ECHE 571/Fall 2010 Problem Set #9 Impressed Current Cathodic Protection
1) For a steel immersed in corroding acidic solution, calculate
a) Corrosion current and corrosion potential b) Impressed current density I app required to bring down the corrosion to 10-6A/cm2. Also calculate the minimum Impressed current density required for no corrosion to occur. c) Cathodic polarization required to bring down the corrosion rate from 460 mpy to 0.46 mpy.
Given Io anodic = 10 -6 Io cathodic = 10 -8
bc = -0.10 V
ba = 0.040 V
2) Iron corrodes in a solution saturated with oxygen. The pH of the solution has a significant role in the corrosion of iron and hence the applied current needed to protect the iron. To analyze the effect of pH calculate the following: a) Calculate the corrosion current and corrosion potential when the solution has a pH of 14, 10 and 7 b) To bring down the corrosion current to the order of 10-5, calculate the impressed current needed to be applied? c) Does the impressed current needed increases / decreases with pH? If so, why? Given
ba = 0.1, bc= -0.1 pO2 = 1 atm, [Fe2+] = 0.5 M -7 Exchange current density for oxygen reduction io = 10 - Exchange current density for iron dissolution io= 10 3) During cathodic protection in sea water conditions, the generation of the OH-, leads to the calcareous deposits by reaction with calcium and magnesium ions? a) List down the reactions for the scale formation. Does the limiting current density of oxygen increases or decreases with scale formation? b) After 6 months, the limiting current for oxygen reduction changes by an order of two due to the formation of deposits, Calculate the current density required for the protection of the substrate, before and after formation of the calcareous deposit? c) If we don’t change the applied current density, will it lead to over protection or under protection?
Given: The current for oxygen reduction i= 10-3 A/cm2. In the absence of calcareous deposit. -6 2 ba = 0.04 V, exchange current density for Fe dissolution is io= 10 A/cm .