Phase diagrams, transformations and processing history

Specify what a phase diagram holds fixed

A phase is a region with a consistent structural and compositional character under the conditions considered. Liquid and solid are phases, and two different solid structures can also be distinct phases. Grains with different orientations can belong to the same phase. A phase diagram maps stability against variables such as temperature, pressure and composition.

In a binary temperature–composition diagram, two chemical components define the composition axis and pressure is held fixed. Our invented example uses mass percent B, from pure A to pure B. A point combines the overall composition and temperature. Labels identify a liquid region, a solid-solution region α and a two-phase region L + α between liquidus and solidus boundaries.

Read phase compositions before calculating amounts

Inside a two-phase region, draw a horizontal tie line at the chosen temperature. Its intersections with the boundaries give the compositions of the phases that coexist at equilibrium. These generally differ from the overall composition. Do not mistake the fraction of B in one phase for the fraction of that phase in the specimen.

At the constructed 800 °C tie line, liquid contains 20 mass % B and solid α contains 60 mass % B. For an overall 40 mass % B alloy, the amounts must combine to reproduce the overall composition. The lever rule expresses this balance: solid fraction = (40 − 20)/(60 − 20) = 0.5 by mass. Liquid is the remaining 0.5.

The figure is an original schematic phase diagram with invented boundary coordinates and straight segments. It teaches how to read a two-phase field and balance composition. It is not a measured diagram, a named alloy or a heat-treatment recipe.

Derive the amount from two mass accounts

Let m be the total mass and let f_s be the fraction of that mass in solid α. Solid mass is then mf_s and liquid mass is m(1−f_s). Let C_s, C_l and C₀ denote the mass fractions of component B in solid, liquid and the whole specimen. Use the same basis for all three: 60 mass % becomes 0.60, not 60 grams or 60 % solid.

Conserving B gives mf_sC_s + m(1−f_s)C_l = mC₀. Divide by m and collect the solid-fraction terms: f_s(C_s−C_l)=C₀−C_l. Therefore f_s=(C₀−C_l)/(C_s−C_l), provided the two phase compositions are distinct and C₀ lies between them in this two-phase field. The liquid fraction is 1−f_s. Component A is the remaining mass, so its balance follows when total mass and B are both conserved.

In the drawn example, C_l=0.20, C_s=0.60 and C₀=0.40. Moving the overall composition towards the solid endpoint increases the solid amount. Its fraction uses the interval from the liquid endpoint to the overall composition, opposite the solid endpoint on the tie line. At the liquid endpoint f_s is zero; at the solid endpoint it is one. The opposite-distance construction is the mass balance expressed geometrically, not an extra rule about lever hardware.

Equilibrium does not specify the route or speed

A transformation needs a way to begin and continue. Nucleation forms new regions; growth enlarges them. Diffusion is atomic movement that redistributes composition. Temperature affects both the driving forces and the rates of movement, so the same final temperature can be reached with different structures after different holding times and cooling paths.

Rapid cooling can prevent an equilibrium structure from forming or retain a metastable arrangement. Slower cooling can permit redistribution, but “slower means softer” is not a universal rule. The consequence depends on the material, phases and mechanisms. Ageing can form fine precipitates; extended treatment can change their size and spacing, altering strengthening.

Mechanical working also leaves history through orientation, shape and dislocation structure. Heating a worked metal can allow recovery, recrystallisation or grain growth under appropriate conditions. A useful material description therefore includes composition, processing path and final condition, together with measurements that check what actually resulted.

Check a phase fraction by conserving component mass

Take 100 g of the hypothetical 40 mass % B alloy at the stated equilibrium tie line. Fifty grams of solid at 60 % B contains 30 g B. Fifty grams of liquid at 20 % B contains 10 g B. Their sum is 40 g B in 100 g, which matches the overall composition.

If the overall composition were 30 mass % B at the same tie line, solid fraction would be (30 − 20)/(60 − 20) = 0.25. This means 25 g solid and 75 g liquid per 100 g, not 30 g solid. The result applies only within this two-phase field and on the mass-composition basis used.

The same derivation also checks the original 30 mass % B comparison: 0.25×0.60 + 0.75×0.20 = 0.30. The phase compositions stay at the endpoints for this chosen equilibrium tie line; the overall composition changes their amounts. Densities would be needed to convert those mass fractions into volume fractions.

Try the reasoning

1. Does an equilibrium diagram determine how long a transformation takes?

   Show the answer and reasoning

   No. It describes equilibrium phase information. Rates, nucleation, diffusion and processing path require additional evidence or models.

2. At the stated tie line, does 40 mass % B mean 40 % solid?

   Show the answer and reasoning

   No. Overall component composition and phase fraction are different quantities. The mass balance here gives 50 % solid.

3. Can two specimens with identical composition have different properties after processing?

   Show the answer and reasoning

   Yes. Phase distribution, grain structure, defects, orientation and retained metastable structures can differ.

4. At another invented equilibrium tie line, liquid contains 10 mass % B and solid contains 70 mass % B. A 120 g specimen has overall 45 mass % B. Find both phase masses and check component B.

   Show the answer and reasoning

   The solid fraction is (0.45−0.10)/(0.70−0.10)=7/12. Solid mass is 70 g and liquid mass 50 g. They contain 70×0.70=49 g and 50×0.10=5 g B: 54 g total, equal to 120×0.45. These are mass amounts, not volume amounts or a prediction of transformation time.

5. A learner applies that same two-phase tie line to an overall 80 mass % B specimen and gets more than 100 % solid. Should the result be clipped to 100 %?

   Show the answer and reasoning

   No. An overall composition outside the 10–70 mass % endpoints cannot be formed from a nonnegative mixture of those two endpoint compositions. Recheck the equilibrium field and applicable tie line. Clipping hides a model-domain error.

What to carry forward

Use phase diagrams for equilibrium information, then add kinetics and processing history before predicting actual structure.