---
title: "Lever rule in Cu-Ni alloy"
subject: "Material Science"
chapter: 5
key: "semester_3/material_science:lever-rule-copper-nickel-alloy"
source: https://www.bctengineer.com/q/semester_3/material_science/lever-rule-copper-nickel-alloy
site: BCT Engineering
---

# Lever rule in Cu-Ni alloy

Material Science · Chapter 5: Solidification, Phase Relations and Strengthening Mechanism

- Asked in: year not given, 6 marks
- Video: [Lever Rule Concept | cu-Ni phase Diagram | tie line rule | 1-2-1 principal in material science |](https://www.youtube.com/watch?v=yDsGUXfEeuM) — Pathshala Club Jsr
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/lever-rule-copper-nickel-alloy
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-5

## Question

Explain the Gibbs phase rule and the lever rule. A Cu-35 wt% Ni alloy of mass 5 kg is held at 1250 °C. From the Cu-Ni phase diagram, at this temperature the liquid contains 31.5 wt% Ni and the alpha phase contains 42.5 wt% Ni. Determine the phases present, their compositions, mass fractions and masses.

## Short answer

Liquid (31.5% Ni) mass fraction = (42.5-35)/(42.5-31.5) = 0.682, mass = 3.41 kg; α (42.5% Ni) mass fraction = (35-31.5)/(42.5-31.5) = 0.318, mass = 1.59 kg.

## Answer

#### 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:

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

| Phase | Composition (wt% Ni) | Mass fraction | Mass (kg) |
|---|---|---|---|
| Liquid | 31.5 | 0.682 | 3.41 |
| Alpha (solid) | 42.5 | 0.318 | 1.59 |

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a binary alloy at constant pressure in two-phase region, F =
   A. 0
   B. 1
   C. 2
   D. 3
   Answer: B. F = C - P + 1 = 2-2+1 = 1. (Key point: Phase rule)

2. Mass fraction of liquid W_L =
   A. (C0 - CL)/(Ca - CL)
   B. (Ca - C0)/(Ca - CL)
   C. C0/Ca
   D. CL/C0
   Answer: B. Fraction of a phase is the opposite lever arm over the tie line. (Key point: Lever rule)

3. W_L for C0 = 35?
   A. 0.318
   B. 0.5
   C. 0.682
   D. 0.75
   Answer: C. 7.5/11 = 0.682. (Key point: Result)

4. Mass of alpha in 5 kg?
   A. 1.59 kg
   B. 3.41 kg
   C. 2.5 kg
   D. 0.318 kg
   Answer: A. 0.318 x 5 = 1.59 kg. (Key point: Mass)

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