Unit 03 PHASE DIAGRAM | Materials Engineering important questions and Answer AKTU

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

Ans. A. Solid Solutions:  

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

2. When the solvent’s crystal structure is unaltered by the addition of the solutes and the combination stays in a single homogenous phase, the mixture is regarded as a solution rather than a compound.

3. Solid solutions easily form when the atom sizes and electron structures of the solvent and solute are comparable, resulting in chemical homogeneity and the inability of the component atoms of the elements to be physically or mechanically differentiated.

4. Two or more components are distributed uniformly in the solid state to produce a single phase or solid solution.

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

a. Substitutional Solid Solution:

• 1. A solute atom may occupy two alternative positions in the lattice of solvent (matrix) metal. 
• 2. If the two atoms have similar sizes, one of the matrix atoms in the crystal lattice will be randomly replaced by the solute atom. A substitutional solid solution is the name given to this type of structure.
• 3. For instance, brass, an alloy of copper and zinc, frequently forms solid solutions due to the comparable sizes and electron structures of these metals atoms.
• 4. Substitutional solid solutions are of two types: 
  • i. Random substitutional solid solutions, and 
  • ii. Ordered substitutional solid solutions. 

b. Interstitial Solid Solutions:

1. To create an interstitial solid solution, a few relatively small atoms can fit in the spaces between solvent atoms.

2. Carbon in iron is the example of such a type of solution which is the basis of steel hardening (Fig.)

3. Interstitial solid solutions are typically very poorly soluble and are only given minor consideration.

4. Both interstitial and substitutional solid solutions are generated to a significant amount in some alloys.

Q2. Discuss Gibbs phase rule. 

Ans. 1. The Gibbs phase rule offers a theoretical framework for describing a system’s chemical state and forecasting the equilibrium relationships of the phases that are present as a function of external factors like pressure and temperature.  

2. It is expressed mathematically as follows: 

3. The quantity of independent external or internal variables, such as temperature, pressure, and concentration, that may be adjusted without causing a phase to dissolve or a new phase to form in the system is known as the number of degrees of freedom.

4. When researching chemical equilibrium, temperature and pressure are taken into account as outside variables affecting the system’s condition.

5. In applying the phase rule of two metal systems the effect of pressure is neglected, leaving only one variable factor- temperature. The equation will then be:

                                                   F = C+1-P 

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

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

7. Therefore, no more than three phases may be in equilibrium in a binary system.

Q3. What do you mean by binary phase diagram ?  

Ans. 1. A graphical representation of a mixture or alloy of two metals instead of pure substance is known as binary phase diagram. 

2. A copper and nickel alloy is a two component system, whereas pure copper is a one component system. A component that is a part of an alloy may occasionally be thought of separately. For instance, two component systems are used to describe ordinary carbon steels that contain both iron and iron carbide.

3. In some binary metallic systems, both the liquid and solid phases of the two constituent elements are entirely soluble in one another.

4. These systems are referred to as isomorphous systems since only one type of crystal structure can be found in them for all component mixtures.

5. The temperature is used as the ordinates in the phase diagram of the copper-nickel system, and the chemical composition in weight percent is used as the abscissa (Fig.)

6. Alloys that have been rapidly cooled through the solidification temperature range are not covered by this figure, which was developed for slow cooling or equilibrium circumstances at atmospheric pressure.

7. In the diagram, the zone of stability for the liquid phase is represented by the liquidus area above the higher line, and the region of stability for the solid phase is shown by the solidus area below the lower line.

8. A two-phase zone where the liquid and solid phases coexist is the space between the liquidus and solidus.

9. The temperature and chemical makeup of the alloy affect how much of each phase is present.

Q4. 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. Determining the phases that are present is quite easy. The temperature-composition point on the diagram is easily found by noting the phase (or phases) that the relevant phase field is designated with.

b. Determination of Phase Compositions : 

• 1. The first step in the determination of phase compositions is to locate the temperature-composition point on the phase diagram.  
• 2. If just one phase is present, its chemical make-up is identical to that of the alloy as a whole.
• 3. The situation is more problematic for an alloy whose composition and temperature are placed in a two-phase area.
• 4. To visualise a sequence of horizontal lines in all two-phase zones, one for each temperature, think of a tie line or, occasionally, an isotherm. These tie lines finish at the phase boundary lines on either side of the two-phase region, which they cross.
• 5. To compute the equilibrium concentrations of the two phases, the following procedure is used: 
  • i. At alloy temperature, a tie line is built across the two-phase region.
  • ii. The phase boundaries and tie line intersections on either side are noted.
  • iii. From these intersections, perpendiculars are lowered to the horizontal composition axis, which is used to read the composition of each of the corresponding phases.

