Question

Question S. (8 points) In the Sn-Pb binary eutectic diagram very slow cooling rate in figure below nbove in question 4, you see a phase transformation (Line A) from a to a +B. Unlike the question 4. Assume you can control the cooling rate freely. To achieve a microstructure like the shows a distribution of fine precipitates insde the a grains, what cooling path do you need to 4, you perform. (the SEM image below is not actualy Sn-Pb but assume that small p phase are residing inside (a) Indicate the cooling path in the CCT diagram. In(time) (b) With this microstructure above, which has fine precipitates inside the grain, compared to the slow cooling rate phase Why? transformed alloy in Question 4, line A material, do you expect that the thermal conductivity increase or decrease? And (c) Do you expect a more efficient soft magnet or hard magnet if you have a microstructure like the one you made in question 5(b), with small precipitates well distributed in the grains? Explain Why? You can use the Magnetization (M) vs Magnetic field graph to explain. Hc Mt- ve

Answer 1

ANSWER:

Sn-Pb is an example of binary eutectic system. In a typical phase diagram, 3 single phase regions are found, they are
α, β,
, and liquid. $\alpha$ represents a solid solution of tin in lead, and $\beta$ represents a solid solution of lead in tin.

From Sn-Pb phase diagram, it is found that maximum  $\alpha$ exists till 18.3 wt% Sn at 183 degrees.

As per question, we need to start forming $\alpha$ grains once formed, it has be converted to $\beta$ so that at the end of cooling we obtain small grains of $\beta$ in $\alpha$ grains.

1834 α 8.3 61.9一 20 pb

Cooling curve is as follows,

193と

Point B is the point where first sign of small grains of lead form and still stays in molten form,

The path BC or BE is decided by the size of Beta grains required to form and cooling as CD or EF respectively.