1. (a) (i) If the interfacial energy at the boundary is isotropic, the second phase particles will be spherical to reduce the interphase boundary energy because spherical particles have the least surface area.

(ii) If the interfacial energy at the boundary is non-isotropic, the second phase particles will appear elongated. This is because different parts of the particle will experience different surface tension. For example, if coherency is possible along some planes but not others, the second phase particle will grow as a plate, extensive along low-energy planes and narrow along high energy ones.

(b) Refer to the answer to Qn 1 (b) in the Oct/Nov 1996 paper.

3. (a) Refer to Calister pg. 245, example problem 9.1.

(b) (i) Composition of hypoeutectic a phase = 18%Sn

Composition of liquid phase = 44% Sn

Weight % of liquid phase = (30-18)/(44-18) = 0.46 = 46%

Weight % of hypoeutectic a phase = (44-30)/(44-18) = 0.54 = 54%

(ii) Weight fraction of a phase = (97.5-30)/(97.5-19.2) = 0.862

Weight fraction of b phase = (30-19.2)/(97.5-19.2) = 0.138

Weight of a phase = 0.862 kg

Weight of b phase = 0.138 kg

4. (a) The 4 Hume-Rothery rules are that the 2 solids must have similar:

• crystal structure (both BCC or both FCC),
• atomic radii (<15%)
• electronegativities (do not form compounds) and
• valence

(b) A terminal phase has a composition that extends to one of the composition extremity of a binary phase diagram, while an intermetallic compound is an intermediate phase (phase that has a composition range that does not extend to either of the pure compounds of the system) that exists over a very narrow range of compositions. An intermetallic compound is a compound of 2 metals, whereas a terminal phase may not be. An intermetallic compound has a distinct chemical formula, whereas a terminal phase does not have (its composition can take a wide range).

5. (a) Composition of alloy chosen is 20wt%Pb and 80wt%Mg.

1. Solution heat treatment

• All solute [Pb] atoms dissolved to form a single-phase solid solution.

• Heat alloy to temperature within a phase field, e.g. 500^oC.
• Wait until all the Pb atoms have dissolved.
• Alloy now consists of only a phase, of composition 20wt%Pb.

• Rapid quenching to room temperature.

• So that any diffusion and the accompanying formation of b phase is prevented, i.e. phase separartion is prevented.

à A non-equilibrium situation in which only the a phase solid solution supersaturated with Pb atoms is present. At this stage, alloy is relatively soft and weak. Because the diffusion rates at room temperature are extremely slow, the single a phase is retained for relatively long periods of time.

1. Precipitation Heat Treatment

• Heat to some intermediate temperature, e.g. 250^oC within the a +b 2-phase region at which diffusion rates become more appreciable.
• Age to precipitate finely sipersed second phase particle.
• Cool (optional)

(b) During overaging, the size of the precipitates has increased until the precipitates have become incoherent. Overaging results in softening and weakening in the alloy as a result in a reduction in the resistance to slip that is offered by the precipitate particles.

(c) Coring refers to the compositional variation in the microstructure as a result of non-equilibrium cooling. There is non-uniform distribution of the two elements within the grains, i.e. there is segregation and concentration gradients are established across the grains. The centre of the grains (first part to freeze) is rich in the high-melting element. The concentration of the low melting element increases with position from the centre to the grain boundary.

Coring results in less than optimum properties. During reheating, the grain boundaries melt first because they are richer in the low-melting component. This results in a sudden loss in mechanical integrity due to the thin liquid film that separates the grains. Also, melting begins at a temperature well below the equilibrium solidus temperature of the alloy.

6. (a) Subst n =5, y =0.3, t = 100 min into the Avrami equation,

k = 3.57 x 10^-11

Subst n = 5, y = 0.5, k = 3.57 x 10^-11 into the Avrami equation,

t _0.5 = 114 min

Rate of recryatllisation = 1/ t _0.5

= 8.76 x 10^-3 min^-1

(b) (i) Martensite

(ii) Martensite

(iii) ferrite, martensite

(iv) ferrite, pearlite, martensite

(v) ferrite, pearlite