installation consisting of nine 32-lb. anodes spaced 7 ft
oxygen reaching the pipe and causing depolarization, the apart in 2,000 ohm-cm soil?
potential drop will be less for zinc than for magnesium Refer to Figure 9. This chart is based on 17# packaged
anodes. With zinc anodes, the current needs could increase anodes in 1,000 ohm-cm soil. For nine 32-lb. anodes, the
by as much as 50% and the pipe polarization potential would resistivity will be:
still be about 0.8 volts. The polarization potential would
drop to about 0.8 volts with only a 15% increase in current 1 x 2,000/1,000 x 0.9 = 1.8 ohm
needs if magnesium were used.
The current efficiency for zinc is 90%, and this value holds See Figure 10 for a table of multiplying factors for other
over a wide range of current densities. Magnesium anodes size anodes.
Anode Bed Resistance, RA (ohms)
Anode
Spacing
Number of Anodes
Figure 9. Anode bed resistance versus number of anodes; 17# packaged magnesium anodes.
20.
Next Page
Chart based on 17-Ib. magnesium anodes installed in 1,000 ohm-cm soil in groups of
10 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
Number of Anodes per 1,000 ft
Number of Anodes per Mile
Coating Conductivity (micromhos/sq.ft)
Figure 10. Number of anodes required for coated line protection.
Example. A coated pipeline has a coating conductivity Source
of 100 micromhos/sq. ft and is 10,000 ft long, and the
diameter is 10/4 in. How many 17-lb magnesium anodes will Pipeline Corrosion and Cathodic Protection, Gulf Publishing
be required to protect 1,000 ft? Refer to Figure 7 and read 2 Company, Houston, Texas.
anodes per 1,000 ft. A total of twenty 171b. anodes will be
required for the entire line.
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: life in years = the Ib. of galvanic anode material that will be
required.
• 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/linear Multiplier = OD x lb./amp.-yr x (12 PI/144)
ft x % of bare pipe x 3.0 m.a./sq. ft x lb./amp.-year x design x 30 x 0.05 x 0.003 x 1,000