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Unit 1 Phase Diagrams in Material Science: Important Questions for AKTU B.Tech.

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Important Questions For Material Science : 
*Unit-01     *Unit-02    
*Unit-03    *Unit-04 
*Unit-05    *Short-Q/Ans
*Question-Paper with solution 21-22 

Q1. What do you understand by solid solution ? Explain with neat sketches the substitutional solid solution and interstitial solid solution.

Ans. 1. A solid solution is a solution of one or more solutes in a solvent in a solid state

2. When the crystal structure of the solvent stays unaffected by the addition of the solutes and the combination remains in a single homogenous phase, the mixture is called a solution rather than a compound.

What do you understand by solid solution ? Explain with neat sketches the substitutional solid solution and interstitial solid solution.

3. Solid solutions form easily when the solvent and solute atoms have comparable sizes and electron configurations, resulting in a chemically homogenous solution in which the constituent atoms of the elements cannot be distinguished physically or mechanically separated.

4. A single phase or solid solution is formed by the homogenous distribution of two or more elements in the solid state.

B. Types of Solid Solutions : These are of following two types :

a. Substitutional Solid Solution :

  • 1. A solute atom can take one of two locations in the solvent (matrix) metal lattice.
  • 2. If the two atoms are similar in size, the solute atom will randomly substitute for one of the matrix atoms in the crystal lattice. A substitutional solid solution is the name given to this type of structure.
  • 3. Brass, for example, is an alloy of copper and zinc that readily forms solid solutions because the atoms of these elements have similar sizes and electron structures.
  • 4. Substitutional solid solutions are of two types:
    • i. Random substitutional solid solutions, and
    • ii. Ordered substitutional solid solutions.
What do you understand by solid solution ? Explain with neat sketches the substitutional solid solution and interstitial solid solution.

b. Interstitial Solid Solutions :

1. To generate an interstitial solid solution, a few relatively tiny atoms can be accommodated in the interstices between solvent atoms.

2. Carbon in iron is an example of such a solution, which serves as the foundation for steel hardening (Fig.)

What do you understand by solid solution ? Explain with neat sketches the substitutional solid solution and interstitial solid solution.

3. Interstitial solid solutions often have relatively limited solubility and are regarded as secondary in importance.

4.Some alloys form significant amounts of both interstitial and substitutional solid solutions.

Q2. Discuss Gibba phase rule.

Ans. 1. The Gibbs phase rule offers the theoretical basis for describing a system’s chemical state and forecasting the equilibrium relationships of the phases present as a function of physical circumstances such as pressure and temperature.

2. It is expressed mathematically as follows :



      ∵                             n = number of external factors

                                        = 2 (temperature and pressure)

3. The number of degrees of freedom is the number of independent external or internal variables, such as temperature, pressure, and concentration, that can be modified without causing a phase to disappear or a new phase to arise in the system.

4. Temperature and pressure are regarded external elements impacting the condition of the system when researching chemical equilibrium.

5. The influence of pressure is ignored when applying the phase rule to two metal systems, leaving just one variable factor – temperature. The equation will then be :


6. In equilibrium all factors have definite values, hence the degree of freedom cannot be less than zero

                                                    C-P+1 ≥ O

then                                             P ≤ C + 1

which shows that the number of phases in a system cannot exceed the number of components plus one.

7. As a result, in a binary system, no more than three phases can be in equilibrium.

Q3. What kind of interpretation can be obtained from a binary phase diagram ?

Ans. A binary phase diagram is helpful in obtaining the following three type of information :

a. Phases Present :

1. It is relatively straightforward to determine which phases are present. The temperature-composition point on the figure is simply located, and the phase (s) with which the related phase field is labelled is noted.

b. Determination of Phase Compositions :

  • 1. The temperature-composition point on the phase diagram is the first step in determining phase compositions.
  • 2. If only one phase exists, its composition is simply the same as the alloy’s overall composition.
  • 3. The situation is more complicated for an alloy whose composition and temperature are in a two-phase area.
  • 4.Imagine a sequence of horizontal lines, one at each temperature, in all two-phase zones; each of these is known as a tie line, or sometimes as an isotherm. These tie lines run through the two-phase region, terminating at the phase boundary lines on either side.
  • 5. To compute the equilibrium concentrations of the two phases, the following procedure is used:
    • i.  At the alloy’s temperature, a tie line is built across the two-phase region.
    • ii. On either side, the intersections of the tie line and the phase boundaries are noted.
    • iii. From these intersections, perpendiculars are dropped to the horizontal composition axis, from which the composition of each of the individual phases is read.

c. Determination of Phase Amounts :

