Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-10 GTBL042-Callister-v2 August 13, 2007 18:16

350 • Chapter 10 / Phase Diagrams

Concept Check 10.2

A copper–nickel alloy of composition 70 wt% Ni–30 wt% Cu is slowly heated from

a temperature of 1300◦C (2370◦F).

(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 http://www.wiley.com/college/callister (Student Companion Site).]

Concept Check 10.3

Is it possible to have a copper–nickel alloy that, at equilibrium, consists of anαphase

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 http://www.wiley.com/college/callister (Student Companion Site).]

EXAMPLE PROBLEM 10.1

Lever Rule Derivation

Derive the lever rule.

Solution

Consider the phase diagram for copper and nickel (Figure 10.3b) and alloy

of compositionC 0 at 1250◦C, and letCα,CL,Wα, andWLrepresent the same

parameters as above. This derivation is accomplished through two conservation-

of-mass expressions. With the first, since only two phases are present, the sum

of their mass fractions must be equal to unity; that is,

Wα+WL= 1 (10.3)

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

WαCα+WLCL=C 0 (10.4)

Simultaneous solution of these two equations leads to the lever rule ex-

pressions for this particular situation, Equations 10.1b and 10.2b:

WL=

Cα−C 0

Cα−CL

(10.1b)

Wα=

C 0 −CL

Cα−CL

(10.2b)