Advanced Corrosion Engineering Questions and Answers on “Corrosion Rate Expression”.
1. What is meant by MPY in corrosion?
a) Mils penetration per year
b) Meter penetration per year
c) Milli meter penetration per year
d) Micro meter penetration per year
Answer: a
Clarification: MPY in corrosion is abbreviated as Mils penetration per year (1 mil = 1/1000 inch). It is usually used in the U.S. measurement system in order to estimate the corrosion rate of pitting corrosion.
2. Which of the following expression is used to calculate MPY?
a) MPY=(543*W)/DAT
b) MPY=(534*W)/DAT
c) MPY=W/DAT
d) MPY=DAT/(534*W)
Answer: b
Clarification: W = weight loss in milligrams, D=Density of a specimen (g/cm3), A=Area of specimen(sq.in), T=time of exposure in hours and 534 is a conversion factor. MPY means mils penetration per year I.e. milli inches penetration per year.
3. MDD in corrosion is abbreviated as ______
a) milligrams/sq. decimeter/decade
b) millimeters/sq. decimeter/day
c) milligrams/sq. decimeter/day
d) microns/decimeter/decade
Answer: c
Clarification: MDD in corrosion known as milligrams/sq.decimeter/day. It is the common unit to measure weight loss of metal per unit area. It is usually used for uniform and galvanic corrosion.
4. MPY is used to measure weight loss per unit area.
a) True
b) False
Answer: b
Clarification: Usually mils per year is used to measure the penetration rate. Penetration is the average depth of metal loss in the specimen. MPY is usually used for pitting corrosion. MDD (milligrams/sq. Decimeter/day) is used to measure the weight loss per unit area.
5. _____ is/are used to measure the extent of pitting corrosion.
a) MPY
b) MDD
c) Both MPY and MDD
d) MPK
Answer: a
Clarification: Pitting corrosion involves loss of metal at the selected spots on the surface of a metal. It penetrates into various depths based on the concentration of corrosive medium and nature of metal. MPY measures penetration range.
6. Which of the following law is used to derive the corrosion rate expression?
a) Newton’s law
b) Henry’s law
c) Raoult’s law
d) Faraday’s law
Answer: d
Clarification: Faraday’s law is used in order to derive the corrosion rate expression [W/At=I*am/n*F]. W/At denotes the corrosion rate, I=current density, am=molecular weight, n=number of electrons transferred, F=faraday constant (96500).
7. Which following expression is used to find the corrosion rate from current density of a corroding specimen?
a) W/At = [(I × am) / (n × F)]
b) W/At = [(I × am) / (n × d × F)]
c) W/At = [534 × (I × am) / (n × F)]
d) W/At = [(I × am) / (n × F)] and W/At = [(I × am) / (d × n × F)]
Answer: d
Clarification: W/At denotes the corrosion rate, I= current density, am=molecular weight, n=number of electrons transferred, F=faraday constant (96500), d=density of a specimen. [(I*am) / (n*F)] is used to calculate the MDD whereas the [(I*am) / (n*d*F)] is used to calculate the MPY.
8. Rate of oxidation is equals to the rate of reduction in corrosion.
a) True
b) False
Answer: a
Clarification: It is the second statement in the mixed potential theory that the number of electrons generated and consumed are same. Anode produces the same number of electrons that require for the cathode.
9. Iron is corroding at a current density of 1.69*10-4 amp/cm2. What would be the corrosion rate in MDD?
a) 773
b) 723
c) 423
d) 473
Answer: c
Clarification: 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/cm2
Atomic weight of iron = am = 55.86 g/mol
n = number of electrons = 2 [Fe ==> Fe+2 +2e–]
F = faraday’s constant = 96500 coulombs
= (1.96×10-4coulombs/sec*cm2) × (55.86) ÷ (2×96500 coulombs)
= [(1.96×10-4)*(60*60*24)*(100)*(55.86 *103)] / [2*96500]
= 422.53 MDD.
10. Iron is corroding at a current density of 1.69*10-4 amp/cm2. What would be the corrosion rate in MPY?
a) 423
b) 77
c) 473
d) 97
Answer: b
Clarification: We know MDD = W/At = (I*am) / (d*n*F). We should convert the given unit values in milli inches per year.
Given, I = 1.69*10-4 amp/cm2
Atomic weight of iron = am = 55.86 g/mol
n = number of electrons = 2 [Fe ==> Fe+2 +2e–]
F = faraday’s constant = 96500 coulombs
d=density of iron = 7.875 g/cm3
= [(1.96×10-4 coulombs/sec*cm2) × (55.86)] ÷ [(7.874 g/cm3×2×96500 coulombs)]
= [(1.96×10-4coulombs/sec*cm2) × (55.86)] ÷ [(7.874 g/cm2*cm×2×96500 coulombs)]
= [(1.96×10-4 1/sec) × (55.86)] ÷ [(7.874 g/cm ×2×96500)]
= [1.69×10-4 × (60*60*24*365) × (55.86) × 103] / [2 × 96500 × 7.875 × 2.54]
= 77.117 MPY.
11. _____ is used to measure the extent of uniform corrosion.
a) MDD
b) MPY
c) Both MPY and MDD
d) MPE
Answer: a
Clarification: Uniform corrosion is the corrosion that occurs throughout the entire exposed surface of a metal and it is uniform in depth of corrosion layer. Weight loss method I.e. MDD is used to know the extent of uniform corrosion.
12. Corrosion rate of aluminum is _________ when compared with corrosion of mild steel in open oxygenated atmosphere.
a) small
b) medium
c) large
d) very small
Answer: d
Clarification: In oxygenated atmosphere aluminum has high corrosion resistant than the mild steel due to its oxide layer formation. Hence the corrosion rate of aluminum is very small when compared to the corrosion rate of mild steel.
13. ___________ will occur, if current pass from an electrode to electrolyte.
a) Oxidation
b) Reduction
c) Oxidation and reduction
d) Anion formation
Answer: a
Clarification: The direction of current flow is exactly opposite to the movement of electrons. We know electrons starts at the anode and travels towards cathode whereas current will flow from anode to electrolyte. This results in oxidation or corrosion on that electrode.
14. Oxidizing impurities in electrolyte increases the corrosion rate.
a) True
b) False
Answer: a
Clarification: Mixed potential theory states that any electrochemical reaction can be divided into two or more partial reactions (oxidation and reduction reaction). Presence of oxidizing impurities such as Fe+3 increases the number of reduction reactions. It correspondingly increases the oxidation rate or corrosion rate.
15. What are the uses of corrosion rate estimation of materials in daily life?
a) To predict the life time of a component
b) To compare the corrosive-resistant of materials
c) To increases the corrosion rate
d) To predict the life time and to compare the corrosive resistant of materials
Answer: d
Clarification: Corrosion rate expression is used to quantify the corrosion process. This helps us to predict the life time of a component and to compare with different materials. This also helps us to change the process variables in order to decrease the corrosion rate.
To practice advanced questions and answers on all areas of Corrosion Engineering,