Phase Transformation Pune University MCQs
Phase Transformation Pune University MCQs
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Thermodynamics and Phase Diagrams – Equilibrium”.
1. The relative stability of a system for transformations that occur at constant temperature and pressure is determined by its ____________
a) Exergy
b) Helmholtz free energy
c) Entropy
d) Gibbs free energy
Answer: d
Explanation: Gibbs free energy decides the stability of a system for transformations that occur at constant temperature and pressure. Helmholtz free energy decides the stability of a system for transformations that occur at constant volume and temperature. The amount of exergy a system has is not dependent on whether or not it’s an isothermal or isobaric process.
2. Based on the Gibbs phase rule, how many degrees of freedom are present at the triple point of water?
a) 1
b) 2
c) 0
d) 3
Answer: c
Explanation: According to the Gibbs phase rule, a single component system has no degrees of freedom when the three phases exist in equilibrium and the system is invariant.
3. Solid phases are most stable at low temperatures.
a) True
b) False
Answer: a
Explanation: From the concept of Gibbs free energy it is clear that the state with highest stability will be that with the best compromise between low enthalpy and high entropy. Thus, at low temperature solid phases are more stable since they have the strongest atomic binding and therefore the lowest internal energy .
4. Which of the following is a necessary criterion for any transformation ?
a) G2-G1<0
b) G1=G2
c) G2>G1
d) G2=0
Answer: a
Explanation: When ΔG = G2-G1 is less than zero the reaction is spontaneous in the forward direction. Any transformation that results in a decrease in Gibbs energy is possible.
5. Which of the thermodynamic functions can be considered as an intensive property?
a) Volume
b) Pressure
c) Enthalpy
d) Entropy
Answer: b
Explanation: Pressure is an intensive property. Intensive properties are those which are independent of the size of the system such as temperature and pressure.
6. Which among the following points represents metastable equilibrium?
phase-transformation-questions-answers-equilibrium-q6
a) A
b) B
c) C
d) D
Answer: a
Explanation: Point A represents the metastable state. Point A is a local minimum of free energy and satisfies the condition ΔG=0, but do not have the lowest value among these points hence it is in the metastable state.
7. When the system is at local equilibrium, its Gibbs free energy function has reached its minimum value.
a) True
b) False
Answer: b
Explanation: When the system is at complete equilibrium or global equilibrium, its Gibbs free energy function has reached its minimum value. Local equilibrium, on the other hand is defined in such a way that the equilibrium exists only at the interfaces between the different phases present in the system.
8. Graphite and diamond at room temperature and pressure are examples of _____________
a) Unstable and stable equilibrium states
b) Stable and unstable equilibrium states
c) Metastable and stable equilibrium states
d) Stable and metastable equilibrium states
Answer: d
Explanation: Graphite and diamond at room temperature and pressure are examples of stable and metastable equilibrium states. Given time, diamond under these conditions will transform to graphite.
9. The quantum of energy in an elastic wave is called a ___________
a) Debye model
b) Sonar
c) Phonon
d) Photon
Answer: c
Explanation: Phonon is the name for the quantum of energy in an elastic wave. Sonar is method or technique to detect object on or under the surface of water. Debye model is a technique to estimate the phonon contribution to the specific heat.
10. What happens to the entropy in a system with constant volume and constant internal energy during a spontaneous process?
a) Remains same
b) Decreases
c) Increases
d) First decreases then increases
Answer: b
Explanation: At either constant internal energy or constant entropy .
In the Clausius inequality when we apply the above mentioned condition, finally we get the equation which states that the entropy in a system decreases when the volume and internal energy remains constant during a spontaneous process.
11. Two crystal structures are in equilibrium and their Gibbs energies are the same. Thus, the driving force for the transformation is_________
a) Less than zero
b) Zero
c) Greater than Zero
d) One
Answer: b
Explanation: Since the Gibbs energies are the same that means, ΔG is 0, hence the driving force will be zero, if the driving force is not zero then it violates the basic rule of phase transformation.
This set of Phase Transformation online quiz focuses on “Thermodynamics and Phase Diagrams – Single Component System”.
1. The slope of the H-T curve is ________
a) Specific heat CV
b) Pressure
c) Volume
d) Specific heat CP
Answer: d
Explanation: The slope of the graph gives the value of specific heat Cp. As dH/dt for pure metal gives the CP value. CP is actually the specific heat of the gas at constant pressure and interestingly solids and fluids are having only a single value for specific heat.
2. Reason why in many metals, we find the phase with a relatively closely packed structure is stable at a lower temperature, whereas a relatively loosely packed structure is stable at higher temperature?
a) Enthalpy
b) Volume
c) Entropy term
d) Temperature
Answer: c
Explanation: The increasing importance of the entropy term that is –TS is the reason behind the stability of closely packed structure at relatively low temperature and as the temperature increases the entropy term increases because of increase in T hence the randomness increases and hence the movement of individual particle increases which creates a kind of instability in closely packed structure at high temperature.
3. At all temperatures the liquid has a higher enthalpy than solid.
a) True
b) False
Answer: a
Explanation: Liquid absorbs more heat and they covet their gained heat into higher vibrational and translational molecular motion, hence at temperatures the liquid has a higher enthalpy than solid.
4. Iron exist as BCC ferrite at temperature____________
a) Above 910 degree Celsius
b) Below 910 degree Celsius
c) Below 1000 degree Celsius
d) Above 1000 degree Celsius
Answer: b
Explanation: At atmospheric pressure iron can exist as either BCC ferrite below 910 degree Celsius or FCC austenite above 910, and at 910 degree Celsius both phases can exist in equilibrium.
5. Which of the following is a spontaneous exothermic process?
a) Rusting
b) Ice melting
c) Removal of sugar from the coffee
d) Dissolution of sand in water
Answer: a
Explanation: Rusting is spontaneous exothermal process .Melting of ice is a spontaneous endothermic process and the last two are non-spontaneous in nature.
6. A liquid metal is undercooled by ΔT below Tm before it solidifies, solidification will be accompanied by _____
a) A constant free energy
b) An increase in free energy
c) A decrease in free energy
d) ΔG=0
Answer: c
Explanation: There will be a decrease in free energy. The decrease in free energy provides the driving force for solidification. To get the solidification process started, the liquid phase must be undercooled, cooled to a temperature below the freezing point.
7. Which rule states that the entropy of fusion is a constant ==<R (8.3 J mol -1 K -1 ) for most metals?
a) Richards rule
b) Stefan’s Rule
c) Charles law
d) Fermat’s rule
Answer: a
Explanation: Richards rule states that the entropy of fusion is a constant for most metals. Fermat’s rule helped in finding the greatest and smallest ordinates in curved lines and Charles law states that pressure remaining constant, the volume of a given mass of gas increases or decreases by 1/273 part of its volume at 0 degree Celsius for each degree Celsius rise or fall of its temperature.
8. The enthalpy of a phase which is stable at higher temperature is higher than that of a phase which is stable at lower temperature.
a) True
b) False
Answer: a
Explanation: The enthalpy of a phase which is stable at higher temperature is higher because ΔH value is greater than 0 that is H1-H2>0.
9. What happens to the internal energy of a system with constant volume and constant entropy during a spontaneous process?
a) Increases first then decreases
b) Decreases
c) Remains same
d) Increases
Answer: b
Explanation: Internal energy of the system decreases. This in fact a statement about entropy since it states that if the entropy of the system remains unchanged in the transformation, there must be an increase in the entropy of the environment caused by the outflow of heat from the system.
10. Which of the following structure is stable at low temperature?
a) α-TI with BCC structure
b) β-TI with HCP structure
c) β-TI with BCC structure
d) α-TI with HCP structure
Answer: d
Explanation: The reason for this lies in a more closely packed structure where there is a lesser degree of vibrational freedom, hence it makes α-TI with HCP structure stable at low temperature.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Thermodynamics and Phase Diagrams – Binary Solutions”.
1. When X1 mole of A and X2 mole of B are brought together, the corresponding free energy of system is given by___________
a) X1*G1+X2*G2
b) X1*G2+X2*G1
c) X1*
d) X2*
Answer: a
Explanation: In order to calculate the free energy of the alloy, the mixing can be made in two steps. After the first step the Gibbs free energy of the system is given by X1*G1+X2*G2.
2. If we consider that there are no volume and enthalpy changes caused by mixing, then the only contribution to the entropy will be__________
a) Thermal
b) Configurational
c) Hydrostatic
d) Rotational
Answer: b
Explanation: The only contribution to the entropy will be Configurational. Configurational entropy comes from the possibility of arranging the atoms A and B in different ways for a particular macro state.
3. In an exothermic transformation, the total free energy of the system will change far more drastically with composition at higher temperature compared to a lower temperature.
a) False
b) True
Answer: b
Explanation: At higher temperature the entropy term of mixing will be higher, hence Δg of mix will be higher making more drastic changes at higher temperature. Since exothermic transformations involves the release of heat, very high temperatures are expected. As the entropy term –TS depends on the value of T, higher the value of T more will be the change in free energy.
4. In endothermic transformation the enthalpy of mixing is __________
a) Less than zero
b) Zero
c) Greater than zero
d) One
Answer: c
Explanation: In endothermic transformation, the enthalpy of mixing is greater than 0. So, if the temperature under consideration is reasonably low, the negative contribution to the Gibbs energy of mixing from entropy term may be smaller than the positive contribution from the enthalpy of mixing within a certain composition range.
5. If we allow inter diffusion to taken place between the elements A and B, there will be change in the free energy because of mixing, if the total molar free energy g of a purely mechanical mixture is 8 Jmol-1
and the ΔG of mix is 5Jmol -1 , then the total free energy of the system is ____________
a) 3 Jmol -1
b) 13 Jmol -1
c) 40 Jmol -1
d) 15 Jmol -1
Answer: b
Explanation: The total free energy is given by ΔG mix +free energy g. The free energy of the system will not remain constant during the mixing of the A and B atoms and once after mixing free energy of the solid solution G, will depend on the ΔG of mix which comes into picture because of the mixing of the 2 atoms.
6. In case of an ideal system ΔH of mixing is _________
a) Negative
b) Positive
c) Zero
d) Cannot be determined
Answer: c
Explanation: In the case of an ideal solution, ΔHof mixing is 0 and the free energy of the system can be written as sum of free energy of mechanical mixture and change in free energy while mixing.
7. The configurational entropy depends on _____
a) Thermodynamic probability
b) Volume
c) Composition
d) Frequency of mixing
Answer: a
Explanation: From statistical thermodynamics one can conclude that the configurational entropy depends on the thermodynamic probability, a kind of measure of randomness. Configuration entropy is actually a part of a system’s entropy that is related to distinct representative positions of its constituent particle.
8. Partial molar free energy of a pure substance A or alternatively the chemical potential of A in the phase depends on_____
a) Composition
b) Volume
c) Enthalpy
d) Configuration
Answer: a
Explanation: The partial molar free energy or chemical potential of a species in a mixture is defined as the rate of change of free energy of a system with respect to the change in the number of atoms or molecules of the species that are added to the system. If we take the partial derivative into consideration the free energy with respect to the concentration remains constant, hence it depends only on the composition.
