Good Essay On P (Corrosion) Of Anodic Metal Of Area A Coupled To Cathodic Metal Of Area B

[The electron flow will be from the steel on the left to the brass on the right. The steel is undergoing oxidation, not the brass, so the steel will corrode. ]
When two different materials are welded or bolted in an electrolyte (sea water) the uncoupled corrosion rate of the least noble (anodic positive) is increased, and the uncoupled corrosion rate of the noble (cathodic negative) is decreases.

P0 is penetration of anodic metal without coupling
E (cell) = ER - EL Where, E cell is the potential of cell, ER is the right side electrode potential, and
EL is the left side electrode potential. Right side half-cell consistently has the most positive potential. E cell of a reaction should be positive in order to keep the Gibbs Free energy negative.
ΔG = -nFE
a. Table 1.4 in Block 5 Part 1 shows the measured potentials, relative to mild steel, for metals in brine and in beer. For the following combinations of metals, determine the cell potential and hence indicate which metal will corrode.
i. A porthole made from stainless steel set into a mild steel housing on an ocean-going ship.

The mild steel experiences oxidation and the electron flow is from the mild steel housing got the stainless steel.

mild steel = 0 V
stainless steel =
ii. A copper pipe carrying beer that is connected to an aluminium tank.
Copper = +0.34 V
Aluminum = -1.67 V
+0.34 V – (-1.67V) = 2.01V
The copper pipe has the higher cell potential so the aluminum will lose electrons flowing to the copper. The aluminum will oxidize causing corrosion.
iii. Magnesium casting that is being protected by a lead flange, which is located in a brewery.
Magnesium = -2.34 V
Lead = -0.13 V
-0.13 – (-2.34) = +2.21 V
Lead has the higher potential and will pull the electron flow towards the lead. The Magnesium will undergo oxidation and corrode.
Part of the machinery space in a cargo ship; a brass pipe carrying cold sea water passes through an aluminium bulkhead as illustrated.
Aluminum = -1.67 V
Brass = 0.12 V
0.12 – (-1.67) = 1.79 V
i. Identify three features of the design as shown that could lead to corrosion occurring, giving an explanation in each case.
(1) The steel nut and bolt can cause problems, because in a salt water environment, crevice corrosion is likely to occur. The problem is the difference in cell potential between steel and mild steel.
(2) The galvanic effect in the entire set up is bound to lead to corrosion because dissimilar metals are paired and they are in an environment with an electrolyte (sea water brine containing salt)
(3) After twenty years in a marine system the aluminum corrodes and needs continual maintenance. (Broli et al., 1994, p.15)
ii. For two of the areas you identified in (i), suggest means by which the design could be altered to control or prevent corrosion
(1) The better choices for the bolt and nuts are copper alloy (400) or titanium (if it does not cost too much). Or make sure all the steel used has equal cell potentials.
(2) Choosing aluminum may or may not be cost effective compared to other materials like steel. But, although, steel has a longer life cycle in sea water, steel is also a heavy material; so a corrosion resistant aluminum alloy is a better choice.
iii. Identify two areas where good design against corrosion has already been applied, giving an explanation in each case.
(1) The plastic insert between the brass pipe and the aluminum pipe keeps the aluminum and brass separated.
(2) The base of the structure is a continuous weld of steel to steel faces so that surface faces will not be exposed and so corrosion will not occur
iv. If the brass pipe was to have a valve inserted to control the flow rate, briefly explain how this might affect corrosion resistance of the pipe.
A brass pipeline is the best choice for salt water because steel piping would be corroded and not very durable A control valve inserted into the brass pipe to control the flow rate therefore should never be made of steel. Alloys or even specific plastics are designed for the job. The tricky part is making sure that valves do not come with metal gaskets.

Question 2

a. Describe the role of dislocations with respect to each of the processes used in the strengthening of metals.

Single-phase

a. Grain reduction size takes into account the crystallographic orientation and how it suddenly changes from one side of a grain boundary to the other side of the grain boundary. When the misalignment is very large or a double grain boundary exists, then further dislocation across boundaries is not as likely, instead when dislocations take place the pile up against the boundary. That characteristic is taken advantage of by modifying the grain size. One way to alter plastic deformation and modify the grain sizes is to use a heat treatment.
b. The mechanism behind the solid solution strengthening solution is the fact that the alloy compensates for the dislocation. This works because a single metal is softer than its alloy.
c. Strain hardening is the method of strengthening materials that are affected by heat treatment. The trade off is that strain hardening can cause strengthening, but give up some of the ductility of the material.

