Microsoft Word - Materials

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Requires a knowledge for diffusion,phase diagram, phase transformation,corrosion,oxidation, ceramic, po


































lymers & composites.




Microsoft Word - Materials
Answered Same DayDec 31, 2021

Answer To: Microsoft Word - Materials

David answered on Dec 31 2021
111 Votes
Solution
Rom the given expression we have,
D = D0exp(-Qd/RT)
On taking natural log both sides we get,
lnD = lnD0 – Qd/RT
where,
T1 = 1273K (1000
0C)
T2 = 1473K (1200
0C)
Hence,
lnD1 = lnD0 – Qd/RT1
ln 9.4 x 10-16 = lnDo – Qd/ 8.31*1273
similarly for other temperature we have,
ln 2.4 x 10-14 = lnDo – Qd/ 8.31*1473
for calculating Diffusion coefficient we need to c
alculate both Do and Qd first then it can be calculated.
On Solving for Qd and Do we get,
Qd = 252400J/mol
And D0 = 2.2 x 10
-5 m2/s
Hence Diffusion coefficient at temperature 8000C will be,
lnD = lnD0 – Qd/RT
ln D = ln 2.2 x 10-5 – 252400/8.31*1073
so D = 2.4 x 10-15 m2/s (Answer)
Question 2
Answer
Hardening is the inherent property of material and depends upon quenching conditions.
The above graph shows the hardness variation with respect to the distance from the quenched end given on Horizontal
axis of the graph.
As per the fact greater the distance of the sample from the quenched end on the graph harder is the sample. So,
hardening order could be,
1040 < 5140 < 8640 < 4140 < 4340
4340 is hardest because it is at maximum distance from the quenched end.
TTT Curve for 5140
TTT Curve for 4140
In a 5140 alloy steel Carbon manganese and chromium are present in composition 0.4% , 1% and 0.9% respectively while
4140 alloy steel carbon , manganese , chromium and molybdenum is present in composition 0.4% , 1%, 1% and 0.2%
respectively. In TTT diagram of 5140 it is clearly visible that curve shifted towards right due to increase in composition
and in 4140 it shifted more to the right because of additional Molybdenum. With increased alloy P+B nose shifted
towards right which more hardened material. Hence 4140 will be harder than 5140 and the molecules in a resulting
microstructures are more closely packed in 4140 than 5140. Microstructures of 4140 will be more complex than 5140
due to increased hardening.
Question 3
Answer
From the above diagram, a vertical line is drawn at the composition of 80 – 20 % as shown
(a) For determining the composition of alloy we need to draw a horizontal line which cuts the vertical temperature
axis at 6300C hence first solid phase exits at 6300C.
And its composition can be determined by drawing a vertical line from the point of intersection of the straight
line,
4% Pb and 96% Mg (Answer)
(b) For calculating the mass fraction in each case that is α and β phase, look at the parallel line on the phase
diagram which shows that phase α does not exist at room temperature while phase β has 100% composition of
Pb and 0% composition of Mg.
Hence mass fraction of primary phase α = 0 while for β is 1
(c) As per the diagram eutectics are absent at room temperature.
(d) Since only 100% of Pb present at room temperature hence its microstructure will be,


Question 4
Answer
Heat treatment route for the given case;
Cools slowly from 7000C to Room temperature
(b)
1.
Brinell No = 180
Ductility = 61%
2.
Brinell No = 650
Ductility = 41%
3.
Brinell No = 700
Ductility = 38%
4.
Brinell No = 540
Ductility = 67%
5. Exactly as 3
Question 5
Answer
(a) A piece of copper is connected with a piece of zinc. From the table of reduction potentials we can see that zinc is
more reducing than copper hence copper will reduce and zinc will oxidize in this combination,
So the reactions are,
Anodic reaction
Zn = Zn+2 + 2e-1
Cathodic Reaction
Cu+2 + 2e-1 = Cu
Now since Reduction potentials for the two electrodes is ,
E0(Cu) = + 0.34 V
E0 (Zn) = - 0.76 V
Hence electrode potential between the two will be,
E0 ( Cell) = E0 ( Cathode) – E0 (Anode)
= 0.34 + 0.76
= 1.10V
(b) Here we need to plot a graph with all quantities well marked,

(c) Here we need to calculate the Pilling – Bedworth ratio of the layer, so
P – B ratio = Voxide/Vmetal
P-B Ratio =...
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