|5. Cathodic Protection|
5.3 Cathodic Protection Systems [7/8]
Current Requirement for Coated Pipes
It is to be remembered that the current requirement test for a coated structure is different from that of a bare pipe in soil, the reason being a very high contact resistance between the external surface of the coated pipe and the earth, which is not the case with the bare (uncoated) pipe.
A temporary station is set up for current requirement test at the same location as specified for the permanent cathodic system. A temporary layout for cathodic protection current requirement test is shown below.
Storage batteries or portable DC generators can be used as a current source.
|Typical portable current sources:|
A temporary ground bed of steel bars driven in ground and of parallel configuration can be made.
With the copper switch closed, the current is gradually increased until the voltmeter shows a potential of -0.85 volts with respect to a Cu-CuSO4 electrode. After the output current has been adjusted to a satisfactory value, the pipe to soil potential is measured at selected locations. Measurements are taken in the 'OFF' and 'ON' positions to determine the effect of current. For pipe to soil potential measurements, the reference electrode is placed over the soil over the pipe because of the high contact resistance of the coated pipe in contact with the soil. The total amount of current required for the cathodic protection of pipeline of any length can then be estimated from the current requirement test, once the magnitude of the current required to raise the potential of the experimental pipe to -0.85 V is determined by the test.
Suppose the metal current requirements to protect a 2 mile section of a 6.625 inch pipe is to be determined.
The test set up as described above, test data of current applied (amp) Vs potential measured (against the Cu-CuSO4 electrode) is obtained. Suppose that 0.06 amp of current was required to raise the potential to -0.85 volts.
|First calculate the area of the experimental coated pipe|
|Area = p D L|
|= 3.1415*(6.625/12)*(5280*2) ft2|
|= 18344 ft2|
|The current density required for protection is therefore|
|i = (total current required) / (total area of the pipe)|
|= (0.06) / 18344 A/ft2|
|= 0.00327 mA/ft2|
|which is the estimate of the current required.|