Multi-phase metals

d. Precipitation hardening allows strength to increases over time. The ductile matrix of a material with minute second phase particles are distributed throughout out the matrix by mixing and consolidation or by precipitated in solid state.
e. Fiber strengthening uses fine particles to reinforce materials. The fibers can be either continuous or short (discontinuous) but either way they must have characteristics of high modulus and high strength. The matrix must be non-reactive with the fibers as well as ductile. The main mechanism is the ability of the high modulus fibers to carry the entire load but the matrix works to distribute the load between the fibers.
f. Martensite strengthening does not allow diffusion-controlled invariant transformation because rapid cooling controls the strengthening mechanisms. Extreme temperatures are key to the process. The basic reason the process is successful is because the bonding between carbon atoms is set up replacing dislocations between the carbon atoms with stronger bonds.
b. Briefly explain how the atomic bonding in polymers and ceramics greatly influences the differences in their mechanical properties and why these materials cannot be strengthened in the same way as metals.
Metals and thermoplastic materials are only flexible to particular limits, after which the material will fracture and then, these two types of materials remain permanently deformed; actions can be take to discourage further deformation but not really repaired. On the other hand, ceramics are able to withstand very low or very high temperatures because their material nature is not metallic and their composition is inorganic so Carbon cannot be used in mechanisms for strengthening. The preparation of raw materials for ceramics requires a purification step, because high purity chemical solutions are used to form the end product using precipitation. A crystalline structure or partially crystalline structure, the strength is based on the atomic order of the end product (after solidification). Instead of using the types of strengthening methods written about for metals, ceramics are produced with hydrothermal synthesis or sol-gel synthesis. The great challenge for producing ceramics with optimum strength characteristics is to reduce the size of the grains to a certain limit without going to small and producing an amorphous material.
The sintering method uses temperatures below the melting point of the ceramic to create a soft object that is hardened in an oven. The oven-type heat causes high reduction in porosity, so the material is made more dense and solid. The soft body is larger than the product that comes from the oven. Whereas, in metals carbon to carbon bonds are strengthened to produce a stronger material, in ceramics the pores that allow porosity are eliminated.
The optimum yield strengths are observed when the grain size is in the nanometer region of size, so that is challenging. Contemporary methods are trending to self-assembly mechanisms in materials including ceramics and composite polymers.
c. An aluminium alloy has an average grain size of 575 μm and a 0.2% proof stress of 180 MN m−2. After cold work and recrystallization, the grain size is reduced to 95 μm. Given that the constant s0 in the Hall-Petch equation is 85 MN m−2 for this particular alloy, what proof stress could be expected in the alloy in its worked and recrystallized condition?

Aluminium alloy

Average grain size = 575 μm
0.2% proof stress of 180 MN m−2
after cold work and recrystallization
grain size = 95 μm (reduction)
Hall-Petch = 85 MN m-2
Known yield strength increases as grains diameter is decreased/ σy = σ0 + kd-1/2

What is the expected proof stress?

? = 95 micrometer + (85 MN/m2 multiplied by 575 micrometer)-1/2 = ~ 24,100 micrometer

Question 3

a. A specimen corresponding to a finite plate of width W = 55 mm is made from a steel with a yield strength of 580 MN m−2 contains an edge crack 6.5 mm long. A fracture toughness test is carried out on this specimen, and a failure stress of 345 MN m−2 is measured. Determine the value of fracture toughness, KIC, for this particular steel.

Material is Steel

Width = 55 mm
Yield Strength = 580 MN /m2
Edge Crack = 6.5 mm long
Failure Stress = 345/ MN /m2
1 m = 1000 mm

What is the Value of Fracture Toughness (KIc)?