9. Assuming that A and B mix to form a substitutional solid solution and that all configurations of A and B atoms are equally probable, the number of distinguishable ways of arranging the atoms on the atom sites, if N1=8 and N2=4 is_____ .
a) 467
b) 499
c) 495
d) 500
Answer: c
Explanation: The required equation for the above mentioned problem is ! / *). So substituting the value of N1 and N2 we have 12! / = 495.
10. The variation of activity with composition is given below, out of this which one obeys Raoults law__________
phase-transformation-questions-answers-binary-solutions-q10
a) 1
b) 2
c) 4
d) 3
Answer: a
Explanation: For the Raoults law to satisfy the slope of the line should be 1.Raoults law is a special case of Henrys law. The law takes into account that the vapor pressure of the whole solution will always be less than that of the pure solvent. Therefore, the vapor pressure of the solution will depends on the concentration of the solute.
11. When does the strain energy change can make a difference in the enthalpy of mixing in real solutions?
a) When the temperature is high
b) When the size difference between atoms is very large
c) When the temperature is low
d) When the size difference between atoms is very small
Answer: b
Explanation: When the size difference between the atoms is very large then interstitial solid solutions are energetically most favourable and the temperature doesn’t play a big role in making a difference in the enthalpy of mixing in real solutions when the strain energy gets altered.
This set of Phase Transformation Interview Questions and Answers focuses on “Equilibrium in Heterogeneous Systems”.
1. Which rule gives the relative number of moles of α and β that must be present, if the overall composition is given as some X?
a) Richard’s Rule
b) Raoul’s Rule
c) Henry’s Rule
d) Lever rule
Answer: d
Explanation: If the overall composition of the phase mixture is X, the lever rule gives the relative number of moles of α and β that must be present and the Gibbs free energy can be calculated from the molar energy curve for the phase.
2. Properties of metallic material primarily depends on _____________
a) Processing
b) Type of crystal structure
c) Microstructure
d) Grain size
Answer: a
Explanation: The properties of metallic material primarily depends on chemical composition and processing. This is the property that decides the microstructure that gets developed into the material.
3. The driving force for the phase transformation increases in proportion to__________
a) Microstructure
b) Change in composition
c) Degree of undercooling below critical temperature
d) Melting point
Answer: c
Explanation: It primarily depends on undercooling and that provides for the energy to create surface, enveloping the product phase. Similarly as you cool a single phase material below a solvus the driving force of precipitation increases so cooling is basically the driving force of many processes.
4. In a general sense it is not possible to obtain any information about kinetics or the morphology of the phase mixture from the phase diagram.
a) True
b) False
Answer: a
Explanation: It is not possible to obtain any information about kinetics or the morphology of the phase mixture from the phase diagram. The diagram only gives information about the phases that can be in equilibrium under certain composition–temperature combinations. Although there are three different species present in a system, there are times when the phase diagram is presented as a binary phase diagram.
5. The first step in drawing the free energy curve of the FCC α phase is, to convert the stable BCC arrangement of β atoms into ____________
a) An unstable FCC arrangement
b) A stable FCC arrangement
c) A metastable BCC arrangement
d) An unstable BCC arrangement
Answer: a
Explanation: The first step is to convert it into an unstable FCC arrangement, since was in a stable state it requires the optimum energy to continue in that state but to convert it into an unstable FCC arrangement, we require an increase in free energy.
6. Which among the following condition the most stable state of a system should satisfy so that it can be defined as a state in which pure component has unit activity?
a) Highest free energy
b) Lowest free energy
c) Change in Gibbs energy 0
d) Positive free energy
Answer: b
Explanation: The most stable state, with the lowest free energy, is usually defined as the state in which the pure component has unit activity. The definition of activity can be shown graphically.
7. In heterogeneous systems containing more than one phase the pure components can, at least theoretically, exist in different crystal structures.
a) True
b) False
Answer: a
Explanation: In a heterogeneous system containing more than one phase the pure substance can exist in different crystal structures, the definition of heterogeneous system itself gives an idea about the statement.
8. Calculate the number of phases present in a system in equilibrium, if the degree of freedom is 1, the number of components and non-compositional variables are 1 and 2 respectively?
a) 1
b) 3
c) 0
d) 2
Answer: d
Explanation: The Gibbs phase rule explains the number of phases that will be present in a system in equilibrium and is expressed as F+P = N+C. So here the F value is given as 1 and the N, C as 1 and 2 respectively, hence the P = 2.
9. What happens when the two Gibbs energy curves for the α- and β-phases have a common tangent and the equilibrium has been established?
a) Driving force for diffusion increases
b) Driving force for diffusion decreases
c) Driving force for diffusion remains constant
d) Driving force first decreases then increases
Answer: a
Explanation: As the diffusion proceeds, the driving force for diffusion gradually decreases and vanishes when the chemical potentials of the components become equal in both phases.
10. The condition for equilibrium in a heterogeneous system containing two phases can also be expressed using the_____________
a) Lever’s concept
b) Raoults rule
c) Henrys rule
d) Activity concept
Answer: d
Explanation: The same activity concept defined for homogeneous system can be used to state the above mentioned condition. From the definition of activity coefficient, for an ideal solution, its value is unity for all the species. For a non-ideal solution, however, it may be either greater or less than unity, the larger the departure from unity the greater the non-ideality of the solution.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Binary Phase Diagrams”.
1. Which among the following condition about a simple binary system is not a part of Hume- Rothery rule?
a) Same crystal structure
b) Size difference is less than 15
c) Electronegativity’s have similar values
d) Composition should be almost similar
Answer: d
Explanation: Hume-Rothery rules doesn’t state anything about the composition of the system. Hume-Rothery rule states about the atomic size factor, it states about the similarity of 2 elements based on the solid solubility factor, it tells about the dissolving capacity of a metal based on its valency and it relates the electronegativity with solubility.
2. The Gibbs free energy variation for a binary system of constituents A and B is shown in the figure below. The point of intersection corresponds to____________________
phase-transformation-questions-answers-binary-phase-diagrams-q2
a) The composition of the solid particle that precipitate out
b) The composition of the liquid particle that precipitate out
c) The amount of solid that converts to liquid
d) The amount of liquid that converts to solid
Answer: a
Explanation: For the situation corresponding to the point of intersection of the solidus and liquid line indicate the composition of the solid particles that precipitate out at the initiation of the solid liquid transformation during cooling.
3. Enthalpy of mixing is negative, for such systems melting will be more difficult in the alloys and a maximum melting point mixture may appear.
a) True
b) False
Answer: a
Explanation: In these systems melting will be more difficult in the alloys and a maximum melting point mixture may appear. This type of alloy also has a tendency to order at low temperatures. If the attraction between unlike atoms is very strong the ordered phase may extend as far as the liquid.
4. What happens when the the atoms in the alloy ‘repel’ each other making the disruption of the solid into a liquid phase possible at lower temperatures than in either pure A or pure B?
a) Alloys melt at temperature below melting point
b) Alloys melt at temperature above melting point
c) Alloys melt at room temperature
d) Alloys solidify
Answer: a
Explanation: The effect of a positive enthalpy of mixing in the solid is already apparent at higher temperatures where it gives rise to a minimum melting point mixture so hence the alloys melt at temperature below the melting point.
5. The stable composition range of the phase in the phase diagram need not include the composition with the minimum free energy, but is determined by _____
a) Relative free energy of adjacent phases
b) Enthalpy of phases
c) Free energy of stable phase
d) Relative enthalpy of adjacent phases
Answer: a
Explanation: The Relative free energy of adjacent phases is used to determine the composition with minimum energy, actually this is an interesting result of common tangent construction.
6. According to Gibbs rule the number of phases present in a system in equilibrium is given as_________
a) P=C+N-F
b) P=C+F-N
c) P=F+N-C
d) P=C+N+F
Answer: a
Explanation: P+F=C+N, where F is the degree of freedom, N is the non-compositional variable and C is the number of components. Here the degree of freedom is an intensive variable hence can be varied independently while still maintaining the equilibrium.
7. The equations for free energy and chemical potential can be used to derive the effect of temperature on the limits of solid solubility in a terminal solid solution.
a) True
b) False
Answer: a
Explanation: The equations for free energy and chemical potential can be used to derive the effects and this can be proved by using a phase diagram and from the equation dG= -SdT + VdP + μdN, it is clear that the free energy changes with the change in pressure, temperature and chemical potential.
8. What happens to the internal energy when atoms are moved from their respective sites?
a) Internal energy remains constant
b) Internal energy increases
c) Internal energy decreases first then increases
d) Internal energy decreases
Answer: b
Explanation: The removal of atoms from their sites increases the internal energy of the metal because atoms at their respective positions are stable. If they are disturbed, forced, or moved from their respective sites then due to the broken bonds around the vacancy the internal energy increases.
9. What happens to the configurational entropy when atoms are moved from their respective sites?
a) Remains constant
b) Decreases
c) Increases
d) Become zero
Answer: c
Explanation: Due to the increase in randomness the configurational entropy increases. These configurational entropy can experience some changes during the ordering or during the magnetic phase transitions and they are difficult to calculate when there are partial correlation over small distances.
10. What causes the increase in thermal entropy due to the vacancies present in the site?
a) Collision of the adjacent atoms
b) Extra freedom of vibration
c) Due to restriction in motion
d) Due to close packing of remaining atoms
Answer: b
Explanation: Energies of electrons in atoms depend upon temperature of the host material. More temperature means high amplitude atomic vibrations. More electrons will be in excited/ionized states at a higher temperature. One can notice a variation in the pattern of vibration next to the vacancies and this happens because of the extra free space.
11. In a cooling curve, the invariant reaction is represented by ______
a) Thermal arrest
b) Vertical arrest
c) Horizontal arrest
d) Slope
Answer: c
Explanation: Cooling curve is a kind of line graph that depict the phase change in matter. When you fit the thermocouple in a work piece or test probe, finally you can end up obtaining a cooling curve. Horizontal arrest represents the invariant reaction in a cooling curve.
This set of Phase Transformation Questions and Answers for Freshers focuses on “Influence of Interface on Equilibrium”.
1. The surface of a single crystal is an example of a_________
a) Solid-Vapor interface
b) Solid-Liquid interface
c) Solid-Solid interface
d) Liquid-Liquid interface
Answer: a
Explanation: The surface of a single crystal is an example of a Solid-Vapor interface. These crystals are material whose constituents are arranged in a highly ordered fashion and they form the crystal lattice which extends in all the direction.
2. Grain boundaries are an example of a_________
a) Solid-Vapor interface
b) Solid-Liquid interface
c) Solid-Solid interface
d) Liquid-Liquid interface
Answer: c
Explanation: Grain boundaries are example of homo phase interface which is a type of solid-solid interface. Grain boundaries are strongly affected by the change in orientation involved. If the angles of disorientation are small, less than about 3 degree, they can be readily accommodated by the formation of a two-dimensional network of dislocations.