  • 1. Phase diagrams can also be used to calculate the relative amounts of the phases present at equilibrium.
  • 2. In the single-phase region, the solution is obvious: because only one phase is present, the alloy is wholly composed of that phase, i.e., the phase fraction is 1.0 or, alternatively, the percentage is 100%.
  • 3. Things become more complicated if the composition and temperature position are in a two-phase zone.
  • 4. The tie line must be used in conjunction with a technique known as the lever rule (or the inverse lever rule), which is implemented as follows:
    • i. At the alloy’s temperature, the tie line is built across the two-phase region.
    • ii. On the tie line is the overall alloy composition.
    • iii. The fraction of one phase is calculated by dividing the total tie line length by the length of the tie line from the overall alloy composition to the phase boundary for the other phase.
    • iv. In the same way, the proportion of the other phase is calculated.
    • v. If phase percentages are desired, multiply each phase fraction by 100.
  • 5. Tie line segment lengths can be determined using the lever rule either directly from the phase diagram using a linear scale, preferably graded in millimetres, or by subtracting compositions as taken from the composition axis.

Q4. Explain eutectic phase diagram with its reaction.

Ans. 1. When the melting points of the two components are not significantly different and there is only partial or minimal solid solubility between them, the phase diagram is referred to as a eutectic phase diagram.

2. In (Fig.), we chose lead and tin because their melting points are so close. There are two solid phases present: a and 13.

3. In a-phase very small amount of tin is dissolved in lead and in β-phase very small amount of lead is dissolved in tin.

4. In α+ β phase both tin and lead are dissolved in each other sufficiently.

5. BDE line separates the two phase regions of α + L and α + β, α + β and L, α + β and β + L . This line is called eutectic line and temperature corresponding to this line is called eutectic temperature.

Explain eutectic phase diagram with its reaction.

6. At point D eutectic reaction occurs as liquid phase is in equilibrium with two solid phases.

Explain eutectic phase diagram with its reaction.

7. At point D,

Degree of freedom,D = C – P + 1

                               C = 2,P=3

So ,                         D = 2- 3+1=0

8. At point D, eutectic temperature and eutectic reaction are called invariant temperature and invariant reaction.

Q5. Explain the invariant reactions.

Ans. 1. When a liquid of eutectic composition is steadily chilled to eutectic temperature, the single liquid phase changes into solid forms simultaneously. This is known as an invariant reaction or a eutectic reaction.

2. It is given as,

Explain the invariant reactions.

3. It is so named because it happens under equilibrium conditions at a fixed temperature and alloy composition.

4. Some invariant reactions are shown in Table.

Explain the invariant reactions.

Q6. Explain microstructural changes during cooling.


  • 1. Consider the copper-nickel system, more precisely an alloy with the composition 35 wt% Ni-65 wt% Cu that has been cooled from 1300 °C.
  • 2. Fig.. depicts the region of the Cu-Ni phase diagram near this mixture.
  • 3. Moving down the vertical dashed line corresponds to cooling an alloy with the above composition.
  • 4. At 1300 °C, point a, the alloy is entirely liquid (composition 35 wt% Ni-65 wt% Cu) and exhibits the microstructure shown in the fig. by the circle inset.
  • 5. No microstructural or compositional changes will be observed as cooling begins until we hit the liquidus line (point b. -1260 °C).
  • 6. At this point, the first solid a. begins to form, which has a composition dictated by the tie line drawn at this temperature [i.e., 46 wt% Ni-54 wt% Cu, noted as a.( 46 Ni)]; the composition of liquid is still approximately 35 wt% Ni-65 wt% Cu [L(35 Ni)], which is different
  • from that of the solid α.
  • 7. The compositions and relative amounts of each phase will alter as cooling continues.
  • 8. The liquid and solidus compositions will follow the liquidus and solidus lines, respectively. Furthermore, the fraction of the α. phase will increase with continued cooling.
  • 9. The overall alloy composition (35 wt% Ni-65 wt% Cu) remains unchanged during cooling even though there is a redistribution of copper and nickel between the phases.
  • 10. At 1250 °C, point c in (Fig. ), the compositions of the liquid and α phases are 32 wt% Ni-68 wt% Cu [L(32 Ni)] and 43 wt% Ni-57 wt% Cu [α (43 Ni)], respectively.
Explain microstructural changes during cooling.
  • 11. At around 1220 °C, point d, the solid’s composition is approximately 35 wt% Ni-65 wt% Cu (the total alloy composition), whereas the last remaining liquid’s composition is 24 wt% Ni-76 wt% Cu.
  • 12.When the residual liquid crosses the solidus line, it solidifies, yielding a polycrystalline α-phase solid solution with a homogeneous 35 wt% Ni-65 wt% Cu composition, point e. (Fig.).

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