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 1. Steels used in platform and pipework construction are not pure materials.  Although iron(Fe) forms the major part, steel contains a certain quantity of carbon (C) up to about 1%.  The amount of carbon
determines the type of steel – for example from mild steels up to hard tool steels.  The carbon takes the form of iron carbide (Fe3C), and, because it is not spread quite evenly through the steel, there are adjacent surface areas of pure iron and iron carbide throughout the steel structure. These individual areas are minute in size and are very close together a matter of thousandths of an inch.
2. Iron carbide has a lower potential than pure iron when in the presence of conducting electrolyte (such as seawater or damp soil).  That is to say, a mass of iron/iron-carbide “cells” are formed in electrolyte, with the pure iron areas positive and the iron-carbide areas negative.  Each cell, which can be regarded as a primary battery, is short-circuited by the structure metal work
3. At the iron surface some of the electrons surrounding the iron atoms break away, leaving the atoms of iron positively charged (Fe++).  These are called iron“ions”, and they are, in fact, positively charged atoms of the metal.  They travel through the sea  or soil down the potential gradient from the positive iron areas to the negative iron-carbide areas, forming an electric current through the sea or soil as they do so. This current returns through the main structure metal conducting path, as shown in Figure below:

4. The positive areas from which current flows into the sea or soil are called “anodes” (or anodic areas) and the negative areas into which the current flows are “cathodes” (or cathodic areas).  As the iron metal is removed, atom by atom, from the anode and deposited at the cathode, a corrosion pit is formed as shown above.  Hundreds of thousands of such pits are formed, which appears as general corrosion of the structure member or pipe.  This process can be slowed by painting or by other forms of encapsulation, but it appears increasingly as the  covering wears off.  On the other hand marine growth may tend to inhibit the process with time, but noreliance can be placed on it.

5. Since corrosion is due to the outflow of iron atoms carried by the sea or soil current, it can be reduced or even prevented if such
currents could be stopped or reversed, so causing all the submerged or buried metal parts to receive current instead of giving it out – that is, by making all areas cathodic.  This can be done by placing other, independent electrodes in the sea or soil near the structure and causing them to force current into the structure.

6. There are two quite distinct methods of achieving this, known as the “sacrificial anode” and the “impressed-current” systems.  Either method may be used on an offshore installation and on either type of platform, steel-jacket or concrete; in the latter the well risers and their guide-tube structures still have to be protected.

There are two quite distinct methods of achieving this, known as the “sacrificial anode” and the “impressed-current” systems.  Either method may be used on an offshore installation and on either type of platform, steel-jacket or concrete; in the latter the well risers and their guide-tube structures still have to be protected.

Zery Engineering Company Ltd offers design, consultancy and installation services for Cathodic Protection of Pipelines and Vessels in oil and gas exploration and production.

The following material was researched and was submitted as a guideline for calculating the requirements of sacrificial anodes for your application

(a) Calculate the AREA to be protected.

(b) AMPP corrosion and coatings,
AMPP standard RP-01-69 states the requirement of a polarization to -0.85V versus Cu-saturated CuSO4, for protection of a steel structure in a neutral environment. This is known as the POLARIZED POTENTIAL. The sacrificial anode system design requires the current density to achieve this potential.

(c) The CURRENT DEMAND is calculated by multiplying the required current density by the area. Note that it is important to consider the environment that the metal is exposed to since the "current demand" may vary with different combinations.

d) The next step is to determine the total required mass of the sacrificial
     anodes. The ANODE CONSUMPTION is determined from tabulated consumption rates for the calculated current demand.
(e) Divide the total required mass by an appropriate quantity of anodes that will create a uniform current distribution over the entire area to be     protected.

(f) Calculate the ANODE RESISTANCE from the distribution and quantity of the sacrificial anodes.

(g) From the anode resistance, R and voltage, V from the selected sacrificial anodes, the DESIGN OUTPUT CURRENT, I is calculated using the following formula: I = V/R

Note that the output current should meet or exceed the required current from step C. 


(1)Assume that the calculated Area of the ship's hull/steel surface to be protected is: 480m2.

(2) The current density may be obtained from tabulated values found in the following text: {F.W. Hewes, Cathodic Protection Theory and Practice, V.Ashworth and C.J.L. Booker, eds., Wiley (Horwood), Chichester, West Sussex, 1986} For this example, we will assume a current density requirement of 35mA/m2.

(3) From step C, the Current Demand is: 35mA/m2 X 480m2 = 16800mA.

(4) The output for zinc sacrificial anodes was determined to be 810 Ah/kg, with an efficiency of 90% usually applied, thus yielding an effective output of 729 Ah/kg. Therefore the consumption is calculated as (729 Ah/kg) /(16.80 A) = 43.4 h/kg. Now, there are 8760 hours per year, thus the amount of zinc required is: (8760 h/year) / (43.4 h/kg) = 202 kg/year.

(5) For an anode supply of 4 years, 808 kg will be required. (202 kg/year X 4years)

(6) Next, calculate the anode resistance and determine the output current and compare the output current to that calculated in step 3.

*Remember to distribute the anodes evenly over the entire area to be protected.

**To Calculate from Kilograms to Pounds, multiply the kg's by 2.2. This will give you the weight in pounds. 

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