3. Solidification front is an example of_________
a) Solid-Vapor interface
b) Solid-Liquid interface
c) Solid-Solid interface
d) Liquid-Liquid interface
Answer: b
Explanation: Solidification front is an example of Solid-Liquid interface and it is one among the 3 interfaces concerned with a crystalline structure.
4. For a particular or given diffusion coefficient, which among the following interface moves the slowest?
a) Coherent
b) Semi coherent
c) Incoherent
d) Mixed
Answer: a
Explanation: For a given diffusion coefficient the coherent interface moves the slowest and then the semi coherent and the fastest one is the incoherent interface.
5. If the α/β interfacial energy is given as 10N/mm and the particles are spherical with a radius 2.5mm, ΔP is given approximately by_________
a) 8 N/mm 2
b) 4 N/mm 2
c) 3 N/mm 2
d) 2 N/mm 2
Answer: a
Explanation: If γ is the α/β interfacial energy and the particles are spherical with a radius r, ΔP is given approximately by ΔP = 2*γ/r. So substituting the respective values we get the value as ΔP = 2*10/2.5 = 8 N/mm 2 .
6. There is an increase in free energy due to interfacial energy is known as a capillarity effect or the Gibbs-Thomson effect. Find the increase in this free energy if the interfacial energy, the molar volume and the radius of the spherical particle are given as 10 (kJ/mm 2 ), 6mm 3 /mol and 4mm respectively?
a) 40 kJmol -1
b) 30 kJmol -1
c) 50 kJmol -1
d) 60 kJmol -1
Answer: b
Explanation: This increase in the Gibbs free energy can be calculated using the formula
G = /r
Hence substituting this value of interfacial energy, molar volume and radius in the equation we get the increase in energy as /4 = 30 kJmol -1 .
7. MnS in steel is an example of _____
a) Coherent interface
b) Semi coherent interface
c) Incoherent interface
d) Mixed interface
Answer: c
Explanation: It is an example of incoherent interface because inclusion in alloys have incoherent interface. When the interface plane has a very different atomic configuration in the two adjacent phases, there is no possibility of good matching across the interface, hence it turn out to be an Incoherent interface.
8. The free energy increase due to interfacial energy is known as_____
a) Gibbs-Thomson effect
b) Perkins effect
c) Nano effect
d) Hauls effect
Answer: a
Explanation: Free energy increase due to interfacial energy is known as a capillarity effect or the Gibbs-Thomson effect. And this can be calculated if the interfacial energy, molar volume and the radius of the spherical particles are known.
9. If the surface area of the large particle remains unchanged the increase in free energy will be due to the____
a) Motion of spherical particle
b) Vibration of spherical particle
c) Increase in the interfacial area of the spherical particle
d) Decrease in the interfacial area of the spherical particle
Answer: c
Explanation: This happens because under this condition the increase in free energy is directly proportional to the radius of the spherical particle.
10. Generic high angle grain boundaries are typically___________
a) Coherent interface
b) Semi coherent interface
c) Incoherent interface
d) Mixed interface
Answer: c
Explanation: Definition of the incoherent surface says that when the interface plane has a very different atomic configuration in the two adjacent phases, there is no possibility of good matching across the interface and in case of the high angle grain boundaries are notably more disordered, with large areas of poor fit and a comparatively open structure hence they are typically incoherent in nature.
11. The structure of interface plays a key role in determining its energy. Which among the energy considerations play a major role when their sizes are large and small respectively?
a) Interfacial and elastic
b) Elastic and interfacial
c) Interfacial in both cases
d) Elastic in both cases
Answer: b
Explanation: Let’s first consider the first part that is when their sizes are small, for such cases the interfacial energy plays a leading role and when it comes to higher sizes, the elastic energy plays the leading part. This is because interfacial energy scales with interfacial area while elastic energy scales with the volume of the phase.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Ternary Equilibrium”.
1. The composition of a ternary alloy can be indicated on a ___________
a) Thomson’s triangle
b) Hershel’s triangle
c) Gibbs triangle
d) Harley’s triangle
Answer: c
Explanation: The composition of a ternary alloy can be indicated on an equilateral triangle whose corners represent 100% the individual components.
2. The chemical potentials of A, B and C in any phase are then given by the points where the tangential plane to the free energy surfaces intersects the A, B and C axes in a Gibbs triangle..
a) False
b) True
Answer: b
Explanation: The Gibbs free energy of any phase can be represented by a vertical distance from the point in the Gibbs triangle. If this is done for all possible compositions the points trace out the free energy surfaces for all the possible phases.
3. Which of the following is true about the direction of tie-lines?
a) They must run from liquidus to solidus
b) They must run from solidus to liquidus
c) They can run from solidus to liquidus and vice versa
d) They have no specific direction
Answer: a
Explanation: They must run from the liquidus to solidus. A tie line is actually drawn on the phase diagram to speculate and determine the fraction of each component or mass fraction of each element. At the composition temperature, the tie line is drawn horizontally from one phase to another. .
4. No two tie-lines at the same temperature may ever cross.
a) False
b) True
Answer: b
Explanation: The above mentioned statement is true. Violation of the statement will directly lead to the violation of Gibbs phase rule. And the most important thing is, these tie-lines only have meaning in two-phase region were chemical potentials of both phases are fixed along the line.
5. Three-phase equilibrium in a ternary system exists _____
a) At a particular temperature
b) Within a certain temperature range
c) Over a wide range of temperature
d) Not temperature dependent
Answer: b
Explanation: When there are three phases in equilibrium in a ternary system under constant pressure, there is still one degree of freedom left. Thus, three-phase equilibrium in a ternary system exists within a certain temperature range.
6. Which among the following cannot exist in local equilibrium with solder enriched with Bi at 200 degree Celsius?
a) Cu 6 Sn 5
b) Cu 3 Sn
c) Cu 9 Sn 5
d) CuSn 7
Answer: c
Explanation: Cu 3 Sn can, however, exist in local equilibrium even with pure Bi. Therefore, the Cu 6 Sn 5 should transform into the Cu 3 Sn layer. The formation of porous Cu 3 Sn Intermetallic compounds takes place when high current stressing is done at high temperatures in Low-Bump-Height solder joints.
7. The Cu 6 Sn 5 phase does not nucleate until below___________
a) 222 degree Celsius
b) 275 degree Celsius
c) 250 degree Celsius
d) 300 degree Celsius
Answer: a
Explanation: The Cu 6 Sn 5 phase does not nucleate until below 222 degree Celsius, where the composition of the liquid reaches the eutectic valley.
8. Which of the following can provide useful information about the formation of the reaction layer sequence when used together with the isothermal sections?
a) Activity diagram
b) Triangle rule
c) Cooling curve
d) Phase diagrams
Answer: a
Explanation: The activity diagrams provide useful information about the formation of the reaction layer sequence when used together with the isothermal sections. This can be proved using an example of Si/TaC/Cu.
9. A critical point at which liquid and vapor phase are identical is known as____
a) Bait point
b) Gibbs point
c) Plait point
d) Lever point
Answer: c
Explanation: Plait point is that particular point where the vapor and the liquid part coincides in a binodal curve. These points depends on the critical pressure and temperature. Even the existence of this are dependent on the pressure, temperature, and components.
10. Under what conditions three phases may coexist for some ternary system?
a) Fixed pressure and temperature
b) Fixed volume and pressure
c) Fixed volume and temperature
d) Fixed pressure and concentration
Answer: a
Explanation: According to the phase rule, three phases may coexist at a fixed temperature and pressure for some ternary systems.
11. Gibbs triangle is _______
a) An equilateral triangle
b) An isosceles triangle
c) A scalene triangle
d) Right angled triangle
Answer: a
Explanation: These diagrams are based on the property of equilateral triangles, that the sum of the perpendicular distances from any point to each side of the diagram is a constant equal to the length of any of the sides.
This set of Phase Transformation Questions and Answers for Aptitude test focuses on “Kinetics of Phase Transformation”.
1. Which among the following condition should be satisfied by homogeneous nucleation?
a) Occurs slowly
b) Occurs at the preferential site
c) Occurs in an orderly fashion
d) Occurs randomly
Answer: d
Explanation: Homogenous nucleation takes place randomly and it’s spontaneous in nature and for this nucleation, there are no preferential sites like the ones for heterogeneous nucleation where it occurs at the surface.
2. Which among the following condition should be satisfied by heterogeneous nucleation?
a) Occurs spontaneously
b) Doesn’t have a preferential nucleation site
c) Occurs randomly
d) Occurs at preferential sites
Answer: d
Explanation: It is associated with the surface rather than the bulk and occurs at preferential sites. And in case of homogeneous nucleation it occurs randomly and spontaneously without any preferential site.
3. Nucleation that takes place at the grain boundaries are known as___________
a) Heterogeneous nucleation
b) Homogeneous nucleation
c) Partial nucleation
d) Mixed nucleation
Answer: a
Explanation: Heterogeneous Nucleation occurs at preferential sites such as grain boundaries, dislocations or impurities. They rather prefer the surface than the bulk of the material unlike the homogenous nucleation.
4. Homogeneous nucleation occurs only when liquid is undercooled by___________
a) Critical undercooling for nucleation
b) Critical undercooling for growth
c) Layered undercooling for nucleation
d) Layered undercooling for growth
Answer: a
Explanation: It occurs only when liquid is undercooled by critical undercooling for nucleation. The following assumptions are made before the process. Liquid with nuclei is an ideal solution of various size clusters and each size cluster contains N atoms or molecules.
5. Phase transformations in metals/alloys occur by _____
a) Random motion
b) Explosion
c) Rotation and vibration
d) Nucleation and growth
Answer: d
Explanation: During nucleation new phase appears at certain sites within the metastable parent phase and during growth, nuclei grows into the surrounding matrix.
6. During homogeneous for a spherical, the change in Gibbs free energy involve_______
a) Interfacial energy
b) Volume free energy and surface free energy
c) Rotational and Vibrational energy
d) Surface free energy and vibrational energy
Answer: b
Explanation: The ∆G of homogeneous nucleation depends on the Volume free energy and Surface free energy and both are dependent on radius of sphere. Furthermore, in the curve resulting from the sum of volume free energy and surface free energy, value of the volume free energy will be negative if the temperature is below the equilibrium solidification temperature and the surface free energy is always positive.
7. The magnitude of the contribution of volume free energy of a spherical particle increases with radius.
a) True
b) False
Answer: a
Explanation: The magnitude of the contribution of volume free energy associated with a spherical particle increases with increase in radius because it is directly proportional to the cube of radius. But the value of the volume free energy will be negative if the temperature is below the equilibrium solidification temperature.
