### Binary phase diagram of Ni-Cu alloy

The diagram represents phase diagram of binary alloy i.e. copper-Nickel system.

#### Characteristics of this system:

• There is complete solubility because both copper and nickel have same crystal structure.
• The radii of copper and Nickel is similar.
• Both have same electronegativity.
• The melting point of pure copper is $$1085^\circ$$C and that of Nickel is $$1453^\circ$$C. At 40% composition of Nickel (60 wt % of copper) just above $$1200^\circ$$C, the material starts to melt and begins into new phase (solid + liquid) and just above $$1300^\circ$$C the substances completely melts and becomes liquid.
• The reference point A indicate that when the composition of alloy is 60% Nickel and 4o wt % copper the material is at $$110^\circ$$C in the solid state, In $$1350^\circ$$C it is in liquid $$\alpha + L$$ state. And at$$1500^\circ$$C the substance is in liquid phase. For the same composition ther change in the state represents different microstructure.
• In multicomponent system melting occurs over range of temperature between solidous and liquidous line. Between this temperature separated by two phase boundary the substances is in equilibrium.

#### Determination of composition in two phase region

The amounts (as fraction or as percentage) of the phases present at equilibrium can be computed with the help of phase diagrams. The single and two-phase conditions must be treated separately. The solution is easy in the single-phase region due to only one phase is present. The alloy is composed entirely of that phase; i.e. the phase fraction is 1.0 or, alternatively, the percentage is 100%. From the previous example for the 60 wt% Ni–40 wt% Cu alloy at $$110^\circ$$C, only the $$\alpha$$ phase is present. Hence, the alloy is completely or 100% solid phase. If the temperature and composition position is located within a two-phase region, then the calculations are more complex. So, the tie line must be used to reduce the complexity by the rule that is often called the lever rule (or the inverse lever rule), which is applied as follows;

#### Rules:

1. Locate the composition and temperature in diagram.
2. In two phase region draw tie line or isotherm.
3. Note intersection with phase boundary. Read the composition at phase boundary.

Finding the amount of phase in two phase region

Rules

1. Locate the amount of phase in two phase.
2. In two phase region draw the tie line or isotherm.
3. The fraction of the phase and dividing total length of tie line. The tie line is a mathematical analogous of beam balance .

#### Calculation:

The tie line in two phase region is analogous to lever balance on fulcrum.

Derivation of lever rule;

Consider a binary alloy having three phases $$\alpha, \alpha +L$$ and L as shown in diagram. Let the over all alloy composition along the tie line denoted as $$C_\circ$$ and mass fraction be represented by $$\omega_L$$ and $$omega_\alpha$$ for the respective phase. The lever rule states that,

$$\omega_L=\frac{s}{R+S}$$where S is the distance between solidous line and vertical axes having concentration $$C_\circ$$ and R is length between vertical line to liquidous line in tie line as shown in figure.

Now,

$$\omega_L=\frac{C_\alpha-C_\circ}{C_\alpha-C_L}\dotsm(1)$$

$$\omega_L=\frac{R}{R+S}$$

$$\omega_\alpha=\frac{C_\circ-C_L}{C_\alpha}{C_\alpha-C_L}$$

Proof:

1. All material must be in one phase or other.

$$\omega_L+\omega_\alpha=1\dotsm(3)$$

1. Mass of component that is present in both phase equal to mass of the component of one phase plus (+) mass of component in second phase i.e.

$$\omega_L C_L+\omega_\alpha C_\alpha=C_\circ\dotsm(4)$$

Substitution, $$\omega_\alpha=1-\omega_L$$ in equation (4)

$$\omega_L C_L+(1-\omega_L)C_\alpha=C_\circ$$

$$\omega_L(C_L-C_\alpha)=C_\circ-C_\alpha$$

$$\omega_L=\frac{C_\circ-C_\alpha}{C_L-C_\alpha}=\frac{C_\alpha-C_circ}{C_\alpha-C_L}$$Similarly, when we substitute $$\omega_L=1-\omega_\alpha$$ we got,

$$\omega_\alpha=\frac{C_\circ-C_L}{C_\alpha-C_L}$$

Summary:

Compositions of phases are expressed in terms of weight percents of the components (e.g., wt% Cu, wt% Ni). For any alloy consisting of a single phase, the composition of that phase is the same as the total alloy composition. If two phases are present, the tie line must be used, the extremities of which determine the compositions of the respective phases. With regard to fractional phase amounts, when a single phase exists, the alloy is completely that phase. For a two-phase alloy, the lever rule is used. In which a ratio of tie line segment lengths is take.

References:

Callister, W.D and D.G Rethwisch. Material Science and Engineering. 2nd. New Delhi: Wiley India, 2014.

Lindsay, S.M. Introduction of Nanoscience . New York : Oxford University Press, 2010.

Patton, W.J. Materials in industry . New Delhi : Prentice hall of India, 1975.

Poole, C.P. and F.J. Owens. Introduction To Nanotechnology. New Delhi: Wiley India , 2006.

Raghavan, V. Material Science and Engineering. 4th . New Delhi: Pretence-Hall of India, 2003.

Tiley, R.J.D. Understanding solids: The science of Materials. Engalnd : John wiley & Sons , 2004.

1.Characteristics:

1. There is complete solubility because both copper and nickel have same crystal structure.
2. The radii of copper and Nickel is similar.
3. Both have same electronegativity.
4. The melting point of pure copper is $$1085^\circ$$C and that of Nickel is $$1453^\circ$$C. At 40% composition of Nickel (60 wt % of copper) just above $$1200^\circ$$C, the material starts to melt and begins into new phase (solid + liquid) and just above $$1300^\circ$$C the substances completely melts and becomes liquid.

2. important equations:

$$\omega_\alpha=\frac{C_\circ-C_L}{C_\alpha-C_L}$$

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