Materials for engineering, 3rd Edition - (Malestrom)

Structure of engineering materials 17
quickly, and their composition follows the line of the solidus as they cool
until, at temperature T 2 , their composition is given by b. This crystal growth
occurs by the deposition of layers of atoms which are richer in copper content,
but atomic migration takes place by solid state diffusion within each dendrite
between the new layers and the original nucleus, to enable the composition
to adjust itself to b.
The dendrites we are considering will be at a temperature very close to
their melting point, so that this diffusion process can continue to allow the
dendrites to adjust their composition to follow the line of the solidus as the
temperature continues to fall slowly, the remaining ‘mother liquor’ following
the line of the liquidus. When temperature T 3 is reached, the last liquid (of
composition e) freezes, and the accompanying solid-state diffusion brings
the now completely frozen solid to the composition c once again. The solidified
alloy is now (below T 3 ) in a single-phase field once more and is, thus, stable
at all lower temperatures.
In summary, therefore, we see that in the slow solidification of a solid
solution alloy, although we started with a liquid alloy of composition c and
finished with a set of solid crystals of composition c, the process was more
complicated than in the simple freezing of a pure solid. The initial nuclei
were seen to have a different composition from the liquid in which they
formed and both the liquid phase and the solid phase progressively change
their composition during the process of solidification.
The lever rule
In the temperature range T 1 –T 3 , when the two phases (L + S) were present,
the construction of isothermal lines was shown (Fig. 1.11) to give the
composition of the two phases which were in equilibrium. This same
construction also determines how much of each phase is present at a given
temperature, for a given alloy. Consider again the Monel of composition c;
if, at temperature T 2 , the fraction of the alloy which is liquid is f L , and the
fraction of the alloy which is solid is f S , then
f L + f S = 1.
If the concentration of nickel in the liquid phase = d and the concentration
of nickel in the solid phase = b, then
bf S + df L = c
but
so
f S = 1 – f L
c = b – bf L + f L + df L