cathodic protection

Cathodic Protection

ME 472-061 Corrosion Engineering IME, KFUPM

Dr. Zuhair M. Gasem

Dr. Z. Gasem ME 472-061

KFUPM2

References:ASM Handbook, vol 13, pp. 466-477Corrosion for Science and Engineering, K.R. Trethewey and J. Chamberlain, chapter 16Handbook of Corrosion Engineering, P.R. RobergeCathodic Protection in ARAMCO’s Engineering Encyclopedia


KFUPM3

Fe2+

Fe2+

HH22HH++

HH++ HH++ HH++ HH++

HH++

HH HH

e eMetal

Cathode

Anode

AcidSolution

Electron Flow

Anodic and Cathodic Reactions of Iron in Acids

Corrosion Cell on a Metal Surface


KFUPM4Basic Physics of CP of Iron in Acids

Electrons from an external source are forced to flow into the structure to be protected resulting in:

Increased cathodic reaction (2H+ + 2e- → H2) Decreased anodic reaction and hence reduced corrosion rate

Electrons from external source

eeee

eeeeeee

eeee

Electrolyte

H+H+ H+ H+ H+ H+ H+

eeH+Fe2+

HHHHH H H H

H2 H2 H2 H2

H+ H+

H+

H+

H+H+


KFUPM5Polarization Principle of CP of Iron in Acids

Before CP:ianode = icathode= icorr

E = Ecorr

After CP:icorr = iaicathode = iciapp = ic – iaE = ECP

ECP


KFUPM6Polarization Principle of CP of Iron in Water

The cathodic reaction for corrosion of steel and iron in aerated-water is usually (O2+2H2O+2e→4OH-) under concentration polarization. Before CP:

ianode = icathode = iLE = Ecorr

After CP:icorr = iaicathode = ic = iLiapp = ic – iaE = ECP

ECP


KFUPM7Cathodic Protection

Summary of cathodic protection:CP makes the structure’s potential more negative which promotes cathodic reactions and slows anodic reaction Increases icathode

Decreases ianode

Need to supply iapp = icathode - ianode

Where CP is used?CP is often applied to coated structures, with the coat providing the primary form of corrosion protection and the CP system acts as asupporting protection.

The main applications of CP include: Buried pipelineAcids storage tanksOffshore steel structures such as platforms and oil rigs ShipsConcrete structures exposed to seawater such as bridges


KFUPM8CP of Buried Pipelines

Before CP is applied:Anodes and cathodes are on the same surface of the pipe The soil is the electrolyteIonic current flow b/w the anode and the cathode in the external surfacesElectrons flow in the metal from anode to cathode



After CP is applied:The structure to be protected becomes the cathode The anode is an external electrode:

Amore active metal (sacrificial anode)An inert anode with impressed DC current (Impressed current)

The soil is the electrolyteIonic current flow b/w the anode and the cathode in the external surfacesElectrons flow between the anode and cathode through an insulated copper wire.

cathodeSacrificial anode

e-

+ve ions current in electrolyte



Sources of currentSacrificial anode systemImpressed current system (note the - polarity from the rectifier)


KFUPM11Electrochemical reaction in sacrificial Anode CP System

Cathodic reactions on the steel structure:

In aerated wet soilO2+2H2O+4e- ⇒ 4OH-

In aerated wet acidic soil O2+4H+ + 4e- ⇒ 2H2O

In neutral seawater O2+2H2O+4e- ⇒ 4OH-

In de-aerated soil or water2H2O+2e- ⇒ 2OH- +H2

Anode reactions At active anode in sacrificial anode CP system (Mg, Al, Zn)

M → M+n + ne-


KFUPM12CP of pipelines

Note that cathodic protection current will only protect external surfaces on buried structures, because the anode-electrolyte-cathode is at external surfaces. Above ground, structures cannot be protected by cathodic protection because the current discharged from the current source can not travel through the atmosphere (no electrolyte).CP is not usually used to protect internal surfaces of pipelines because of difficulty in placing anodes.internal surfaces of pipelines can be protected by: inhibitors, coatings, or by using a corrosion resistant alloy.


KFUPM13Protection Criteria

How much current is needed to protect the pipeline?

