Practice Questions and Solution - Set # 5
MECH 321
March 2014
Note: These questions will be discussed during the tutorial sessions on March 28th.
Question 1:
A brine solution is used as a cooling medium in a steel heat exchanger. The brine is circulated within the
heat exchanger and contains some dissolved oxygen. Suggest four methods for reducing corrosion of the
steel by the brine. Explain the rationale for each suggestion.
Solution:
The most common and practical method is using cathodic protection. Other possible methods that may be
used to reduce corrosion of the heat exchanger by the brine solution are as follows:
(1) Reduce the temperature of the brine; normally, the rate of a corrosion reaction increases with
increasing temperature.
(2) Change the composition of the brine; the corrosion rate is often quite dependent on the composition
of the corrosion environment.
(3) Remove as much dissolved oxygen as possible. Under some circumstances, the dissolved oxygen may
form bubbles, which can lead to erosion-corrosion damage.
(4) Minimize the number of bends and/or changes in pipe contours in order to minimize erosioncorrosion.
(5) Add inhibitors.
(6) Avoid connections between different metal alloys
Question 2:
In galvanization and tinning processes, iron (or steel) are protected with a layer of zinc and tin
consequently. Explain how each of these methods protect iron from corrosion. If a galvanized steel
surface is scratched, does zinc still protects steel? What about a tinplate?
Solution:
In galvanizing, steel is protected through cathodic protection. It means that when Zn and Fe are exposed
to a corrosive environment, zinc acts as the anode. Therefore if a galvanized steel is scratched, zinc still
protects steel as a sacrificial layer. However tinning only protects the steel physically, thus when
scratched, the tin layer does not protect the steel anymore. Instead, an electrochemical cell is formed and
iron oxidizes rapidly.
Question 3:
Identify the type of corrosion in the following cases:
a) General rusting of steel and iron when the entire surface is exposed to air;
b) Domestic water heater where copper and steel tubing are joined;
c) Corrosion in firearms, in the bore of the barrel when corrosive ammunition is used and the barrel
is not cleaned soon afterward;
d) Flanges, Washers, O-rings on metal plates immersed in seawater;
e) Weld decay in stainless steel;
f) Stainless steel parts in a combustion engine operated at temperatures between 500 and 800oC for
long time;
g) Elbow fitting in a steam condensate line;
Solution:
a) Rusting of steel Corrosion type: Uniform Attack
b) Domestic water heater where copper and steel tubing are joined; Corrosion type: Galvanic
corrosion where steel will corrode in the vicinity of the junction
c) Corrosion in firearms, in the bore of the barrel when corrosive ammunition is used and the barrel
is not cleaned soon afterward; Corrosion type: Pitting corrosion
d) Flanges, Washers, O-rings on metal pates immersed in seawater; Corrosion type: Crevice
corrosion
e) Weld decay in stainless steel; Corrosion type: Selective leaching
f) Stainless steel parts in a combustion engine operated at temperatures between 500 and 800oC for
long time; Corrosion type: Intergranular corrosion
g) Elbow fitting in a steam condensate line; Corrosion type: Erosion corrosion
Question 4:
An electrochemical cell is constructed such that on one side a pure nickel electrode is in contact with a
solution containing Ni2+ ions at a concentration of 3 × 10-3 M. The other cell half consists of a pure Fe
electrode that is immersed in a solution of Fe2+ ions having a concentration of 0.1 M. At what temperature
will the potential between the two electrodes be +0.140 V?
Solution:
On the basis of their relative positions in the standard emf series (Table 17.1), assume that Fe is oxidized
and Ni is reduced. Thus, the electrochemical reaction that occurs within this cell is just
Ni2+ + Fe → Ni + Fe 2+
Thus, Equation 17.20 is written in the form
RT [ Fe 2 + ]
ln
∆V = (V − V ) −
nF [ Ni 2 + ]

Ni

Fe
Solving this expression for T gives




 
nF ∆V − (VNi − VFe )


T =−
[ Fe 2 + ] 
R 
 ln [ Ni 2 + ]





V
V
Fe = – 0.440 V and 
Ni = – 0.250 V. Therefore,
The standard potentials from Table 17.1 are 



(2)(96,500 C / mol ) 0.140V − {−0.250V − (−0.440V )}


T =−
8.31 J / mol - K 
 0.1M 

ln

−3


 3 × 10 M 


= 331 K = 58°C
Question 5:
Lead experiences corrosion in an acid solution according to the reaction:
Pb + 2H+ → Pb2+ + H2
The rates of both oxidation and reduction half-reactions are controlled by activation polarization.
Compute the rate of oxidation of Pb (in mol/cm2-s) given the following activation polarization data:
For Lead
For Hydrogen
V(Pb /Pb 2+ ) = −0.126 V
V(H + /H ) = 0 V
2
i0 = 2 × 10–9 A/cm2
i0 = 1.0 × 10–8 A/cm2
β = +0.12
β = –0.10
Solution
The first thing necessary is to establish relationships of the form of Equation 17.25 for the potentials of
both oxidation and reduction reactions. Next we will set these expressions equal to one another, and then
solve for the value of i which is really the corrosion current density, ic. Finally, the corrosion rate may be
calculated using Equation 17.24. The two potential expressions are as follows:
For hydrogen reduction
VH = V
(H + /H 2 )

i
+ βH log 
 i0
 H




And for Pb oxidation

 i
+
β
log
Pb
i
(Pb/Pb 2+ )
 0Pb
VPb = V




Setting VH = VPb and solving for log i (log ic) leads to

1
log i c = 
 β Pb − β H
[

 V( H + / H ) −V
− β H log i 0 + β Pb log i 0
( Pb / Pb 2 + )
H
Pb
2

]
And, incorporating values for the various parameters provided in the problem statement leads to


1
−8
−9
log i c = 
 0 − (−0.126) − (−0.10){log(1.0 ×10 )} + (0.12){log(2 × 10 )} = –7.809
 0.12 − (−0.10) 
[
]
ic = 10 -7.809 = 1.55 × 10 -8 A/cm2
And from Equation 17.24
r =
=
ic
nF
1.55 ×10 −8 C / s - cm 2
= 8.03 × 10 -14 mol/cm 2 - s
(2)(96,500 C / mol )
Question 6:
(a) Why is the abrasive wear resistance of a material a function of its hardness?
(b) Why is it difficult to use friction sawing on nonferrous metals? Explain.
Solution:
(a) Higher hardness indicates greater resistance to penetration, hence less penetration of the abrasive
particles or hard protrusions into surfaces, and the grooves produced are not as deep. Thus,
abrasive wear is a function of hardness
(b) Nonferrous metals have a tendency to adhere to the blade, caused by adhesion at the high
temperatures and attributable to the softness of these materials. Note also that these materials
typically have high thermal conductivity, so if the metal has melted, it will quickly solidify and
make the operation more difficult.
Question 7:
How can fatigue wear be reduced?
Solution:
(a) reducing the load and sliding distance and increasing the hardness
(b) improving the quality of the contacting materials, such as eliminating inclusions, impurities, and
voids;
(c) improving the surface finish and integrity during the manufacturing process;
(d) surface working, such as shot peening or other treatments;
(e) reducing contact stresses; and
(f) reducing the number of total cycles.