8. The magnitude of surface free energy of a spherical particle increases with radius.
a) True
b) False
Answer: a
Explanation: Surface free energy increases with the increase in the radius and it is directly proportional to the square of the radius hence it is positive, furthermore, the magnitude of this contribution is the product of interfacial energy and the surface area of the nucleus. Finally, the total free energy change is equal to the sum of the surface free energy and volume free energy.
9. Gibbs free energy of a homogeneous nucleation is _____
a) Greater than heterogeneous nucleation
b) Less than heterogeneous nucleation
c) Greater than or equal to heterogeneous nucleation
d) Less than or equal to heterogeneous nucleation
Answer: c
Explanation: ΔG heterogeneous is equal to some S𝛳*∆G of homogeneous where S𝛳 ≤ 1, is a function of the wetting angle. So hence the product of this will always be less than or equal to the homogeneous nucleation.
10. The rate at which a transformation occurs will depend on the frequency with which atoms________
a) Reach the activated state
b) Deviate from the activated stage
c) Vibrate in the activated stage
d) Rotate in the activated stage
Answer: a
Explanation: According to kinetic theory, the probability of an atom reaching the activated state is given by exp where K is Boltzmann’s constant, G is known as the activation free energy barrier. The rate at which a transformation occurs will depend on the frequency with which atoms reach the activated state.
11. Radius of the new nuclei formed during heterogeneous nucleation depends on___________
a) ∆T
b) S𝛳
c) A
d) V
Answer: a
Explanation: Radius of the new nuclei in case of a heterogeneous nucleation is inversely proportional to ∆T and this is the same when you take homogeneous nucleation but much lesser degree of super cooling is needed for heterogeneous nucleation.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Atomic Mechanism of Diffusion”.
1. The mean vibrational energy possessed by each atom at 300K is given by__________
a) 4.14*10 -23
b) 1.242*10 -20
c) 1.242*10 -23
d) 1.242*10 -21
Answer: b
Explanation: The mean vibrational energy possessed by each atom is given by 3 KT, and therefore it increases in proportion to the absolute temperature and in this case take the value of T as 300K and substitute in the equation and we get the answer as 1.242*10 -20 .
2. Which of the following atoms use the vacancy mechanism to diffuse?
a) Large atoms
b) Surface atoms
c) Substitutional atoms
d) Small interstitial atoms
Answer: c
Explanation: Substitutional atoms usually diffuse by a vacancy mechanism whereas the smaller interstitial atoms migrate by forcing their way between the larger atoms.
3. Vibrational energy is increased by increasing the amplitude of the oscillations.
a) True
b) False
Answer: c
Explanation: Since the mean frequency of vibration is approximately constant the vibrational energy is increased by increasing the amplitude of the oscillations.
4. When considering the flux of particles in a one-dimensional system caused by a concentration gradient, the flux can be expressed as ____________
a) J = dm/dt
b) J = dm/
c) J = dm/dx
d) J = dm/
Answer: b
Explanation: J = dm/dt*A = – D*
Where J (mol/m 2 s) is the flux, dm is the change in the amount of matter in small time dt , A (m 2 ) is the area, D (m 2 /s) is the diffusion coefficient, C (mol/m 3 ) is the concentration of the particles, and x is the position parameter. The negative sign stems from the fact that diffusion occurs in the direction opposite to the increasing concentration gradient.
5. Fick’s first law can be directly applied only in a steady-state condition, where the composition does not change with time.
a) True
b) False
Answer: b
Explanation: Fick’s first law can be directly applied only in a steady-state condition, where the composition does not change with time. In addition, there should not be any external driving forces present other than the concentration gradient.
6. The movement of atom from copper to Nickel is an example of________
a) Intra diffusion
b) Inter diffusion
c) Self diffusion
d) Mixed diffusion
Answer: b
Explanation: Atoms move from higher to lower concentration region. If this movement is from one element to another e.g. Cu to Ni, then it is termed as Inter-diffusion. If the movement is within similar atoms as in pure metals, it is termed self-diffusion.
7. Usually the concentration of interstitial atoms is_________
a) So low that only a fraction of available site is occupied
b) Very high that entire fraction is occupied
c) Low but most of the fraction of available site is occupied
d) High but most of the available sites are vacant
Answer: a
Explanation: Usually the concentration of interstitial atoms is so low that only a small fraction of the available sites is occupied. This means that each interstitial atom is always surrounded by vacant sites and can jump to another position as often as its thermal energy permits it to overcome the strain energy barrier to migration.
8. The distance between two adjacent atoms in an octahedral site is given by _____
a) a/√2
b) a/2
c) 3a/√2
d) 2a
Answer: a
Explanation: In FCC materials the interstitial sites are midway along the cube edges or, equivalently, in the middle of the unit cell. These are known as octahedral sites since the six atoms around the site form an octahedron.
9. The diffusion coefficient is given as 16(m 2 /s) and the concentration gradient is given as 2, the flux of the particle is given by _____
a) 8
b) 32
c) -32
d) -8
Answer: c
Explanation: The flux of particles in a one-dimensional system caused by a concentration gradient is given by .
10. The mean vibrational energy of a particle is given by 1.656*10 -20 . Calculate the temperature at which the particle attains this vibrational energy?
a) 900K
b) 400K
c) 300K
d) 500K
Answer: b
Explanation: The mean vibrational energy possessed by each atom is given by 3 KT, and therefore it increases in proportion to the absolute temperature. Here its value is given as 1.656*10 -20 , so equate it with 3KT, finally we get T=1.656*10 -20 /3K so we get T as 400K.
11. State the reason behind the following
The diffusion coefficient for carbon in fee-Fe at 1000 °C is 2.5 X 10 -11 (m 2 *s -1 ) at 0.15 wt% C, but it rises to 7.7 x 10 -11 (m 2 *s -1 ) in solutions containing 1.4 wt % C.
a) Melting point of carbon
b) C atom strain Fe lattice
c) Melting point of Fe atom
d) Shrinking of C atom
Answer: b
Explanation: The reason for the increase of D that is the diffusion coefficient with concentration is that the C atoms strain the Fe lattice thereby it always try to make the diffusion easier as the amount of strain increases.
12. For a random walk during a diffusion process, at time t the average atom will have advanced a radial distance r from the origin and it is given by_________
a) 3.6/√Dt
b) 2.4/√Dt
c) 4.8√Dt
d) 1.2√DT
Answer: b
Explanation: For a random walk during a diffusion process, at time t the average atom will have advanced a radial distance r from the origin and it is given by 2.4/√Dt. It will be seen that the distance √ is a very important quantity in diffusion problems because only a very few of the atom jumps provide a useful contribution to the total diffusion distance.
13. After 2s, the average atom has an advanced radial distance of 3m from the origin, then calculate the diffusion coefficient__________ (in m 2 /s)
a) 0.32
b) 0.48
c) 0.69
d) 0.96
Answer: a
Explanation: For a random walk during a diffusion process, at time t the average atom will have advanced a radial distance r from the origin and its diffusion coefficient is given by 5.76/(t*r 2 ).
14. Assume that on average an interstitial atom jumps 6 times per second and that each jump is in a random direction, then the probability of the atom jumping to every one of the six adjacent sites is given by___________
a) 1/6
b) 1/3
c) 1
d) 1/2
Answer: c
Explanation: Assume that on average an interstitial atom jumps k times per second and that each jump is in a random direction, then there is an equal probability of the atom jumping to every one of the six adjacent sites.
15. During a diffusion process, at time 1s the average atom will have advanced a radial distance r from the origin and it is given by_________ (D = 2.5*10 -11 (m 2 /s))
a) 48000m
b) 45000m
c) 36000m
d) 24000m
Answer: a
Explanation: For a random walk during a diffusion process, at time t the average atom will have advanced a radial distance r from the origin and it is given by 2.4/√Dt and just substitute the value of D and t in the formula, we get the value of r as 48000m.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Interstitial Diffusion”.
1. If plane contains 5 B-atoms per metre square, calculate the number of atoms that will jump from plane to in 1s?
a) 5
b) 6
c) 1
d) 2
Answer: a
Explanation: Here one can use the formula =1/6* * to obtain the result and in this case, it is 5 because on an average it jumps 6 times in a second.
2. Which of the following equation is related to Fick’s first law of diffusion?
a) J = D*
b) J = D*
c) J = D*
d) J = D*
Answer: a
Explanation: This equation is identical to that proposed by Fick in 1855 and is usually known as Fick’s first law of diffusion. D is known as the intrinsic diffusivity or the diffusion coefficient. The partial derivative has been used to indicate that the concentration gradient can change with time.
3. Assume that on average an interstitial B-atom jumps 10 times per second. The distance between the plane is given by 2m. Then, calculate the coefficient of diffusion of B atoms? (In m 2 /s)
a) 20/3
b) 40/3
c) 10/3
d) 6
Answer: a
Explanation: For this question one can use the formula, Diffusion coefficient = 1/6* * and just substitute the values in this formula we get the coefficient of diffusion as 20/3.
4. The concentration of oxygen at the inner surface of the vessel is maintained at a level C depending on the pressure in the vessel, while the concentration at the outer surface is reduced to zero by the escape of oxygen to the surroundings. A steady state will eventually be reached when the concentration everywhere reaches a constant value. Provided D is independent of concentration there will be a single concentration gradient in the wall and the thickness of the wall is L, the flux is given by________________
a) DC/L
b) DL/C
c) CL/D
d) DC/L + K
Answer: a
Explanation: Here in this equation the concentration gradient is given by /L and the diffusion coefficient is D. Hence the product of this gives the flux.
5. In a hexagonal lattice on average, an interstitial atom jumps 6 times per second and that each jump is in a random direction, then the probability of the atom jumping to every one of the six adjacent sites is given by___________
a) 1
b) 1/6
c) 1/2
d) Cannot be predicted
Answer: d
Explanation: This cannot be predicted because in the hexagonal lattice the diffusion rates of atoms in the perpendicular and parallel directions are not constant and they vary. Generally, in case of non-cubic lattices the jumping probability in different crystallographic directions is not in equal proportion and depends on direction.
6. Particularly violent oscillation of an interstitial atom or some chance coincidence of the movement of the matrix and interstitial atoms will result in a jump.
a) True
b) False
Answer: a
Explanation: When the thermal energy of a solid increases the atoms starts to vibrate about their mean positions. When they acquire enough energy there is a chance that a particularly violent oscillations of an interstitial atom can make an attempt to jump from its mean position and this result in a jump.
7. The rest positions of the interstitial atoms are positions of____________
a) Minimum potential energy
b) Maximum potential energy
c) Minimum kinetic energy
d) Maximum internal energy
Answer: a
Explanation: The rest positions of the interstitial atoms are positions where these atoms have the lowest energy and they are in their comfort zone. In order to push or move an interstitial atom to a nearby or an adjacent lattice we require sufficient energy so that we can force that to that particular high energy position.
8. During decarburization of steel, the surface concentration is reduced to a very low value.
a) True
b) False
Answer: a
Explanation: During decarburization of steel the surface concentration is reduced to a very low value and carbon diffuses out of the specimen and this is how the carbon content is reduced while decarburization.
