5. Cathodic Protection

5.3 Cathodic Protection Systems [8/8]

Current Requirement for Bare Structures

An uncoated structure might take a very long time to attain a potential of -0.85 V as compared to a coated structure which is rapidly polarized. In a bare structure, the potential readings become increasingly more negative with time due to the development of a film due to polarization. A considerable time may therefore be required to achieve the protection potential while measuring the pipe to soil potential of a bare pipe. The reference electrode (Cu-CuSO4) is to be placed at a remote location due to the resistance of the mass of the earth surrounding the pipe. The resistance between the bare pipe and the earth comprises of the resistance of the bare pipe to earth and the surrounding mass of earth. The location is determined by moving the Cu-CuSO4 electrode away from the structure until no change in negative potential reading is observed which is the desired remote location. It is basically an area outside the anode-to-earth and pipe-to-soil influence. The remote distance is generally 100 ft. The set for current requirement is shown below. 

The basic circuitory is the same as for coated pipes. As considerable length of time would be required to achieve  the full polarization of -0.85 volts, the process is shortened by obtaining a a polarization curve until a potential close  to -0.85 volts is obtained (say -0.82 volts) by a known amount of current. The additional current required to achieve -0.85 volts could be calculated by extrapolation of the curve.


The current applied is, for instance, 20 amp and a potential of -0.82 volts is obtained after 40 hours. The original potential before applying cathodic protection was for instance -0.557 volts.

DE = -0.82 - (-0.557) = -0.263 V
The average potential change per ampere is DE/DI = 0.263/20 = 0.013 Volt/amp
The additional current needed to raise the potential to -0.85 Volt is obtained by extrapolation:
Required increase in potential = -0.85 - (-0.82) = 0.03 V
Additional amount of current = 0.03 / 0.013 = 0.2307 amp
The amount of current required = 20 + 0.2307 = 20.2307 amp