ʄ=1.12
Stress Intensity factor K = (geometry factor and flaw * applied stress) π flaw size
K = ʄσ πa = (1.12) (345/ MN /m2) π 6.5 mm long
= (386.4 MN / m2) π 6.5 mm long
=(386.4 MN / m2) 0.0204 m
18.98MN/m
 b.A steel similar to the one used during the fracture toughness test described in part (a) is used to manufacture a large component that can be considered to be equivalent to an infinite plate. The component will be subjected to periodic cyclic loading. For safety, if the maximum stress that the component can be subjected to is 50% of the yield strength determine
o i.Whether a crack that has grown by fatigue to 12.0 mm long is acceptable.
for infinite plate ʄ=1
edge crack = 12.0 mm
yield strength = (580 MN /m2)
Width = 55 mm
Maximum stress is 50 percent of the yield stress (0.5)(580.MN/m2) = 290 MN/m2
Yσ πa=KIc
a/W = (12 mm/ 55 mm) = 0.218 = Y
ii.At what length the growing crack exceeds the safety factor?
stress = 290/580 = 0.5 and the safety factor is 0.218 so there is a problem
c.Figure 2 shows a micrograph of fatigue striations taken from the component described in part (b). If the cyclic loading remains constant calculate how many cycles it would take for the crack to grow to the length you calculated in part (b)(ii) of this question?
cyclic load is constant
? = # cycles for crack to grow to 12 inches
cyclic loading constant continuously
fatigue striations aprox. 2 micrometer apart
every covers about 3 cycles
 d.If the rate of cyclic loading is 0.01 cycles s−1 determine how long it would take (in days) for the crack to grow to this length.

Question 4

a.Briefly and in your own words, explain the following terms in relation to creep
o i.recovery
Recovery does not change the mechanical properties of a material; the dislocations are morphologically changed but do not change in number. The residual stresses are decreased and can be eliminated. The corrosion resistance of a metal can increase the corrosion resistance in a metal ( not 100 percent of the time but in some cases.
Ex. When dislocations are untangled (like with thermal treatment is used to ‘untangle’ dislocations; the dislocations become the boundaries for the plygonized subgrain structure) o
ii.diffusion creep
When atoms change position over time in a stressed crystal the action is called diffusion creep. The atoms movement or preferential migration takes place (when the crystal is under stress) with the result that applied stress is alleviated.

The rate of diffusion creep is temperature dependent.

o iii.dislocation climb.
Dislocation climb coupled with slip causes creep deformation. In diffusion in presence of high temperatures produces diffusion along crystal planes. If the temperature is low, then the rate of diffusion is governed by the “fast paths” like core dislocations.
(9 marks)
 b.Figure 5.11 of Block 5 Part 5 shows creep curves for Nimonic 80A. This alloy is being considered for two components in a high-temperature application. One of the components will experience a temperature of 973 K and a stress of 250 MN m−2; the other will experience a temperature of 1088 K and a stress of 100 MN m−2.

The alloy Nimonic 80A

Can it be used for a high temperature application for 2 components?
Component 1 with environment of 973 K and stress of 250 MN /m2
Component 2 with environment of 1088 K and stress of 100 MN/m2
The specification for the components states that they must not exceed 0.5% creep strain after 1000 hours of service.

Discuss whether Nimonic 80A will meet the criteria for the two components.

Creep ≤ 0.5 % after 1000 hours of service
with 0.5 % creep strain at 1000 hour of service in 1088 K temperatures the Stress handled is approximately 150 MN /m2 and after that as time increases the ability to handle stress deteriorates.
with 0.5 % creep strain at 1000 hour of service in 1023 K temperatures the stress handled at 1000 hours is above 150 NM /m2 As time increases the capability to handle stress decreases and rupture occurs at approximately 50 NM /m2
(6 marks)
c.Suggest two methods by which the creep resistance of the components in part (b) could be improved.

How to improve creep resistance at 1023 K and at 1088 K?

The best alloy agents to gain the best results can be added to improve strength resistance.
The length of time the material is needed to handle the stresses can be shortened to less than 1000 hours.
The temperature environment of the components can be reduced.
Under conditions of reduced Temperature and reduced Time the Nimonic80A can handle more time.
(4 marks)
 d.Give a brief account of the origins of viscoelasticity in polymers and why creep strains can often be recovered.
Viscoelasticity in polymers is the elastic strain that can be described as the recoverable strain and the viscous flow that is non-recoverable. Although vicious and elastic strain are opposite in terms of recovery; the combination works to enhance polymer recovery and allow recovery of part of the strain when the load is removed. Viscoelasticity in polymers occurs after time under continuous stress. Creep strains are recoverd because the stress to strain behavior is nonlinear and is different depending on time, temperature and strain rate. The non-linear stress behavior is governed by the capability molecules have to move around in polymers.

The way the elasticity and the viscous features interact is due to

the carbon to carbon bonds stretching in the polymer cause very little strain
recovery strain is increased with bond rotation is present (rubber elasticity or long range elasticity)
time is a significant factor because the viscous flow is time dependent because in a viscous flow molecules slide by each other
(6 marks)
(Total 25 marks)