c. Determination of Phase Amounts:

• 1 Phase diagrams can be used to calculate the relative amounts of the phases present at equilibrium.
• 2. In the single-phase region, the answer is clear: Since there is only one phase, the alloy is wholly made up of that phase, or the phase fraction is 1.0 or alternatively, 100%.
• 3. Things become more complicated if the composition and temperature position is found within a two-phase zone.
• 4. The tie line must be utilized in conjunction with a procedure that is often called the lever rule (or the inverse lever rule), which is applied as follows: 
  • i. The tie line is constructed across the two-phase region at the temperature of the alloy. 
  • ii The overall alloy composition is located on the tie line. 
  • iii. To calculate the percentage of one phase, multiply the length of the tie line from the overall alloy composition to the other phase border by the total length of the tie line.
  • iv. The fraction of the other phase is determined in the same manner. 
  • v. If phase percentages are desired, each phase fraction is multiplied by 100.   
• 5. When utilising the lever rule, tie line segment lengths can be calculated by subtracting compositions as obtained from the composition axis or by taking a direct measurement from the phase diagram using a linear scale, ideally graduated in millimetres. 

Q5. Write a short note : 

i. Peritectoid phase diagram. 

ii. Monotectic phase diagram. 

Ans. i. Peritectoid Phase Diagram:

1. When two different solid phases change to a single solid phase, phase diagram is called peritectoid phase diagram and temperature and composition with respect to this peritectoid point are called peritectoid temperature and peritectoid composition.

2. Let there are two solid phase 𝜸 and β which are in equilibrium with a single solid phase 𝞪 at point A. Point A is called peritectoid point. 

3. In the given (Fig.), at point A

ii. Monotectic Phase Diagram:

1. Another three phase invariant reaction that occurs in some binary system is monotectic reaction in which a liquid transforms to another liquid and a solid.

2. Cu-Pb system has a monotectic at 36 % Pb and 955 °C as shown in (Fig.)

Q6. Explain microstructural aspects of ledeburite, austenite, ferrite and cementite. 

Ans. a. Microstructural Aspects of Cementite: 

• 1. To create cementite, a definite quantity of iron and carbon are required. Fe,C is the chemical formula for it. By weight, it has a carbon content of 6.67 percent.
• 2. It is a hard, brittle interstitial compound with a high compressive strength but a poor tensile strength (about 5000 psi).
• 3. Its crystal structure is orthorhombic. It is the hardest structure that appears on the iron carbide diagram.

b. Microstructural Aspects of Austenite: 

• 1. It is an interstitial solid solution of carbon dissolved in iron having a face centered cubic (FCC) crystal structure. 
• 2. Average properties of austenite are as under:
  • i. Tensile strength: 150,000 psi. 
  • ii. Elongation: 10 % in 2 inch gauge length.  
  • iii. Hardness: Rockwell C 40.  
  • iv. Toughness: High. 
• 3. At ambient temperatures, austenite is typically unstable. Austenite can be obtained under specific circumstances at room temperature (as in austenite stainless steels).
• 4. Austenite is non-magnetic. 

c. Microstructural Aspects of Ledeburite:

• 1. It is the eutectic mixture of austenite and cementite. It contains 4.3 percent carbon.  
• 2. It exists when the carbon content exceeds 2%, which serves as the boundary between cast iron and steel on the equilibrium diagram.
• 3. Its melting point is 1147 °C.

d. Microstructural Aspects of Ferrite:

• 1. It is an interstitial solid solution of a small amount of carbon dissolved in iron having a body centered cubic (BCC) crystal structure. 
• 2. It exists above 723 °C. 
• 3. It is the softest structure on the iron carbide diagram. 
• 4. Its maximum solubility is 0.022% at temperature of 727C. 
• 5. Average properties of ferrite are as under 
  • i. Tensile Strength: 40,000 psi.
  • ii. Elongation: 40% in 2 inch gauge length. 
  • iii. Hardness: Less than Rockwell C0 or less than Rockwell B 90. 
  • iv. Toughness: Low. 

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