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Transcript of anode cal
7/31/2019 anode cal
http://slidepdf.com/reader/full/anode-cal 1/1
The following material was researched and was submitted as a guideline for calculating the
requirements of sacrificial anodes for your application:
1. a) Calculate the AREA to be protected.
2. b) The national Association of Corrosion Engineers, NACE standard RP-01-69 states the
requirement of a polarization to -0.85V versus Cu-saturated CuSO4, for protection of asteel structure in a neutral environment. This is known as the POLARIZED
POTENTIAL. The sacrificial anode system design requires the current density to achievethis potential.
3. 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 isexposed to since the "current demand" may vary with different combinations.
4. 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.5. 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.6. f) Calculate the ANODE RESISTANCE from the distribution and quantity of thesacrificial anodes.
7. 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.
SAMPLE CALCULATION
1. 1) Assume that the calculated Area of the ship's hull/steel surface to be protected is:
480m2.2. 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 willassume a current density requirement of 35mA/m2.
3. 3) From step C, the Current Demand is: 35mA/m2 X 480m2 = 16800mA.
4. 4) The output for zinc sacrificial anodes was determined to be 810 Ah/kg, with anefficiency 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. 5) For an anode supply of 4 years, 808 kg will be required. (202 kg/year X 4 years)
6. 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 youthe weight in pounds.