Little current will lead to ineffective protection High current will lead to disbonding of coatings and hydrogen embrittlement (more power consumption and higher cost)Experience show that we should keep the pipeline potential less than a protection potential.


KFUPM14Protection Criteria

In less corrosive soil, E< -0.850 mV wrtCu/CuSO4 reference electrode this reference electrode is used because it is less sensitive to temperature variation (0.318 s. SHE)In Saudi’s Aramco, the protection potential for cross-country pipeline is -1.1 V vsCu/CuSO4 (due to highly corrosive soil)More –ve potential means more current required and more operation cost.


KFUPM15Reference Electrodes

Common reference electrodes used in CPCu/CuSO4 in soil

CuSO4 + 2e- ↔ Cu+SO42-

E vs. SHE 0.318 V

AgCl in seawater AgCl + e- ↔ Ag + Cl-

E vs. SHE 0.222 V


KFUPM16Potential Protection

Why <-0.850 mV vs. Cu/CuSO4?From Pourbaix diagram, Fe is stable below -0.6 V vs SHECu/CuSO4 is more +vethan SHE by 0.318 VHence, Fe is stable and corrosion is minimum if potential is (-0.6-0.318= -0.918 V vs Cu/CuSO4)


KFUPM17Potential and Corrosion of Buried Steel

Severe overprotection (disbonding of coatings, hydrogen blistering, HE)

-1.1 to -1.4

Overprotection-0.9 to -1.1

Cathodic protection-0.8 to -0.9

Slow corrosion -0.7 to -0.8

Corrosion-0.6 to -0.7

Intense corrosion-0.5 to -0.6

Corrosion condition Potential (V vs. Cu/CuSO4)


KFUPM18NACE Standards for CP


KFUPM19

Saudi Aramco’s Potential Requirements

Structure Minimum Required Potentials

Buried Cross-Country Pipelines -1.10 volts versus CuSO4 electrode.

Buried Plant Piping, Tank -1.00 volt versus CuSO4 electrode.

Bottom Externals, -850 mV versus CuSO4 electrode.Isolated Buried Casings

Water Tank Interiors -0.90 volts vs. AgCl electrode

Marine structures -0.90 volt or more negative versus AgCl electrode


KFUPM20Soil Corrosivity

Soil is composed mainly of mineral particles (mainly SiO2).Soil is composed of a mixture of:

Fine sand (0.02-0.2 mm)Coarse sand (0.2-2 mm)Slit (0.002-0.02 mm)Clay (< 0.002 mm)

The soil particles are covered with thin surface film of moisture with dissolved salts and gases.The total volume of soil consists of solid particles and pores filled with moisture and air. Soils with a high proportion of sand have very limited storage capacity for water whereas clays are excellent in retaining water Air in the pores contains 10-20 times as much CO2 as atmospheric air. Soils with high moisture content, high electrical conductivity, high acidity, and high dissolved salts will be most corrosive.



Variables affecting soil corrosivity:Water is the electrolyte for electrochemical corrosion reactions Oxygen: the oxygen concentration decreases with increasing depth of soil pH: soils usually have a pH range of 5-8

Soil acidity is produced by decomposition of acidic plants, industrial wastes, and acid rain Alkaline soils tend to have high sodium, potassium, magnesium and calcium contents which form calcareous deposits on buried structures with protective properties against corrosion.

Chloride level: harmful for metals sulfate level: harmful for concreter



For CP design against corrosion, the most important property of a soil in determining its corrosivity is its electrical conductivity. The table shows soil corrosion severity ratings. Soil corrosion causes corrosion in underground petroleum storage tanks, pipelines, and water distribution systems.Soil resistivity is measured by Wenner 4-pin method

Very corrosive

Corrosive

Moderately corrosive

Mildly corrosive

Essentially non-corrosive

Corrosivity Rating

<1,000

1,000 to 5,000

5,000 to 10,000

10,000 to 20,000

Dry sand

Clay with saline water (sabkha)

>20,000

Soil resistivity (ohm cm)



Electrolyte Resistivity (ohm-cm)

Seawater (Gulf) 16Raw water 200-2000Drinking water 2000-5000

10,000

2,000

1,000500

Progressively LessCorrosive

Mildly Corrosive

Moderately CorrosiveCorrosiveVery Corrosive

Ohm-cm

Seawater Resistivity


KFUPM24

Current density required to reach Ecp for steel in moving and standing seawater and in soil

