NATURE OF MATERIALS - Materials for Civil and Construction Engineers - page 71

where

p lB = percent of the liquid phase that is composed of material B

of the solid phase that is composed of material B
of the material that is component B

p sB = percent

p B = percent

From these two equations, the amount of material in the liquid and solid
phase can be derived as

p B m t = p lB m l + p sB 1 m t - m l 2

p B m t = p lB m l + p sB m t - p sB m l

p B m t - p sB m t = p lB m l - p sB m l

1 p B - p sB 2

m l =

2 m t 1 p lB - p sB

m s = m t - m l (2.6)

Sample Problem 2.3

Considering an alloy of the two soluble components A and B described by a phase dia-
gram similar to that shown in Figure 2.14, determine the masses of the alloy that are in
the liquid and solid phases at a given temperature if the total mass of the alloy is 100
grams, component B represents 40% of the alloy, 20% of the liquid is component B,
and 70% of solid is component B.

Solution

m t = 100 g

p B = 40 %

p lB = 20 %

p sB = 70 %

From Equations 2.4 and 2.5,

m l + m s = 100

20 m l + 70 m s = 40 * 100

Solving the two equations simultaneously, we get

m l = mass of the alloy that is in the liquid

of the alloy that is in the solid

phase = 60 g

m s = mass phase = 40 g

The same answer can also be obtained using Equation 2.6.

Insoluble Materials The discussion thus far has dealt with two completely
soluble materials. It is equally important to understand the phase diagram
for immiscible materials, that is, for components that are so dissimilar that