9. To move an interstitial atom to an adjacent interstice, the atoms of the parent lattice must be forced apart into higher energy positions. The work that must be done to accomplish this process causes____
a) An increase in free energy of system
b) A decrease in free energy of system
c) An increase in internal energy
d) A decrease in internal energy
Answer: c
Explanation: The work that must be done to accomplish this process causes an increase in the free energy of the system by ΔGm . Gm is known as the activation energy for the migration of the interstitial atom. In any system in thermal equilibrium the atoms are constantly colliding with one another and changing their vibrational energy.
10. The activation enthalpy for interstitial diffusion increases as the size of the interstitial atom increases.
a) False
b) True
Answer: b
Explanation: The activation enthalpy for interstitial diffusion increases as the size of the interstitial atom increases. This is to be expected since smaller atoms cause less distortion of the lattice during migration.
11. On average, the fraction of atoms with an energy of ΔG or more than the mean energy is given by_____
a) Exp
b) Exp
c) Exp
d) Exp
Answer: a
Explanation: Thus, if the interstitial atom is vibrating with a mean frequency v in the x direction it makes v attempts per second to jump into the next site and the fraction of these attempts that are successful is given by exp .
12. Which of the following equation is related to Fick’s second law_________
a) J = D*
b) = D*(∂ 2 C/∂X 2 )
c) J = D*
d) ∂C/∂T=k*
Answer: b
Explanation: These equations relate the rate of change of composition with time to the concentration profile C. Equation has a simple graphical interpretation as (∂ 2 C/∂X 2 ) is the curvature of the C versus x curve.
13. Diffusion coefficient of B-atom is given as 2.5m 2 /s and L value is given as 2m, calculate the relaxation time?
a) 0.161
b) 0.201
c) 0.321
d) 0.404
Answer: a
Explanation: The relaxation time can be calculated using the formula = L 2 /(π 2 *D), the amplitude of the concentration profile decreases exponentially with time and after a sufficiently long time approaches zero and this rate is determined by using the relaxation time.
14. After a time t=2*Ʈ Ʈ, the amplitude of concentration decreases by__
a) 1/e 2
b) 1/e
c) 1/2
d) 1/3
Answer: a
Explanation: The amplitude of the concentration profile after a time t is given by A=A * exp Ʈ, substituting the value of t=Ʈ gives the resulting solution as A * 1/e 2 .
15. During carburization to obtain a twofold increase in penetration requires a __________
a) Fourfold decrease in time
b) Threefold increase in time
c) Fourfold increase in time
d) Twofold increase in time
Answer: c
Explanation: The thickness of a carburized layer can be calculated using the formula √. Thus, to obtain a twofold thickness either we need to make the value of D as 4D or t as 4t. So, we need a fourfold increase in time to obtain a twofold increase in penetration.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Substitutional Diffusion”.
1. In a substitutional diffusion, the diffusion coefficient can be related to the jump frequency by the equation_______ Ʈ
a) D = *α 2 *Ʈ
b) D = *α 2 *Ʈ
c) D = f**α 2 *Ʈ
d) D = f**α 2 *Ʈ
Answer: d
Explanation: The same way we calculate for interstitial diffusion, but a slight change because once an atom has jumped into a vacancy the next jump is not equally probable in all directions, but is most likely to occur back into the same vacancy. Such jumps do not contribute to the diffusive flux and therefore the equation is multiplied by a correlation factor.
2. In FCC metal, the value of α is given by____________
a) a
b) √2a
c) a/√2
d) 2a
Answer: c
Explanation: The distance between the two adjacent atoms in an octahedral site gives the value of the jump distance and for FCC it is a/√2 and this can be calculated easily if one assumes the octahedral site to be present at the edge center.
3. What is the number of nearest neighbours present in FCC metals?
a) 12
b) 8
c) 6
d) 10
Answer: c
Explanation: The coordination number gives the total number of nearest neighbors for a lattice, here in FCC the coordination number is 12 hence the total number of nearest neighbors present in a FCC metal is 12.
4. Consider the atomic jump. An atom next to a vacancy can make a jump provided it has enough thermal energy to overcome the activation energy barrier to migration. However, most of the time the adjacent site will not be vacant and the jump will not be possible. The probability that an adjacent site is vacant is given by_________.
a) X*z
b) X/z
c) X+z
d) X-z
Answer: a
Explanation: The probability that an adjacent site is vacant is given by z*X, where z is the number of nearest neighbours and X is the probability that any one site is vacant, which is just the mole fraction of vacancies in the metal. Combining all these probabilities gives the probability of a successful jump.
5. Which of the following is known as Darkens equation?
a) D’ = D в *X A + D A *X в
b) D’ = D в *X в + D A *X A
c) D’ = D в *X A
d) D’ = D в *X в
Answer: b
Explanation: Interdiffusion coefficient D’ for substitutional alloys depends on D A and D в whereas in interstitial diffusion Dв alone is needed. This equation was first derived by Darken, hence it is known by his name.
6. When atoms migrate by the vacancy process the jumping of an atom into a vacant site can equally well be regarded as the jumping of the vacancy onto the atom.
a) False
b) True
Answer: b
Explanation: When atoms migrate by the vacancy process the jumping of an atom into a vacant site can equally well be regarded as the jumping of the vacancy onto the atom. In other words one can say that, if there is a net flux of atoms in one direction there is an equal flux of vacancies in the opposite direction.
7. Vacancy diffusion is slower than interstitial diffusion, because_______
a) Bonding of Interstitial to the surrounding atom is weaker
b) Bonding of Interstitial to the surrounding atom is stronger.
c) More vacancy sites than interstitial sites
d) Intermolecular forces are stronger in interstitials
Answer: a
Explanation: There are mainly two reason for this. The first one, the weak bonding of interstitial to surrounding atoms and the second one is due to many more interstitial sites than vacancy sites to jump to.
8. Correlation factor in interstitial diffusion is _____
a) 0.45
b) 0.78
c) 0
d) 1
Answer: d
Explanation: Correlation factor in interstitial diffusion is 1, since we consider that the concentration of interstitial atoms is very low to make random jump possible because of availability of free sites to jump.
9. Correlation factor of a simple cubic crystal structure is _______
a) 0.65549
b) 0.72149
c) 0.78145
d) 0.932
Answer: a
Explanation: The correlation factor of a simple cubic crystal structure is 0.65549, 0.72149 and 0.78145 are the correlation factors of BCC and FCC respectively.
10. In general it is found that in any given couple, atoms with the lower melting point possess a higher D. .
a) True
b) False
Answer: a
Explanation: In general, it is found that in any given couple, atoms with the lower melting point possess higher D. The exact value of D, however, varies with the composition of the alloy.
11. If the mole fraction of A and B are given as 1/3 and 2/3 respectively and their respective diffusion coefficient are 6 and 3. Calculate the inter-diffusion coefficient (in m 2 /s)
a) 3
b) 4
c) 6
d) 1
Answer: b
Explanation: The inter-diffusion coefficient is given by, D’ = D B *X A + D A *X B , (where X A , X B are the mole fractions of A and B and Dᴀ, Dв are their respective diffusion coefficient. Substituting their respective values, we get, D’ = 2 + 2 = 4.
12. If the inter-diffusion coefficient is given as 3m 2 /s and the L value is given as 2m. Calculate the relaxation time?
a) 0.449
b) 0.567
c) 0.789
d) 0.345
Answer: a
Explanation: The relaxation time can be calculated using the formula = L 2 /(π 2 *D) (Here D is the inter diffusion coefficient. Just substitute the values in this equation T= 2 2 / π 2 *3 = 0.449 sec.
13. Calculate the value of D at 1768K from the given parameters? (In 10 -6 m 2 /s)
Self-diffusion
D° = 83mm 2 /s
Q = 283.4 KJ/mol
a) 0.29
b) 0.35
c) 0.49
d) 0.67
Answer: b
Explanation: This can be calculated using the equation D° = D*exp . This equation is similar to that in interstitial diffusion except that the value of activation energy for self-diffusion has an extra term.
14. Calculate the total flux of A-atoms across a stationary plane with respect to the specimen from the following parameters?
Diffusive flux is given as 3 (moles/m 2 s)
Flux due to velocity of the lattice is given as 4(moles/m 2 s)
a) 7
b) 1
c) 12
d) 8
Answer: a
Explanation: The total flux is the sum of diffusive flux and velocity flux . So the total flux is given as 3+4=7.
15. T/Tm is known as____________
a) Homologous temperature
b) Analogous temperature
c) Hector temperature
d) Hetero temperature
Answer: a
Explanation: An immediate consequence of these correlations is that the diffusion coefficients of all materials with a given crystal structure and bond type will be approximately the same at the same fraction of their melting temperature.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Diffusion – Atomic Mobility”.
1. The potential energy of the atom will then be relatively high due to the strain in the surrounding matrix. However, this strain energy can be reduced if the atom is located in a position where it better matches the space available.
a) False
b) True
Answer: b
Explanation: This strain energy can be reduced if the atom is located in a position where it better matches the space available, e.g. near dislocations and in boundaries, where the matrix is already distorted.
2. The diffusion of carbon or nitrogen to dislocations in mild steel is responsible for strain ageing and blue brittleness.
a) True
b) False
Answer: a
Explanation: The technological importance in separating and dividing the atoms to grain boundaries, interfaces and dislocations is really immense.In mild steel, the main reason for the strain ageing and blue brittleness is caused by the diffusion of carbon or nitrogen to dislocations.
3. The segregation of impurities such as Sb, Sn, P and As to grain boundaries in low-alloy steels produces____________
a) Ageing
b) Strain hardening
c) Temper embrittlement
d) Red brittleness
Answer: c
Explanation: Temper embrittlement is the result of above phenomenon. Segregation to grain boundaries affects the mobility of the boundary and has pronounced effects on recrystallization, texture and grain growth.
4. The value of Atomic mobility is given as Mв and the potential gradient is given as .Calculate the value of drift velocity?
a) Vв = -Mв*
b) Vв = Mв/
c) Vв = Mв+
d) Vв = Mв*
Answer: a
Explanation: Since atoms always migrate so as to remove differences in chemical potential it is reasonable to suppose that the drift velocity is proportional to the local chemical potential gradient, and the drift velocity is given by Vв = -Mв*, where Mв is a constant of proportionality known as the atomic mobility.
5. The atoms diffusing towards regions of high concentration can be found when diffusion occurs in the presence of an electric field or a temperature gradient. These are known as_____
a) Electro-migration and Thermo-migration
b) Electric-migration and Thermal-migration
c) Electro-migration and Thermal-migration
d) Electric-migration and Thermo-migration
Answer: a
Explanation: These are known as electro-migration and thermos-migration, thus it can be seen that in addition to the effect of the concentration gradient the diffusive flux is also affected by the gradient of strain energy.
