SME
Materials Science & Material Properties
Phase diagrams
Use this for “A 35 wt% Ni copper–nickel alloy at 1250 °C holds liquid at 31.5 wt% Ni and solid at 42.5 wt% Ni. Find each phase’s share.”
Diagram
Lever
rule:
Wα
=
C₀ − CL
Cα − CL
=
3.5
11
=
0.318.
WL = 1 − Wα = 0.682.
Each phase’s share is the arm on the far side of C₀.
This is an example. Type your own values to start a new problem.
WL
=
Cα (wt%)
−
C₀ (wt%)
Cα (wt%)
−
CL (wt%)
Wα
Solid share
calculated
Formulas
WL
=
Cα −
C₀
Cα − CL
0.6818
=
42.5 − 35
42.5 − 31.5
Wα = 1 − WL
0.3182 = 1 − 0.6818
Assumptions
•
The alloy is inside the two-phase region, so liquid and solid α are both present.
•
CL and Cα are read where the tie line at T meets the liquidus and the solidus.
•
Compositions are wt% of the second metal (nickel); shares are by mass, from 0 to 1.
•
The tie line is drawn at the temperature of the use line; read CL and Cα off your own diagram at your T.
Step-by-step
WE KNOW
C₀ = 35 wt%, CL = 31.5 wt%, Cα = 42.5 wt%
FIND
liquid share (WL), solid share (Wα)
Step 1 · Find liquid share (WL)
Use WL = (Cα − C₀) ÷ (Cα − CL)
The
lever
rule:
the
liquid
share
is
the
arm
from
the
far
end
to
C₀
,
over
the
whole
tie
line.
WL
=
42.5 − 35
42.5 − 31.5
WL
=
7.5
11
WL = 0.6818
Step 2 · Find solid share (Wα)
Use Wα = 1 − WL
The two shares add to 1, so the solid is the rest.
Wα = 1 − 0.6818
Wα = 0.3182
Check
Put the numbers back in. Both sides should match.
0.6818
=
42.5 − 35
42.5 − 31.5
✓
0.3182 = 1 − 0.6818 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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