Engineering Materials Week 7

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Engineering Materials Week 7
10.4 (a) For the solidification of nickel, calculate the critical radius r* and the activation free energy ∆G* if nucleation is homogeneous. Values for the latent heat of fusion and surface free energy are -2.53 X10⁹ J/m3 and 0.255 J/m2.

(b) Now calculate the number of atoms found in a nucleaus of critical size. Assume a lattice parameter of 0.360 nm for solid nickel at its melting temperature.

(116 unit cless/critical nucleus)(4 atoms/unit cell)=464 atoms/critical nucleus
10.16 Briefly cite the diffrences between pearlite, bainite, and speroidite relative to microstructure and mechanical properties.
The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers that alternate with one another. Bainite consists of very fine and parallel needles of ferrite that are separated by elongated particles of cementite. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of spheroidal-shaped particles. Bainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite.
10.24 Figure 10.40 shows the continuous cooling transformation diagram for a 0.35 wt% C iron-carbon alloy. Make a copy of this figure and then sketch and label continous cooling curves to yield the following microstrucutres:
(a) Fine peralite and proeutectoid ferrite
(b) Martensite
(c) Martensite and proeuctoid ferrite
(d)Coarse pearlite and proeutectoid ferrite
(e) martensite, fine pearlite, and proeutectoid ferrite

10.28 Briefly describe the simpliest continous cooling heat treatment procedure that would be used in converting a 4340 steel from one microstructure to another.
(a) (Martensite + ferrite + bainite) to ( martensite + ferrite + pearlite + bainite).
In order to convert from (martensite + ferrite + bainite) to (martensite + ferrite + pearlite + bainite) it is necessary to heat above 720°C, allow complete

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