Lever rule in Cu-Ni alloy

Gibbs phase rule

$$F = C - P + 2$$

where $F$ = degrees of freedom, $C$ = number of components, $P$ = number of phases. For a binary alloy at constant (atmospheric) pressure it becomes $F = C - P + 1$. In a two-phase region of a binary diagram, $F = 2 - 2 + 1 = 1$, so once the temperature is fixed, the compositions of both phases are fixed.

Lever rule

In a two-phase region, draw a horizontal tie line at the temperature. Its ends give the compositions of the two phases, and the mass fractions are inversely proportional to the distance of the alloy composition from each end:

$$W_{L} = \frac{C_\alpha - C_0}{C_\alpha - C_L}, \qquad W_\alpha = \frac{C_0 - C_L}{C_\alpha - C_L}$$

Calculation

Given $C_0 = 35$, $C_L = 31.5$, $C_\alpha = 42.5$ wt% Ni. Since 35 wt% Ni at 1250 °C lies between the liquidus and solidus, both liquid and $\alpha$ are present.

$$W_L = \frac{42.5 - 35}{42.5 - 31.5} = \frac{7.5}{11.0} = 0.682$$ $$W_\alpha = \frac{35 - 31.5}{42.5 - 31.5} = \frac{3.5}{11.0} = 0.318$$

Check: $0.682 + 0.318 = 1$.

Masses for 5 kg:

$$m_L = 0.682 \times 5 = 3.41\ \text{kg}, \qquad m_\alpha = 0.318 \times 5 = 1.59\ \text{kg}$$

Answer: liquid (31.5% Ni) = 68.2% = 3.41 kg; α (42.5% Ni) = 31.8% = 1.59 kg.