6. What can be done to the strain energy if the atom is located near the dislocations?
a) Can be reduced
b) Can be increased
c) Can be maintained at a constant value
d) Cannot say
Answer: a
Explanation: If we consider the regions near dislocations or near boundaries, the matrix there is really distorted and the strain energy there can be reduced. This happens because it has a better match available at its new position.
7. Diffusion in the vicinity of the defects is affected by______
a) Concentration gradient and gradient of interaction energy
b) Velocity
c) Interaction energy and Speed of diffusion
d) Gradient of potential energy and Composition
Answer: a
Explanation: Actually the atoms can reduce their free energies if they shift or migrate to the some defects but at the equilibrium, it is found that their concentration compared to the matrix is higher at the vicinity of the defects and one of the reason is that both the concentration gradient and the gradient of the interaction energy plays an important role during the time when the diffusion takes place in the vicinity of these defects.
8. If the atomic mobility is given as 5 and the potential gradient is given as 4², Then calculate the drift velocity from the given data?
a) 1.25
b) 9
c) -20
d) 20
Answer: c
Explanation: The drift velocity is the negative of the product of the atomic mobility and the driving force F. The force F can be either an elastic energy gradient, interfacial energy gradient or chemical potential gradient, it depend on the situation, here in this case we consider the local potential gradients as the drift force, hence the product of this and the drift velocity gives the required solution that is V=-4*5=-20 and they are negative.
9. Mobility in terms of drift velocity and driving force can be described as __________
a) V/F
b) F*V
c) F+V
d) F/V
Answer: a
Explanation: The simple and basic definition of mobility is, M =V/F, where V is the velocity of the entity and F the force or the driving force. The force F can be either an elastic energy gradient, interfacial energy gradient or chemical potential gradient, it depend on the situation.
10. For an ideal solution the diffusive coefficient in terms of atomic mobility is given as_______
a) D = MRT
b) D = M/RT
c) D = M/T
d) D = M+RT
Answer: b
Explanation: For an ideal solution the diffusion coefficient is given by the equation MRT. But in the case of non-ideal solution a thermodynamic factor is also included along with this equation so there will be a deviation in the diffusion coefficient of the real solution.
This set of Phase Transformation online test focuses on “Tracer Diffusion in Binary Alloys”.
1. With increase in annealing time the penetration distance __________
a) Remains constant
b) Increases
c) Decreases
d) First decreases then increases
Answer: b
Explanation: Normally when the experiment is performed at a certain temperature, the diffusion takes place in the radioisotope impurities and the extent to which they diffuse depend on the random walk of atoms. As the annealing time increases the penetration distance will also start to increase.
2. Pre-exponential factor and the activation energy does not depend on ___
a) Elastic modulus of material
b) Lattice parameter
c) Interstitial atoms
d) Composition
Answer: d
Explanation: The pre-exponential factor and the activation energy depend on a number of factors, such as size of the interstitial atoms, elastic modulus of the material, lattice parameter and the crystal structure.
3. The slope of a ln vs x 2 graph is given as -2.Calculate the value of the impurity coefficient of diffusion at time 3sec?
a) 0.7869
b) 0.4167
c) 0.9865
d) 0.5769
Answer: c
Explanation: The slope of this graph gives the value of -1/4Dt. Hence relating the slope to the given values gives the value of D, which is the required one. So here -2 = 1/ which when calculated gives the value of D as 0.9865.
4. Tracer method is followed to calculate the impurity diffusion coefficient utilizing radioisotopes, which are easy to detect.
a) True
b) False
Answer: a
Explanation: Here what matters is the concentration of the diffusing impurity element and in this case, it will be so small hence this method called tracer method is considered to calculate the impurity diffusion coefficient utilizing radioisotopes, because in this case it is easy to detect.
5. After the experiment for certain time t=2sec, at a desired temperature, the specimen is sliced at different known distances along the x direction and is given as 2m. Calculate the specific intensity if the impurity diffusion coefficient is given as 2m 2 /sec?
a) 0.987
b) 0.568
c) 0.219
d) 0.789
Answer: b
Explanation: After the experiment for certain time at a desired temperature, the specimen is sliced at different known distances along the x direction and the concentration of the radioisotopes is measured by measuring intensities of the emitted rays. Following the solution for thin film from Fick’s second law, we can write the relative intensity as I = )*exp (-x 2 /4Dt). D is the diffusion coefficient of impurity.
6. Calculate the probability of exchanging positions between the vacancy and the tracer atom, if the coordination number is given as 12?
a) 1/12
b) 1/6
c) 1/2
d) 1/4
Answer: a
Explanation: We can say that the probability of exchanging positions between the vacancy and the tracer atom is 1/Z . So, here substituting the value of Z as 12 we get 1/12 as the probability factor.
7. We can say roughly that the fraction of jumps that will give successful diffusion for the tracer atom is_______ .
a) F = 1-
b) Z
c) 1/Z
d) 1-
Answer: a
Explanation: We can say roughly that the fraction of jumps that will give successful diffusion for the tracer atom is F= 1-2/z. Note here that we add the probabilities since these two steps are separate events. This is very rough estimation since we have not considered the possibilities of coming back of the tracer atom to its initial position after few or many jumps.
8. What is the correlation factor of a body centered cubic crystal structure?
a) 0.65549
b) 0.72149
c) 0.78145
d) 0.98765
Answer: b
Explanation: The correlation factor of a body centered cubic crystal structure is 0.72149. Correlation factor of simple cubic is 0.65549 and for FCC the correlation factor is 0.78145.
9. Calculate the distance travelled along the x direction at time t=3sec, if the impurity diffusion coefficient is given as 2m 2 /sec and the relative intensity as 1(m -1 )? (Take the impurity diffusion coefficient as 2.5(m 2 /sec))
a) 0
b) 1.7
c) 2.8
d) 6.98
Answer: a
Explanation: After the experiment for certain time at a desired temperature, the specimen is sliced at different known distances along the x direction and the concentration of the radioisotopes is measured by measuring intensities of the emitted rays. Following the solution for thin film from Fick’s second law, we can write the relative intensity as I = )*exp (-x 2 /4Dt). D is the diffusion coefficient of impurity.
10. The measured tracer diffusion coefficient is exactly the same as the self-diffusion coefficient.
a) True
b) False
Answer: b
Explanation: The tracer diffusion coefficient is not exactly the same as the self-diffusion coefficient. They are related by a correlation factor, f. Sometimes, the tracer diffusion coefficient is stated as the same as self-diffusion coefficient, which is not really correct.
11. The diffusion coefficients are strongly dependent on ________
a) Crystal structure
b) Composition
c) Grain boundaries
d) Correlation factor
Answer: b
Explanation: The diffusion coefficients are composition dependent. There is a difference of about three orders of magnitude across the composition range and it doesn’t depend on the crystal structure or the grain boundaries.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Diffusion in Ternary Alloys”.
1. The addition of a third diffusing species to a solid solution produces some effects. Fe-Si-C alloys are particularly instructive for some reasons. Which among the following is a reason for the above-mentioned statement?
a) Carbon raises the chemical potential of silicon
b) Silicon raises the chemical potential of carbon
c) Ferrous becomes inactive
d) Mobility of carbon and silicon are almost same
Answer: b
Explanation: Firstly silicon raises the chemical potential of carbon in solution, carbon will not only diffuse from regions of high carbon concentration but also from regions rich in silicon. Secondly the mobility’s of carbon and silicon are widely different.
2. In Fe-Si-C alloys, why does carbon diffuse far more rapidly than the substitutionally dissolved silicon?
a) Because it is large in size
b) Because it has more mass
c) Because it is an interstitial solute
d) Because it is more volatile
Answer: c
Explanation: If we consider the mobilities of carbon and silicon, they are actually widely different. The difference comes because silicon is a substitutionally dissolved solute and carbon is an interstitial solute hence carbon has more advantage in the diffusion process in terms of rate.
3. The inter-diffusion flux for a multicomponent system can be calculated using____
a) Matano-Boltzmann analysis
b) Henry’s laws
c) Raoult’s Equation
d) Perkins equation
Answer: c
Explanation: Using the Matano-Boltzmann analysis for the multicomponent system, the inter-diffusion fluxes can be calculated from the experimental diffusion profile. They actually covert the complex differential equation into more easily solvable ODE’s.
4. Less striking effects can arise in ternary systems where all three components diffuse substitutionally if their diffusivities are unequal.
a) True
b) False
Answer: a
Explanation: The redistribution of an atom in the ternary system is particularly interesting when the mobilities of the two different atoms are so different. Something which is almost identical but not so common effect can arise in ternary systems. Here in this case all the 3 components can diffuse substitutionally but under the condition that their diffusivities are unequal.
5. In ternary alloy systems, because of the additional degree of freedom in the composition we may see composition profile which do not change monotonically through solid solution ranges and stable non- planar phase interfaces.
a) True
b) False
Answer: a
Explanation: Yes the statement is true. Even though two phase regions may lie between the terminal compositions of a diffusion sample on the constitutional diagram, they will not necessarily appear on the diffusion profile.
6. Which of the following relate the flux of one component to the composition gradient of the other independent constituent in ternary system?
a) Off-diagonal
b) Side-diagonal
c) Triangle rule
d) Fick’s law
Answer: a
Explanation: Actually, by using linear extension of Fick’s law we introduce the cross or off-diagonal diffusion coefficient which relate the flux of one component to the composition gradient of the other independent constituent in ternary system.
7. Who made the extension of Fick’s law to multicomponent system in the year 1945?
a) Onsager
b) Kirkland
c) Brown
d) Raoult
Answer: a
Explanation: He proposed that, most generally if there are n diffusing species then we may consider the motion of each to be linearly dependent on all gradients.
8. In order to describe the transport process in a ternary diffusion system, we need 4 inter-diffusion coefficients. Why?
a) Doppler principle
b) Catherine cross diagonal
c) Linear extension of ficks law
d) Shear pressure
Answer: c
Explanation: By using linear extension of Fick’s law we introduce the cross or off-diagonal diffusion coefficient which relate the flux of one component to the composition gradient of the other independent constituent. Hence we require 4 diffusion coefficient to describe the transport process.
9. What was the new feature introduced in the study of diffusion by Onsager?
a) Barrel effect
b) Cross effect
c) Multi phase effect
d) Side effect
Answer: b
Explanation: The generalization of Fick’s law by Onsager introduces a significant new feature into the study of diffusion, the presence of “Cross Effects”. In a ternary system, for example the flux of element 1 will depend not only on its own gradient but also on the gradient of element 2.
10. If a system has 3 components, the number of independent coefficients present if we utilize the ORR is given by____________
a) 3
b) 6
c) 2
d) 4
Answer: 3
Explanation: The number of independent coefficients present for the above system can be calculated using the formula */2, where n denotes the number of components present in the system. Substituting the value of n as 3 we get the required solution.