5-1111-3245-75160-270

Stagnant seawater

1.1-0.545.4-1111-1643-54soil

11-1632-5475-105325-375

Moving seawater

Applied CP current

Initial CP current

Applied CP current

Initial CP current

Coated steel (mA/m2)Bare Steel (mA/m2)Environment

ASM Handbook Vol#13 p.476


KFUPM25

Given a coated offshore structure with a surface area of 5,000 m2, 2/3 is immersed in seawater, calculate the amount of initial current and applied current necessary to cathodically protect the structure.The initial and applied current density requirement for coated seawater structures is 35 and 10 mA/m2.iinitial = 5,000*2/3*35= 116,666 mA = 117 A iapplied = 5,000*2/3*10 = 33,333 mA = 34 A

Current Calculation for Design


KFUPM26

Given a 150 m section of 0.75 m diameter steel pipe coated with fusion bonded epoxy (FBE), calculate the total amount of current required. Assume that 10% of the coating was damaged during installation. Assume that the required current density for FBE coated buried pipeline is 0.1 mA/m2 while for uncoated steel is 1 mA/m2 .

surface area = πDL=354 m2

iinitial = bare area*1 mA/m2 + coated area*0.1 mA/m2

= 354*0.1*1 + 354*0.9*0.1 = 67.3 mANote that if the whole pipe is not coated, then the

current requirement would be= 354*1=354 mA (5 times more than above)

Example


KFUPM27Coatings in CP Systems

Bare structures require more current than Bare structures require more current than coated structures coated structures Economical applications of CP for buried pipelines applied only for coated pipelines. Always assume 5-10% of coated area as bare due to damage during pipe installation.


KFUPM28Sacrificial Anode CP Systems

a more active metal than steels can act as a sacrificial anode.The galvanic series indicate that Mg, Zn, and Al are more active than steels.A number of anodes are electrically connected to the steel structure to be protected to provide the needed current.The amount of current output is increased by increasing the number of anodes.Usually applied in:

Low current requirement application Soil resistivity < 10,000 (Ω.cm)


KFUPM29Sacrificial Anode CP Systems

Advantages of sacrificial CP:No external power source neededEase of installation, low maintenance, low costProvides uniform distribution of current

DisadvantagesLimited current and power outputHigh resistivity environments or large structures require a large number of anodesPeriodic replacement of anodes


KFUPM30

VoltsMetal (referenced to Cu/CuSo4)

Commercially pure magnesium -1.75Magnesium alloy (6% Al, 3%, Zn, 0.15% Mn) -1.6Zinc -1.1Aluminum alloy (5% Zn) -1.1Commercially pure aluminum -0.8Mild steel (clean and shiny) -0.5 to -0.8Mild steel (rusted) -0.2 to -0.5Cast Iron -0.5Lead -0.5Mild steel in concrete -0.2Copper, brass, bronze -0.2High silicon cast iron -0.2Mill scale on steel -0.2Carbon, graphite, coke +0.3

Galvanic Series in soil and seawater


KFUPM31Sacrificial anodes

In soil, special backfills are used with sacrificial anodes to improve anode efficiency.Anodes are packaged in porous bags prefilled with backfill materials such as clay. Clay:

absorbs moisture from the soil and reduce anode resistance of anode/electrolyte distribute the anodic reaction all over the anodeIncrease the life of the anode


KFUPM32Sacrificial anodes

Anodes are packaged in bags filled with backfill material Commercial anodes (

60 inch in length4 Kg

Anodes for buried structures (pipes, tanks):

Pure Mg Mg alloy (Mg+6Al+3Zn+0.2Mn)Pure Zn

For marine applications Al alloy containing 5% Zn is used Zn alloy


KFUPM33Design of Sacrificial Anode CP

Select Protection Criterion (depends on the environment)For buried steel (NACE standard gives protection potential = -850 mV vs. Cu/CuSO4 )

Measure resistivity of environment (slide#34):Soil resistivity ranges from (500-20,000 Ω*cm) Seawater ranges from ( 10-50 Ω*cm)

Estimate cathodic current requirement which depends on the environment and the surface area to be protected using either:

Current requirement table (see slide#24)Current requirement test (see slide#35)

