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CORROSION ENGINEERING. PLEASE ANSWER A AND B ASAP. I WILL LEAVE A GOOD RATING IMMEDIATELY IF CORRECT! A corrosion pit in SS316 exposed to chloride
CORROSION ENGINEERING. PLEASE ANSWER A AND B ASAP. I WILL LEAVE A GOOD RATING IMMEDIATELY IF CORRECT!
A corrosion pit in SS316 exposed to chloride solution grows to a depth of 1.5mm in about one month. The density of the alloy is 7.90g/cm3.SS316 is composed of approximately 18%Cr,12%Ni,3%Mo,2%Mn,1%Si, balance Fe. Please assume that the corrosion current density remains constant. a) Please calculate the current density ipit in the inner pit of SS316 b) The electrochemical potentials of alloying metals in chloride solution are listed in the table below. Please evaluate which alloying metals might increase the resistance to pitting in chloride solution of plain steel and which alloying metal would eventually increase the susceptibility for pitting corrosion. Table: Electrochemical potentials of chloride formation Cu+2Cl=CuCl2+2eNi+2Cl=NiCl2+2eFe+2Cl=FeCl2+2eCr+3Cl=CrCl3+3eV+3Cl=VCl3+3eTi+3Cl=TiCl3+3eZr+4Cl=ZrCl4+4eMn+2Cl=MnCl2+2eBe+2Cl=BeCl2+2eMg+2Cl=MgCl2+2eECuCl2=+0.203V,ENiCl2=0.298V,EFeCl2=0.452V,ECrCCl3=0.593V,EVCl3=0.621V,ETiCl3=1.025V,EZrCl4=1.152V,EMnCl2=1.174V,EBCCl2=1.310V,EMgCl2=1.956V. A corrosion pit in SS316 exposed to chloride solution grows to a depth of 1.5mm in about one month. The density of the alloy is 7.90g/cm3.SS316 is composed of approximately 18%Cr,12%Ni,3%Mo,2%Mn,1%Si, balance Fe. Please assume that the corrosion current density remains constant. a) Please calculate the current density ipit in the inner pit of SS316 b) The electrochemical potentials of alloying metals in chloride solution are listed in the table below. Please evaluate which alloying metals might increase the resistance to pitting in chloride solution of plain steel and which alloying metal would eventually increase the susceptibility for pitting corrosion. Table: Electrochemical potentials of chloride formation Cu+2Cl=CuCl2+2eNi+2Cl=NiCl2+2eFe+2Cl=FeCl2+2eCr+3Cl=CrCl3+3eV+3Cl=VCl3+3eTi+3Cl=TiCl3+3eZr+4Cl=ZrCl4+4eMn+2Cl=MnCl2+2eBe+2Cl=BeCl2+2eMg+2Cl=MgCl2+2eECuCl2=+0.203V,ENiCl2=0.298V,EFeCl2=0.452V,ECrCCl3=0.593V,EVCl3=0.621V,ETiCl3=1.025V,EZrCl4=1.152V,EMnCl2=1.174V,EBCCl2=1.310V,EMgCl2=1.956VStep by Step Solution
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