Callister - An introduction - 8th edition

292 • Chapter 9 / Phase Diagrams
It is easy to confuse the foregoing procedures for the determination of phase
compositions and fractional phase amounts; thus, a brief summary is warranted.
Compositions of phases are expressed in terms of weight percents of the components
(e.g., wt% Cu, wt% Ni). For any alloy consisting of a single phase, the composition
of that phase is the same as the total alloy composition. If two phases are
present, the tie line must be employed, the extremities of which determine the compositions
of the respective phases. With regard to fractional phase amounts (e.g.,
mass fraction of the or liquid phase), when a single phase exists, the alloy is completely
that phase. For a two-phase alloy, on the other hand, the lever rule is utilized,
in which a ratio of tie line segment lengths is taken.
Concept Check 9.2
A copper–nickel alloy of composition 70 wt% Ni–30 wt% Cu is slowly heated from
a temperature of 1300C (2370F).
(a) At what temperature does the first liquid phase form?
(b) What is the composition of this liquid phase?
(c) At what temperature does complete melting of the alloy occur?
(d) What is the composition of the last solid remaining prior to complete melting?
[The answer may be found at www.wiley.com/college/callister (Student Companion Site).]
Concept Check 9.3
Is it possible to have a copper–nickel alloy that, at equilibrium, consists of an
phase of composition 37 wt% Ni–63 wt% Cu, and also a liquid phase of composition
20 wt% Ni–80 wt% Cu? If so, what will be the approximate temperature of
the alloy? If this is not possible, explain why.
[The answer may be found at www.wiley.com/college/callister (Student Companion Site).]
EXAMPLE PROBLEM 9.1
Lever Rule Derivation
Derive the lever rule.
Solution
Consider the phase diagram for copper and nickel (Figure 9.3b) and alloy of
composition C 0 at 1250C, and let C a , C L , W a , and W L represent the same parameters
as given earlier. This derivation is accomplished through two
conservation-of-mass expressions. With the first, because only two phases are
present, the sum of their mass fractions must be equal to unity; that is,
W W L 1
(9.3)
For the second, the mass of one of the components (either Cu or Ni) that is
present in both of the phases must be equal to the mass of that component in
the total alloy, or