Select a suitable sacrificial anode and calculate the theoretical capacity and the driving voltage (slide#37 and 38)Estimate the number of anodes needed based on groundbed resistance (slide#41 and 42)Estimate anode life and replacement period


KFUPM34Wenner 4-pin Method to Measure Soil Resistivity

This method is done by placing four pins at equal distances from each other. A current is passed through the two outer pins using a power supply. the voltage across the two inner pins is measured using a voltmeter. the resistance can be calculated using Ohm's law (Resist = ∆V/I). Soil resistivity = 191.2*∆V/I*d (d in feet) ohm-cm, where R is the soil resistance and d is the pin spacing in feet.

Power supply

voltmeter


KFUPM35Current Requirement Test

The current may be increased gradually until the voltmeters at positions A and B reaches -0.85 V with respect to a copper sulfate reference electrode placed directly above the pipe. Current requirement test:

A small DC power system is used (10 A)A temporary anode ground bed is installedPotential loggers are installed at selected test locations to monitor potentialsA current is applied and the potential is measuredThe current that brings the potential of the whole pipe below the protective potential is used the required current for protection


KFUPM36Sacrificial Anode

Anodes must have:High driving potential to generate sufficient current Stable operating potentials over a range of current outputs (Eanode does not vary a lot with i) High capacity to deliver current per unit massDoes not passivate Theoretical capacity: the total charge in coulombs produced by the corrosion (dissolution) of a unit mass of the anode material [units in (A* hr)/Kg]. High Efficiency (efficiency = actual capacity/theoretical capacity*100)


KFUPM37Calculating the theoretical capacity

MW of Mg is 24.3 g/mol, density=1.74 g/cm3Mg→Mg+2 + 2e- (one mole of Mg produces 2 moles of electrons)Take 1 Kg of Mg as a basis:

1000g * mole/24.3g=41.2 mole of Mg# of e- mole= 2*41.2=82.4 moles of e-

82.4 moles of e- *96500 Coulomb/(mole e)= 795,1600 Coulomb/(Kg of Mg) 795,1600 Coulomb/Kg *1 hr/3600s=2,200 (A.hr)/Kg


KFUPM38Calculating driving potential

ED = Eanode – Ep + Epolar

ED = driving potentialEanode = anode potentialEp = protection potentialEpolar = change in potential of anode due to current flow (polarization); usually taken as 0.1 V

ED


KFUPM39Driving Potential

Example: calculate the driving potential for Mg in soil assuming:

Eanode = -1.75 vs Cu/CuSO4Ep (buried pipeline cross country) = -1.0 VEpolar = 0.1 VED = -1.75 – (-1.0) +0.1 = -0.65 V

Example: calculate the driving potential for Al alloy (5%Zn) in soil assume Ep = -0.85 V and Epola=0 and Eanode=-1.1V

ED = -1.1- (-0.85) = -0.25 V


KFUPM40Sacrificial Anodes

-1.1 -1.1-1.75 V (vsCu/CuSO4)

potential

>90%>90%50-60%Efficiency

26407801232Actual capacity

29808102200 (A*hr)/Kg

Theoretical capacity

AlZnMg


KFUPM41Anode bed (groundbed) resistance

In both systems, the flow of current is analogous to a simple resistive circuit.The highest resistance to current flow is due to the anode/electrolyte resistance (Rab)RS (structure/electrolyte) resistance. RLW (lead wire) resistance. Rab = Resistance of anode/electrolyte; depends on the anode shape and the resistivity of the environment. Rtotal = Ra+RLW+RS

(RS and RLW) can be neglectedRtotal ≈ Rab

R

Battery

Resistor

E

Electric CircuitI

Rtotal


KFUPM42Anode bed (ground bed) resistance

Dwight’s equation for single vertical anodes:Ra = ρ/(2πLa)*(ln(8La/Da) -1)(for slender anodes mounted at least 0.3 m away from the steel structure)

La= length of anode (cm)ρ = soil resistivity (Ω.cm)Da = anode diameter (cm)

Anode current output i = ED/Ra

C

A

Rab


KFUPM43Anode bed (ground bed) resistance

Example: Calculate the maximum current output for zinc anodes (L = 150 cm and D = 15 cm) in sacrificial anode CP system. Assume very corrosive soil (ρ=1000 Ω*cm)

