For an alloy in a two-phase region of a phase diagram, find what percentage of each phase is present from the overall composition C0 and the compositions at the two ends of the tie line at that temperature (phase α and phase β). The defaults are a Pb–40 wt% Sn alloy just above the eutectic temperature (α phase 18.3 wt% Sn, liquid 61.9 wt% Sn).
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Results
Mass fraction of phase α, Wα
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Mass fraction of phase β, Wβ
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Check
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Mass of each phase
Mass of phase α
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Mass of phase β
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Entering wt% gives mass fractions; entering at% gives fractions by number of atoms (mole fractions). Make sure all compositions use the same unit.
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What the lever rule is
In a two-phase region of a binary phase diagram, a horizontal line (tie line) drawn at a given temperature has ends that give the compositions of the two coexisting phases. For an alloy of overall composition C0, the mass balance C0 = Wα·Cα + Wβ·Cβ together with Wα + Wβ = 1 gives Wα = (Cβ − C0)/(Cβ − Cα) and Wβ = (C0 − Cα)/(Cβ − Cα). It is called the lever rule because, like a balance (lever) with its fulcrum at C0, the length of the opposite arm corresponds to each phase's amount. For the default Pb–40 wt% Sn alloy, Wα = (61.9 − 40)/(61.9 − 18.3) ≈ 50.2 %, and the liquid is about 49.8 %.
Fractions of primary phase and eutectic structure
When a hypoeutectic alloy is cooled to just below the eutectic temperature, all the liquid present just above the eutectic temperature becomes eutectic structure (a lamellar structure of α and β). The fractions of primary α and eutectic structure are therefore equal to the fractions of α and liquid just above the eutectic temperature, and can be calculated by entering the eutectic composition Ce as Cβ. If, on the other hand, you want the fractions of “all α phase” and “all β phase” just below the eutectic temperature, enter the compositions of the α and β phases at the two ends (about 18.3 wt% and about 97.8 wt% Sn for Pb–Sn). Distinguishing which fraction you are after is the most common source of mistakes in this calculation.
Mass fractions and volume fractions
With compositions entered in wt%, the lever rule gives mass fractions. The area fractions seen in micrographs are close to volume fractions, and when the two phases differ in density they do not match the mass fractions. Convert to volume fraction with Vα = (Wα/ρα)/(Wα/ρα + Wβ/ρβ). With compositions entered in at%, the result is the fraction by number of atoms (mole fraction). Always use the same unit for all three compositions you enter.
Use in the Fe–C system
The lever rule is also widely used in calculating steel microstructures. Just below the eutectoid temperature (727 °C), the fractions of ferrite and pearlite in a 0.40 wt% C hypoeutectoid steel are found by entering 0.022 wt% C for the ferrite end and the eutectoid composition 0.76 wt% C for the other end: about 48.8 % proeutectoid ferrite and about 51.2 % pearlite. For the fractions of all ferrite phase and all cementite phase, use 0.022 wt% and 6.70 wt% C for the two ends.
With the lever rule, the fraction of a phase is the distance from the alloy composition to the composition of the opposite phase, divided by the total length of the tie line. The liquid fraction is W_L = (Cα − C₀)/(Cα − C_L) = (42.5 − 35)/(42.5 − 31.5) = 7.5/11.0 ≈ 0.68. About 0.32 is the mistake of taking the distance on the same side, (35 − 31.5)/11.0, which is the fraction of α. About 0.82 is the ratio of the alloy composition to the solid composition, 35/42.5, which is not the lever rule. The fractions of the two phases always sum to 1, which you can use as a check once you have found one of them.
The eutectic structure is what the liquid (61.9 wt% Sn) remaining just above the eutectic temperature turns into through the eutectic reaction. So apply the lever rule in the α + L region just above the eutectic temperature: W_eutectic = W_L = (30 − 18.3)/(61.9 − 18.3) = 11.7/43.6 ≈ 0.27. About 0.73 is the fraction of the remaining primary α, the mistake of taking the opposite distance. About 0.15 is the fraction of all β phase just below the eutectic temperature, (30 − 18.3)/(97.8 − 18.3). All the β is inside the eutectic structure, but the eutectic structure also contains α, so the two do not match. The key to this kind of problem is distinguishing microconstituents (primary and eutectic) from phases (α and β).
Proeutectoid ferrite is the ferrite already formed in the α + γ region just above 727 °C, and all the remaining γ becomes pearlite. From the lever rule, W_proeutectoid α = (0.76 − 0.40)/(0.76 − 0.022) = 0.36/0.738 ≈ 0.49. About 0.51 is the fraction of the remaining pearlite, the mistake of taking the opposite distance. About 0.94 is the fraction of all ferrite treating the steel as the two phases α + Fe₃C, (6.70 − 0.40)/(6.70 − 0.022), which includes the ferrite inside the pearlite as well. The area ratio of proeutectoid ferrite to pearlite in a micrograph is a clue for estimating the carbon content of a hypoeutectoid steel.
FAQ
When can the lever rule be used?
It can be used for an alloy in a two-phase region of a phase diagram at equilibrium. In a single-phase region there is only one phase, so its fraction is 100 %; on the horizontal line exactly at the eutectic temperature, where three phases coexist, the two-phase lever rule alone cannot fix the fractions.
Why is the length on the opposite side used?
The closer C0 is to Cα, the more α phase there should be, so the fraction of α is proportional to the distance from C0 to Cβ (the opposite arm). Solving the mass balance equations produces this form naturally.
Does it apply when cooling is fast?
No. The lever rule assumes equilibrium. With fast cooling, as in real casting, diffusion in the solid cannot keep up and segregation (coring) occurs, giving compositions and fractions different from equilibrium.