Binary Phase Diagrams

Materials Science · FE Reference Handbook section

Materials Science

0 formulas

10 exam-style examples

~45 min

All Materials Science lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Allows determination of (1) what phases are present at equilibrium at any temperature and average composition,
(2) the compositions of those phases, and (3) the fractions of those phases.
Eutectic reaction (liquid → two solid phases)
Eutectoid reaction (solid → two solid phases)
Peritectoid reaction (two solid phases → solid)

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Lever rule phase fractions — Binary Phase Diagrams

An alloy of overall composition 54 wt% B lies in a two-phase field bounded by 11 wt% and 90 wt% B. What fraction of each phase is present?

Given

$C_{0} = 54 wt%$
$C_L = 11 wt%$
$C_S = 90 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (90 - 54)/(90 - 11) = 0.456$
Solid fraction
$W_S = 1 - 0.456 = 0.544$
Check
$45.6% + 54.4% = 100% ✓$

Answer:

≈ 45.6% liquid, 54.4% solid

Why the other options are there

54.4% liquid (arms swapped)
54% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 2

Lever rule phase fractions — Binary Phase Diagrams (2)

An alloy of overall composition 53 wt% B lies in a two-phase field bounded by 13 wt% and 74 wt% B. What fraction of each phase is present?

Given

$C_{0} = 53 wt%$
$C_L = 13 wt%$
$C_S = 74 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (74 - 53)/(74 - 13) = 0.344$
Solid fraction
$W_S = 1 - 0.344 = 0.656$
Check
$34.4% + 65.6% = 100% ✓$

Answer:

≈ 34.4% liquid, 65.6% solid

Why the other options are there

65.6% liquid (arms swapped)
53% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 3

Lever rule phase fractions — Binary Phase Diagrams (3)

An alloy of overall composition 59 wt% B lies in a two-phase field bounded by 20 wt% and 73 wt% B. What fraction of each phase is present?

Given

$C_{0} = 59 wt%$
$C_L = 20 wt%$
$C_S = 73 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (73 - 59)/(73 - 20) = 0.264$
Solid fraction
$W_S = 1 - 0.264 = 0.736$
Check
$26.4% + 73.6% = 100% ✓$

Answer:

≈ 26.4% liquid, 73.6% solid

Why the other options are there

73.6% liquid (arms swapped)
59% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 4

Lever rule phase fractions — Binary Phase Diagrams (4)

An alloy of overall composition 56 wt% B lies in a two-phase field bounded by 10 wt% and 79 wt% B. What fraction of each phase is present?

Given

$C_{0} = 56 wt%$
$C_L = 10 wt%$
$C_S = 79 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (79 - 56)/(79 - 10) = 0.333$
Solid fraction
$W_S = 1 - 0.333 = 0.667$
Check
$33.3% + 66.7% = 100% ✓$

Answer:

≈ 33.3% liquid, 66.7% solid

Why the other options are there

66.7% liquid (arms swapped)
56% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 5

Lever rule phase fractions — Binary Phase Diagrams (5)

An alloy of overall composition 48 wt% B lies in a two-phase field bounded by 16 wt% and 78 wt% B. What fraction of each phase is present?

Given

$C_{0} = 48 wt%$
$C_L = 16 wt%$
$C_S = 78 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (78 - 48)/(78 - 16) = 0.484$
Solid fraction
$W_S = 1 - 0.484 = 0.516$
Check
$48.4% + 51.6% = 100% ✓$

Answer:

≈ 48.4% liquid, 51.6% solid

Why the other options are there

51.6% liquid (arms swapped)
48% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 6

Lever rule phase fractions — Binary Phase Diagrams (6)

An alloy of overall composition 45 wt% B lies in a two-phase field bounded by 15 wt% and 82 wt% B. What fraction of each phase is present?

Given

$C_{0} = 45 wt%$
$C_L = 15 wt%$
$C_S = 82 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (82 - 45)/(82 - 15) = 0.552$
Solid fraction
$W_S = 1 - 0.552 = 0.448$
Check
$55.2% + 44.8% = 100% ✓$

Answer:

≈ 55.2% liquid, 44.8% solid

Why the other options are there

44.8% liquid (arms swapped)
45% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 7

Lever rule phase fractions — Binary Phase Diagrams (7)

An alloy of overall composition 39 wt% B lies in a two-phase field bounded by 11 wt% and 71 wt% B. What fraction of each phase is present?

Given

$C_{0} = 39 wt%$
$C_L = 11 wt%$
$C_S = 71 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (71 - 39)/(71 - 11) = 0.533$
Solid fraction
$W_S = 1 - 0.533 = 0.467$
Check
$53.3% + 46.7% = 100% ✓$

Answer:

≈ 53.3% liquid, 46.7% solid

Why the other options are there

46.7% liquid (arms swapped)
39% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 8

Lever rule phase fractions — Binary Phase Diagrams (8)

An alloy of overall composition 47 wt% B lies in a two-phase field bounded by 11 wt% and 88 wt% B. What fraction of each phase is present?

Given

$C_{0} = 47 wt%$
$C_L = 11 wt%$
$C_S = 88 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (88 - 47)/(88 - 11) = 0.532$
Solid fraction
$W_S = 1 - 0.532 = 0.468$
Check
$53.2% + 46.8% = 100% ✓$

Answer:

≈ 53.2% liquid, 46.8% solid

Why the other options are there

46.8% liquid (arms swapped)
47% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 9

Lever rule phase fractions — Binary Phase Diagrams (9)

An alloy of overall composition 46 wt% B lies in a two-phase field bounded by 25 wt% and 73 wt% B. What fraction of each phase is present?

Given

$C_{0} = 46 wt%$
$C_L = 25 wt%$
$C_S = 73 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (73 - 46)/(73 - 25) = 0.563$
Solid fraction
$W_S = 1 - 0.563 = 0.438$
Check
$56.3% + 43.8% = 100% ✓$

Answer:

≈ 56.3% liquid, 43.8% solid

Why the other options are there

43.8% liquid (arms swapped)
46% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams

Example 10

Lever rule phase fractions — Binary Phase Diagrams (10)

An alloy of overall composition 40 wt% B lies in a two-phase field bounded by 17 wt% and 86 wt% B. What fraction of each phase is present?

Given

$C_{0} = 40 wt%$
$C_L = 17 wt%$
$C_S = 86 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (86 - 40)/(86 - 17) = 0.667$
Solid fraction
$W_S = 1 - 0.667 = 0.333$
Check
$66.7% + 33.3% = 100% ✓$

Answer:

≈ 66.7% liquid, 33.3% solid

Why the other options are there

33.3% liquid (arms swapped)
40% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Binary Phase Diagrams