1. Calculate the driving potential assuming Epola = 0.1V, EZn = -1.1 vs Cu/CuSO4, Ep (buried pipeline) = -0.85 V

ED = -1.1 – (-0.85)-0.1 = -0.15 V

2. Calculate the ground bed resistanceRa = ρ/(2πLa)*(ln(8La/Da) -1)Ra = 1000/(2π*150)*(ln(8*150/15)-1) = 3.6 Ω

3. Calculate the maximum current output from each anodei = ED/Ra = 0.15/3.6 = 0.042 A = 42 mA


KFUPM44Anodes distribution

Example: suppose you need 1 A to protect a steel structure by using Zn anodes as in the previous example. Each anode provides 0.042 A.Then, you need to have 1/(0.042) ≈ 24 anodes to give sufficient protection. Distance of anode to cathode:

Too far: high resistance in the soil leads to voltage dropToo short: current distribution is not uniform Needs experience (usually less than a meter)


KFUPM45Design Example

Example: design a CP system for a section of coated steel buried pipe assuming that the current required to shift the pipeline potential to the EP was approximated by a current requirement test to be 500 mA. Zn anodes (L=150 cm, D=15 cm) are available. (Density of Zn = 7.14 g/cm3). Assume Zn efficiency is 90% and ρ=1000 Ω*cm.

From the previous example, each Zn anode produces 42 mA. Then, #anodes = 500/42 = 12 anodes.Total mass of anodes = 12*vol*density =12*3.14*D2*L/4*7.14 g/cm3= 2270 KgTotal charge available = efficiency*theoretical capacity*mass= 0.9*810*(A.hr/Kg)*2270 Kg= 1,593,540 (A*hr)Replacement period = 1,593,540 (A*hr)/0.5A= 189540 hr = 364 years


KFUPM46Sacrificial Anode CP System

Install 12 anodes evenly distributed along the pipeline. Keep each anode 30-100 cm away from the pipe. The system design life is indefinite. Monitoring Sacrificial CP

Measure the potential of the pipe and make sure it is -850mV Monitor the current flow from each anode at the junction box


KFUPM47Applications: buried tanks, pipelines, internal protection of heat

exchangers and vessels, ship hulls, marine structures

Aluminumalloy anode

AA-036348

Anodes


KFUPM48Impressed Current CP (ICCP) System

ICCP is used if: high current is required high resistance electrolyte


KFUPM49ICCP

Components of ICCP system

A transformer: to reduce the voltage from high to low voltageA rectifier to convert AC to DC A current distributor (junction box)Anodes with backfills


KFUPM50Reactions

The anode potential is set at high +ve potential and the following oxidations reactions become possible:

2H2O→ O2 + 4H++4e- (in water or in wet soil)2Cl-→Cl2 + 2e- (in salt or brackish water)C+2H2O→ CO2 + 4H++4e-

(in graphite anodes)Reactions at the cathode:

In aerated wet soilO2+2H2O+4e- ⇒ 4OH-

In aerated acidic solutionO2+4H+ + 4e- ⇒ 2H2O

In neutral seawater O2+2H2O+4e- ⇒ 4OH-

In de-aerated soil or water2H2O+2e- ⇒ 2OH- +H2

anodeCathode


KFUPM51Advantages and Disadvantages of

Impressed Current SystemsAdvantages

Higher Current and power outputsAdjustable protection levels (controlled current)Large areas of protectionLow number of anodesCan be used to protect poorly coated structure

DisadvantagesComplex equipment and installation costsHigher maintenance costsPossible interference problems with foreign structuresRisk of incorrect polarity connections


KFUPM52Components of ICCP

An external power sourceAC transmission linesA solar power system

Anodes are not necessarily more active than the structure to be protected

Two types of anodes inert or non-consumable anodes: platinized anodes (a few micrometers thick coating of platinum on Ti or Niobium), graphite, consumable anodes (scrap steel, high-Si Cr cast iron)


KFUPM53

Anode to coke resistance

Coke-to-earthResistance

SoilCokebreeze

Anodes for ICCP

Anodes are used with carbonaceous backfill called coke-breeze to:

increases the effective size of the anode lowers the anode-to-ground resistance.extends the life of the anode.