11. According to the model based on atomic mobility , if a system has 3 mobilities then the number of independent coefficients is given by_____
a) 6
b) 5
c) 3
d) 2
Answer: c
Explanation: For the model based on atomic mobility , if a system has n mobilities then the number of independent coefficients is given by n.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Diffusion – High-Diffusivity Path”.
1. In general, at any temperature the magnitudes of Db and Ds relative to the diffusivity through defect-free lattice D1 are such that_________
a) Db>Ds>D1
b) Db<Ds<D1
c) Db>D1>Ds
d) Ds>Db>D1
Answer: d
Explanation: In general, at any temperature the magnitudes of Db and Ds relative to the diffusivity through defect-free lattice D1 are such that Ds>Db>D1. This mainly reflects the relative ease with which atoms can migrate along free surfaces, interior boundaries and through the lattice.
2. In an average metallic specimen the total grain boundary area is_______
a) Much greater than surface area
b) Same as the surface area
c) Relatively less than surface area
d) Much less than surface area
Answer: c
Explanation: Surface diffusion can play an important role in many metallurgical phenomena, but in an average metallic specimen the total grain boundary area is much greater than the surface area so that grain boundary diffusion is usually most important.
3. The fluxes of solute through the lattice J1 and along the boundary Jb are 3 and 4 mol/mm 2 sec. If the grain boundary has an effective thickness 2mm and the grain size is 2.5mm. Calculate the total flux?(In mol/mm 2 sec)
a) 6.2
b) 4.2
c) 3.2
d) 2.2
Answer: a
Explanation: The total flux is given by the equation /α, where α represents the grain size and β represent the effective thickness of grain boundary. So substituting the values in this equation we get the required solution that is /2.5 = 6.2.
4. In a diffusion couple made by welding together two metals, A and B.A-atoms diffusing along the boundary will be able to penetrate much deeper than atoms which only diffuse through the lattice.
a) True
b) False
Answer: a
Explanation: A-atoms diffusing along the boundary will be able to penetrate much deeper than atoms which only diffuse through the lattice. In addition, as the concentration of solute builds up in the boundaries, atoms will also diffuse from boundary into the lattice.
5. The diffusion through the lattice can be ignored under certain circumstances. Which among the following condition should be satisfied for the same if the diffusivity through a defect free lattice is given as 2.5 mm 2 /sec and the grain size is given as 5mm? (Consider the diffusivity through the grain boundary as 2mm 2 /sec)
a) When the effective thickness of grain boundary is 6.25mm
b) When the effective thickness of grain boundary is 1000mm
c) When the effective thickness of grain boundary is 0.001mm
d) When the effective thickness of grain boundary is 10mm
Answer: b
Explanation: In this case 2*1000>>2.5*5. It can be seen that the relative importance of lattice and grain boundary diffusion depends on the ratio Db*ȣ/ D1*d. When Db*ȣ>>D1*d, diffusion through the lattice can be ignored in comparison to grain boundary diffusion.ȣ.
6. The relative magnitudes of Db*ȣ and D1*d are most sensitive to_______ȣ
a) Pressure
b) Temperature
c) Volume
d) Composition
Answer: b
Explanation: The relative magnitudes of Db*ȣ and D1*d are most sensitive to temperature. Note that although Db > Dȣ, at all temperatures, the difference increases as temperature decreases. This is because the activation energy for diffusion along grain boundaries is lower than that for lattice diffusion .
7. The jump frequency for atoms migrating along the defects is higher than that for diffusion in the lattice.
a) True
b) False
Answer: a
Explanation: It can be shown experimentally that the jump frequency for atoms migrating along these defects is higher than that for diffusion in the lattice. It will become apparent that under certain circumstances diffusion along these defects can be the dominant diffusion path.
8. In FCC metals it is generally found that ___
a) Qb >> 0.5Q1
b) Qb << 0.5Q1
c) Qb = Q1
d) Qb ̴̴ Q1
Answer: d
Explanation: This means that when the grain boundary diffusivity is scaled by the factor ȣ/d the grain boundary contribution to the total, or apparent, diffusion coefficient is negligible in comparison to the lattice diffusivity at high temperatures, but dominates at low temperatures.
9. The rate at which atoms diffuse along different boundaries is_________
a) Same
b) Different
c) Constant
d) Cannot be predicted
Answer: b
Explanation: The rate at which atoms diffuse along different boundaries is not the same, but depends on the atomic structure of the individual boundary. This in turn depends on the orientation of the adjoining crystals and the plane of the boundary. Also, the diffusion coefficient can vary with direction within a given boundary plane ȣ.
10. The dislocations effectively act as pipes along which atoms can diffuse with a diffusion coefficient 2.5mm 2 /sec. Calculate the apparent diffusivity if the cross-sectional area of pipe per unit area of matrix is given as 10 -1 ? is given as 5mm 2 /sec)
a) 5.45 mm 2 /sec
b) 5.25 mm 2 /sec
c) 6.35 mm 2 /sec
d) 5.75 mm 2 /sec
Answer: b
Explanation: It can easily be shown that the apparent diffusivity through a single crystal containing dislocations, Dapp, is related to the lattice diffusion coefficient by Dapp=D1 ), where g is the cross-sectional area of ‘pipe’ per unit area of matrix. The contribution of dislocations to the total diffusive flux through a metal will of course depend on the relative cross-sectional areas of pipe and matrix.
11. Find the frequency factor of the following grain boundary diffusion, if the activation energy is given as 90KJ/mol and the diffusivity at the grain boundary at 1071K is given as 49*10 -6 m 2 /sec?
a) 1.2mm 2 /sec
b) 1.2m 2 /sec
c) 2.2mm 2 /sec
d) 4.2mm 2 /sec
Answer: a
Explanation: It is found experimentally that diffusion along grain boundaries and free surfaces can be described by Db=Db’*exp , where Db the grain boundary diffusivity and Db’ the frequency factor. Qb is the experimentally determined values of the activation energy for diffusion. So in this case substituting the respective values we get, Db = 49*10 -6 *exp =1.2mm 2 /sec.
12. At high temperatures diffusion through the lattice is rapid and g*Dp / D1 is very small so that the dislocation contribution to the total flux of atoms is_______ .
a) Extremely high
b) Negligible
c) Low
d) High
Answer: b
Explanation: The total flux of atoms will be negligible. But it is found that the activation energy in lattice diffusion is higher than that of pipe diffusion. D1 decreases more rapidly than Dp and this happens with the decrease in temperature.
13. Grain boundary diffusion makes a significant contribution to the total flux when___ȣ.
a) ȣ > 1
b) ȣ < 1
c) ȣ = 1
d) ȣ < 0
Answer: a
Explanation: Grain boundary diffusion makes a significant contribution to the total flux when the following condition is satisfied, Db*ȣ> D1*d. The effective width of a grain boundary in many cases is 0.5nm. Grain sizes on the other hand can vary from ̴ 1 to 1000 (10 -6 m) and the effectiveneβ of the grain boundaries will vary accordingly.
This set of Phase Transformation Interview Questions and Answers for freshers focuses on “Diffusion in Multiphase Binary Systems”.
1. Which of the following is an example of Diffusion in multi-phase binary system?
a) Alcohol in water
b) Aldehyde in water
c) Ketone in water
d) Galvanized iron and hot-dipped tin plate
Answer: d
Explanation: In case of a hot dipped tin plate and galvanized iron the welding takes place between the two metals that are not completely miscible in each other and in this diffusion couples the multi-phase diffusion arises.
2. Non equilibrium diffusion couples, a situation often encountered at ________
a) High temperature
b) Low temperature
c) When nucleation is perfect
d) Annealing time is too long
Answer: c
Explanation: This is situation often encountered when the annealing time is short, or when the nucleation problems exist or when the interface energy plays a significant role and it can also arise when the temperature is low.
3. If the diffusion couple technique is not used in a proper way which of the following error is most likely to occur?
a) The necessity to use the decremented couples
b) Small roles played by impurities
c) Steep concentration gradient may occur in the growing phases
d) No errors are likely to occur
Answer: c
Explanation: If the diffusion couple technique is not used in a proper way then steep concentration gradient may occur in the growing phases. Unleβ the role played by the impurities are large it doesn’t create an error in the system. The necessity to use the incremental couples because of the problem of missing plane.
4. Which among the following can give an extra information on layer growth mechanism and concentration limit of various phases?
a) Edges and free surface of diffusion couple
b) Positions of dislocation
c) Composition ratio
d) Shape of the interface
Answer: a
Explanation: Generally, the edges and free surface of diffusion couple can give an extra information on layer growth mechanism and concentration limit of various phases. Often the surface diffusion or evaporation/condensation give rise to the formation of larger number of phases rich in substrate material compared to regular diffusion zone because of impeded supply of diffusion material.
5. Under some circumstances the interface migration is said to be interface controlled.
a) True
b) False
Answer: a
Explanation: Under some circumstances the interface migration is said to be interface controlled. In some cases the interface migration can be slow and there are virtually no concentration gradient in the two phases. These are some extreme cases.
6. The flux of atoms to the interface balance the________
a) Rate of accumulation due to the boundary migration
b) Mass of the system
c) Volume of the system
d) Stability of the grain boundary
Answer: a
Explanation: Here the rate of transfer across the interface can be balanced along with the rate of accumulation due to migration and the rate of diffusion away into the other phase, this is done by the flux of atoms to the interface.
7. Accurate microbe analysis near the interfaces are very difficult if the fluorescent effect affects the concentration measurements at distance as large as 40*10 -6 m from the interface.
a) True
b) False
Answer: a
Explanation: Accurate microbe analysis near the interfaces are very difficult if the fluorescent effect affects the concentration measurements at distance as large as 40*10 -6 m from the interface. In binary couples, usually one of the elements does not affect from the fluorescent effect and one could measure the concentration of that element by the microbe analysis to describe the penetration plot.
8. The direction of the longest crystallographic axis is invariably oriented __________
a) Parallel to diffusion direction
b) Perpendicular to interface
c) Perpendicular to diffusion direction
d) Cannot be predicted
Answer: c
Explanation: The direction of the longest crystallographic axis is invariably oriented perpendicular to the diffusion direction, which is parallel to the interface. Very often the crystals of phases in the diffusion-grown layers have a preferred orientation in the growth direction.
9. If a single crystalline layer of γ is growing between single crystalline substrate α and β, one might expect a completely_____
a) Straight α/γ and γ/β interfaces
b) Curved α/γ and γ/β interfaces
c) Straight γ/α and γ/β interfaces
d) Curved γ/α and γ/β interfaces
Answer: a
Explanation: If a single crystalline layer of γ is growing between single crystalline substrate α and β, one might expect a completely straight α/γ and γ/β interfaces, provided that no orientation related effect occur which might cause a faceted interface. In a polycrystalline layer grown between polycrystalline end members, the interfaces are not completely straight.
