EXAMPLE Design of a Galvanic Anode Cathodic …

1 Begun on 2 Aug 04 Revised on 9 Aug 05 Printed on9/4/20051 EXAMPLED esign of a Galvanic Anode Cathodic protection system For a PipelineAssumptions:Multiple Anode Groundbed system with Magnesium AnodesPipe length is 2500 diameter is 6 in. OD = quality is 98%DesiredCurrent densityis2 Resistivity is 1000ohm-cmAnode type is standard potential size is 17 length is 2 diameter is spacing is 20 life is 20 and Design :(Blue itemsin this examplerefer to thecorresponding items as detailed in the companion document titled Design of a Galvanic Anode Cathodic protection system For a Pipeline.)

2 Referenced formulas refer to formulas listed inthe same document)(Step 1)Surface Areais 4334 (Step 2)Current Requirement is Amps(Step 3)Dwight s Formula for a Single Vertical anodeyields a groundbed resistance of (Step 4)Current output is . which is less than therequired current.(Step 5)Divide the current requirement by the above current output to obtain the number ofanodes necessary to produce a current greater than or equal to the required current. In from which we conclude that two anodes are required.(Step6)Sunde s Formula for two anodes produces a groundbed resistance of ohms.

3 (Step7)Then the current output is Amps.(Step8)Expected life is years.(Step9)An expected life of8 years and Formula 5aproducesa driving voltage of volts.(Step10)ThenFormula 5bproducesa polarized potential of volts.(Step11)Using Formula 7, current output is calculated to Amps.(Step12)Formula 3a produces an years.(Comment 3)Therefore aminumum oftwo anodeswill protect the pipe. However, a two anodesystem will provide protectionforno more than8 years.(Comment4)The next phase of the process is designed to determine the number of anodesrequired to satisfy both the current requirement criteria and the Design life criteria.

4 This part ofthe process will result in another set of output extend the life of the number of anodes to add in afirstattempt is calculated withBegun on 2 Aug 04 Revised on 9 Aug 05 Printed on9/4/20052(Step14)(20)(2)58from Step 5from Step 12( Design life)(number of anodes)(years of expected life)x (Step15)For a five Anode system Sunde s Formula yields a groundbed resistance of ohms.(Step16)Formula 5awith Design lifethen yieldsadriving voltage of volts.(Step17)Formula 5b producesa polarized potential of volts.(Comment5)According to much of the literature and the official position of NACE technicalcommittee the maximum polarized potential of a steel pipe will be Volts.

5 (Step18)We therefore restrict the polarized potential to volts.(Comment6)To achieve a life equal to the Design life, the pipe would have to polarize beyond themaximum polarized potential for steel of volts. Therefore all following calculations arebased on polarized potential for the pipe of volts. This will result in a system life shorterthan the Design life.(Step19)With Formula 5b the driving voltage isrecalculatedto be volts.(Step20)Formula 6 is used to calculatecurrent output to be Amps which exceeds thecurrent requirement.(Step21)Expected life is then calculated to be a five Anode system should protect the pipe for 16 years.

6 The output current shouldbe Amps and the pipe should polarize to volts.(Step22)Because the output current is higher than required and the system wanted to polarize the pipe higher than the maximum we consider a system offouranodes to see if protection canbe achieved for the same length of time.(Loop through Steps 15-22)For a four Anode system Sunde s Formula gives a groundbed resistanceof gives a driving voltage of and a polarized potential of polarized potential is reset to the maximum of volts and driving voltage isrecalculated to be voltsCurrent output in this system is andexpectedlife a four Anode system will protect the pipefor the same number of years (16), the pipewill polarize to the same potential ( volts),and the current output will be greater than butcloser to the required current.