Anode


KFUPM54Anodes in ICCP

applicationsConsumption rate (Kg/(A*yr)

Material

Seawater, concrete,

8x10-6Platinized niobium (inert)

Marine, soil 1-0.25High Si-cast iron

Marine, soil 7-9Scrap steel

Soil,Potable water

0.1-1Graphite (inert)

= =8x10-6Platinized Ti (inert)


KFUPM55Applications

Ships (with coatings)Offshore platformsBuried pipelines (prefmethod)Oil well casingConcrete Structures (offshore bridges)


KFUPM56Buried pipelines


KFUPM57

Producing Zone

Remote Surface Anode Bed

Junction Box Rectifier

Perforations

Well Casing


KFUPM58ICCP Design

1. Evaluation of electrolyte resistivity2. Estimating the current requirement

Current requirement testCurrent requirements theoretical estimation

3. Selecting anode material and current distribution

Uniform current distribution Avoid interference (stray current)

4. Determine the anode bed ground resistance 5. Determine number of required anodes6. Select the power source capacity


KFUPM591. Electrolyte Resistivity Survey

Measure soil resistivity along the pipeline The data from a soil resistivity survey along a 6 km section of pipeline is shown below. The lowest effective soil resistivity points are the most favorable anode bed locations.


KFUPM602. Interruption Test for Bare Pipeline in Impressed

Current System

In impressed current system for bare structure, the applied current is high and IR drop can not be neglected. Thus, protection Criteria for bare pipeline must check I*RΩ effect.The protection criterion for bare steel pipeline uses interruption test where a negative (cathodic) change in potential of >300 mV must take place immediately after CP current is applied. potential

time

CP Power off

CP power on

300 mV


KFUPM613. Current distribution

Variation of electrolyte resistivity b/w anode and cathode (largest current flows along least resistant path)Defects in coatings: current concentrates at defects


KFUPM62Stray Currents

Stray Currents: currents flowing in the electrolyte from external sources other than the applied CP. Sources of stray currents:

Subway systemInterference with another CP systemWelding equipmentElectrical power transmission lines


KFUPM634. Number of Required

Anodes

N = number of impressed current anodesY = design life in yearsI = total current required in amperesC = anode consumption rate in kg/A-yrW = weight of a single anode in kg

N = Y*C*I/W


KFUPM645. Resistance calculation

Dwight’s equation for single vertical anode:

Ra = ρ/(2πLa)*(ln(8La/Da) -1)A group of vertical or horizontal anodes (buried 6 ft below ground):

Vertical anodes ( Rv = ρ*F/537)Each anode is 8-12 in in diameter and 10 ft in length

Horizontal anodes (RH = ρ*F/483)Each anode is 10 ft in length and 6 ft below surface

F is called adjusting factor (F=1 for single anode)


KFUPM65F : Adjusting Factor


KFUPM664. Power Source Selection

The size of the power source is determined by:

the amount of current required to protect the structure (I)the voltage required to force the current through the anode ground bed resistance (R)E=I*R


KFUPM67Example: Design for ICCP

Design an impressed current system to protect a buried pipeline coated with fusion bonded epoxy (FBE) using the following information:

Horizontal anodes 6 ft below ground with 20 ft spacing

Anode material: High silicon-cast iron ( C=0.5 Kg/(A*yr))

Anode dimensions with backfill: 25 cm dia. x 300cm, weight=50 Kg

Pipeline length: 500 m and 115 mm in diameter

the anode to soil resistance is 0.24 ohm

Neglect cable resisitivity.

Soil resistivity: 2,000 ohm-cm

Required current density is 0.2 mA/m2 for FBE.

Design life of 20 years


KFUPM68Example: Design for ICCP

Current required = surface area * current density= π d L *i= 3.14*0.115*500 *0.2 = 36.1 A

# anodes= Y*I*C/W = 20*36.1*0.5/50 = 7.2 anodes . Then use 8 anodes.To calculate the resistance:

RH = ρ*F/483 (for 8 horizontal anodes F=0.184)= 2000*0.184/483 = 0.76Ω

Total R= 0.76+0.24=1 ΩE = I*R=1*36.1 = 36.1 V Hence, use a DC power with a minimum current supply of 40 A and a minimum voltage of 40 V (1600 Watt rating).