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Iron is corroding at a current density of 1.69*10^-4 amp/cm^2. What would be the corrosion rate in MDD?(a) 773(b) 723(c) 423(d) 473The question was asked in final exam.My question is based upon Corrosion Rate Expression in portion Corrosion Principles of Corrosion Engineering

Answer»

The CORRECT answer is (C) 423

The explanation is: We know MDD = W/At = (I*am) / (n*F). We should convert the given unit values in milligrams/sq. Decimeter/day.

Given, I = 1.69*10^-4 amp/cm^2

Atomic weight of IRON = am = 55.86 g/mol

n = number of ELECTRONS = 2 [Fe ==> Fe^+2 +2e^–]

F = faraday’s constant = 96500 coulombs

= (1.96×10^-4 coulombs/sec*cm^2) × (55.86) ÷ (2×96500 coulombs)

= [(1.96×10^-4)*(60*60*24)*(100)*(55.86 *10^3)] / [2*96500]

= 422.53 MDD.



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