Anodes for ships, bridges, marine structures, industrial process equipment and municipal water supply lines are examples of anodes being effectively installed as part of a cathodic protection system.
How do I calculate the number of anodes for my system? There are some fundamental steps, which must be applied in the design of a sacrificial anode system. 1) Area to be protected needs to be calculated. 2) Polarized potential. NACE Standard RP0169 states that for a steel structure in a neutral environment, polarization to 0.85 V versus Cusaturated CuSO_{4}, is required to protect the structure. The current density needed to achieve this potential is used in the design of the system. 3) Current demand. This is calculated as a product of required current density and area. The current demand varies for steel in various environments, thus it is important to assess the different environments steel may be exposed to and complete separate calculation for each environment. 4) Anode consumption. Total required weight of anodes is determined from known consumption rates for the calculated current demand. Consumption is necessary in order to supply current for protection. 5) Anode number and distribution. The weight of anode material consumed must be divided into an appropriate number of anodes, which are distributed around the structure to give as uniform current distribution as possible. 6) Anode resistance needs to be calculated from the number and distribution of anodes. 7) Design output current. From the anode resistance, R, and driving voltage, E, of the selected sacrificial anodes, the design output current can be calculated from Ohm's law, E = IR. In a proper design, the output current will at least match or exceed the required current calculated in Step 3.
Example: Calculations for a buried 6" diameter steel pipeline. 1) If we assume the pipeline length to be 100 metres, and the O.D. of the pipe to be 0.17 m. The area to be protected is the outside area of the pipe. We will assume the pipeline is uncoated, but coating will alter the calculations. The area is: 53.4 m^{2}. 2) Tables of current density requirements have been found to be in the range of 10  60 mA/m^{2}. (F.W. Hewes, Cathodic Protection Theory and Practice, V. Ashworth and C.J.L. Booker, eds., Wiley (Horwood), Chichester, West Sussex, p. 226, 1986.) For our example we will assume a current density requirement of 40mA/m^{2}. 3) Current demand = 53.4 m^{2} X 40mA/m^{2} = 2136 mA 4) The output for zinc anodes has been determined at 810Ah/kg, and the efficiency is normally taken as 90%. Thus, the useful output of zinc is 729 Ah/kg. Thus the consumption is: =729 Ah/kg / 2.136 A = 341 h/kg. Since, there are 8760 h/yr the weight of zinc per year is: 8760 h/yr / 341 h/kg = 25.7 kg/yr. 5) The anodes would need to be distributed equally to distribute the current along the pipeline as evenly as possible. For a system life of 20 years the zinc required would be 514 kg or 1134 lbs. Therefore, a 12 lb zinc anode every metre would provide 20 years of cathodic protection to the pipeline. 6) The final step in the design would be to calculate the anode resistance and then determine the output current from Ohm's Law, which should be equal to or greater than the output calculated in Step 3.