10. The shortest repeat-distance is aligned parallel to the direction of diffusion . This statement is known as____
a) Perkin rule
b) Thumb rule
c) Henrys rule
d) Marks direction principle
Answer: b
Explanation: According to the rule of Thumb the shortest repeat-distance is aligned parallel to the direction of diffusion . These textures are most pronounced at the growth front of the reaction layer. In the oldest part of the layer sometimes change in texture are found due to recrystallization.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Interfacial Free Energy”.
1. The free energy of a system containing an interface of area 5mm 2 and free energy 6 KJ/mol*mm 2 per unit area is given by_______.
a) 40 KJ/mol
b) 50 KJ/mol
c) 20 KJ/mol
d) 10 KJ/mol
Answer: a
Explanation: The free energy of a system containing an interface of area A and free energy ᵞ per unit area is given by, G = Go + Aᵞ where Go is the free energy of the system assuming that all material in the system has the properties of the bulk- ᵞ is therefore the excess free energy arising from the fact that some material lies in or close to the interface.
2. Consider for simplicity a wire frame suspending a liquid film. If one bar of the frame is movable it is found that a force 5N per unit length must be applied to maintain the bar in position. If this force moves a small distance so that the total area of the film is increased by 0.5m 2 . Interfacial free energy per unit area is given as 4 J/m 2 . Calculate the change in interfacial energy? (Take the area of interface as 2m 2 )
a) 0.25 J
b) 0.50 J
c) 8J
d) 0.90 J
Answer: c
Explanation: If one bar of the frame is movable it is found that a force F per unit length must be applied to maintain the bar in position. If this force moves a small distance so that the total area of the film is increased by dA the work done by the force is FdA. Then the force is given as F = ᵞ + dᵞ/dA ᵞ. That is 4*2=8J.
3. In the case of a liquid film which of the following is true? ᵞ.
a) ᵞ > 0
b) ᵞ < 0
c) ᵞ = 0
d) Cannot be predicted
Answer: c
Explanation: In the case of a liquid film the surface energy is independent of the area of the interface and hence the value of ᵞ=0. This leads to the well-known result F = ᵞ i.e. a surface with a free energy ᵞ J m -2 exerts a surface tension of ᵞ Nm -1 .
4. Why is the value of ᵞ =0 in case of a liquid film?
a) Liquid is unable to support shear stresses
b) Liquid develops viscous forces
c) Liquid can support shear stresses
d) Intermolecular forces in liquid is quite high
Answer: a
Explanation: For a liquid the shear stress is something which is unbearable and it won’t be able to support that, so the atoms within the liquid can rearrange and redistribute at the time when they stretch and thereby maintain a constant surface structure. Solids are much more viscous and here the transfer of atoms occurs from the bulk to the surface.
5. What is the order of interfacial free energy found in Grain Boundary?
a) 10 -1
b) 10 -2
c) 10 -0
d) 10 -6
Answer: a
Explanation: Interfacial free energies are found in the order of 10 -1 in the grain boundaries, 10 -2 in Twin boundaries and 10 -0 in free surface .
6. Interfacial free energy is anisotropic in crystalline solids?
a) True
b) False
Answer: a
Explanation: In a crystalline solid the surface formed with different planes will have different energies because the number of broken bonds per unit areas are different on different planes. Hence, the interfacial is also anisotropic in crystalline solids.
7. Solids are much more viscous than liquid films and there is transfer of atoms from the bulk to the surface.
a) True
b) False
Answer: a
Explanation: Solids are much more viscous and transfer of atoms from the bulk to the surface, which is necessary to maintain an unchanged surface structure and energy, will take much longer. If this time is long in comparison to the time of the experiment then ᵞ is not equal to 0 and the surface free energy and surface tension will not be identical.
8. The anisotropy in crystalline solid is prominent at ___
a) Lower temperature
b) Higher temperature
c) Zero degree Celsius
d) Cannot be predicted
Answer: a
Explanation: This anisotropy is more prominent at lower temperatures. As temperatures rise, the entropy contribution becomes dominant and makes the interfacial energy less anisotropic and if this happens the surface formed with different planes will have same energies in crystalline solids.
9. Which among the following has the typical energy (J/m 2 ) in the order 10 -1 ?
a) Coincident site lattice boundary
b) Anti-phase boundary
c) Solidification front
d) Twin boundary
Answer: c
Explanation: The typical energy in Solidification front is in the order of 10 -1 , whereas in the Anti-phase boundary, Twin boundary and CSL boundary it is in the order of 10 -2 .
10. In the equation dG = A*dᵞ+ ᵞ*dA, if dᵞ = 0, what does that mean? ᵞ.
a) The interfacial excess free energy is less than surface tension
b) The interfacial excess free energy is same as surface tension
c) The interfacial excess free energy is greater than surface tension
d) Cannot be predicted
Answer: b
Explanation: If dᵞ = 0, the change in free energy is equal to the work done in increasing the interface area; and, hence, the interfacial free energy is the same as surface tension. On the other hand, in general, in solids, the surface tension is not the same as the interfacial free energy.
11. In a typical fluid-vapor interface, the interfacial excess free energy is same as ________
a) Internal energy
b) Surface tension
c) Kinetic energy
d) Enthalpy
Answer: b
Explanation: Let’s take the case of a typical fluid-vapor interface, here actually the interfacial excess free energy equals the surface tension but this is the case only for only for liquids and gases and this force tries to reduce the surface area.
12. In case of coherent interface elastic energy associated with the interface along with the interfacial energy determines its_______
a) Enthalpy
b) Equilibrium
c) Size
d) Surface tension
Answer: b
Explanation: The coherent and semi coherent interfaces has been associated with some strain energies. The elastic energy associated with the interface and the interfacial energy is the factor that determines the equilibrium shape in case of a coherent interface.
13. Which of the following can be used to calculate surface energy of a solid?
a) Bond making model
b) Bond breaking model
c) Solid bond theory
d) Interface creation model
Answer: b
Explanation: Bond breaking model can be used to calculate the surface energy of a solid. We make a lot of assumption before creating this model which include that the solid is in contact with its own vapor, we also assume that the temperatures are low enough that the primary contribution to the surface energy comes from the broken bonds.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Solid and Vapour Interfaces”.
1. Which among the following plane has the least density of atoms?
a) {110}
b) {200}
c) {220}
d) {111}
Answer: c
Explanation: {220} has the least density among the following, the density of atoms in these planes decreases as (h 2 +k 2 +l 2 ) increases . Similarly substituting other values we get {220} as the highest.
2. If 1 mol of solid is vaporized 12Na broken bonds are formed. Therefore L = 12 Na*E/2. Consequently the energy of a {111} surface is given as ___
a) 0.50*L/Na
b) 0.75*L/Na
c) 0.25*L/Na
d) 1.25*L/Na
Answer: b
Explanation: The energy of a {111} surface is given as 0.25*L/Na . This result will only be approximate since second nearest neighbours have been ignored and it has also been assumed that the strengths of the remaining bonds in the surface are unchanged from the bulk values.
3. Surface free energy is given by_____
a) γ= E +PV+TS
b) γ= E +PV/TS
c) γ= E +PV*TS
d) γ= E +PV-TS
Answer: a
Explanation: From the definition of Gibbs free energy the surface free energy will be given by γ= E +PV+TS
Here even if the ‘PV’ term is ignored surface entropy effects must be taken into account. It might be expected that the surface atoms will have more freedom of movement.
4. Considering the nearest neighbours it can be seen that the atoms on a {111} surface, for example, are deprived of three of their twelve neighbours.
a) True
b) False
Answer: a
Explanation: It can be seen that the atoms on a {111} surface, for example, are deprived of three of their twelve neighbours. If the bond strength of the metal is E each bond can be considered as lowering the internal energy of each atom by E/2. Therefore every surface atom with three ‘broken bonds’ has an excess internal energy of 3E/2 over that of the atoms in the bulk.
5. Extra configurational entropy can also be introduced into the surface by the formation of surface vacancies. Entropy effects γ is slightly dependent on temperature and is given as, γ = -2T+2. Calculate the entropy S from the given condition?
a) T
b) -2
c) 2
d) T 2
Answer: b
Explanation: The entropy S can be determined using the equation, ∂γ/∂T = -S. Measured values of S are positive and vary between 0 and 3mJ m -2 K -1 .Hence once we differentiate it with respect to temperature we get the required value.
6. Melting point of Sn, Ag, Al, and Au are 232, 961, 660, and 1063 given in their respective order. Which among the following values mentioned below corresponds to the surface energy of Sn? (Options corresponds to the surface energy of the above mentioned 4 elements in mJ m -2 )
a) 680
b) 1080
c) 1390
d) 1120
Answer: a
Explanation: Metal with high melting point temperatures have high value of latent heat of sublimation and high surface energies hence from the given value of melting point one can predict that Sn has the least value of surface energy among the four.
7. A crystal plane is at an angle 45 to the close-packed plane. Calculate the energy of the surface, attributing E/2 energy to each broken bond? )
a) E/2
b) E
c) 2E
d) E
Answer: d
Explanation: For a crystal plane is at an angle 45 to the close-packed plane, the Surface energy is given by the equation *(E/2a 2 ), where φ here is 45 and therefore substituting the values we get the surface energy as E.
8. For an isolated crystal bounded by the planes A1, A2, with energies γ1 and γ2 the total surface energy will be given by _______
a) –
b) /
c)
d) +
Answer: d
Explanation: When several planes are bounds a single system with their respective energies, the total surface energy is given by the sum of their product with the respective system. That is + + ……. it goes on depending on the number of planes. Actually the sum of Ai*γi is minimum in the equilibrium shape .
9. If we plot φ vs E , all the low index planes should be located at________
a) Low energy cusps
b) High energy cusps
c) In between the low energy cusps and high energy cusps
d) Cannot be determined
Answer: a
Explanation: It should be noted that the close-packed orientation lies at a cusped minimum in the energy plot. Similar arguments can be applied to any crystal structure for rotations about any axis from any reasonably close-packed plane. All low-index planes should therefore be located at low-energy cusps.
10. The equilibrium shape has the property that the summation of Ai*γi is a minimum and the shape that satisfies this condition is given by the following, so-called as _________
a) Wulff construction
b) Marks construction
c) Henrys diagram
d) Twin diagram
Answer: d
Explanation: Wulff construction is a technique or a method to determine the equilibrium shape of a crystal or a droplet of a constant volume inside a separate phase usually its saturated solution or a vapor.
11. A convenient method for plotting the variation of γ with surface orientation in three dimensions is to construct a surface about an origin such that the free energy of any plane is equal to the distance between the surface and the origin when measured along the normal to the plane. This type of polar representation is known as_____
a) γ-plot
b) Surface plot
c) Energy plot
d) α-γ plot
Answer: a
Explanation: This type of polar representation of γ is known as a γ-plot and has the useful property of being able to predict the equilibrium shape of an isolated single crystal and free energy of any plane is equal to the distance between the surface and the origin when measured along the normal to the plane.
12. Equilibrium shapes can be determined experimentally by________
a) Annealing large void inside a crystals
b) Annealing large crystals at high temperature
c) Annealing small crystals at low temperature
d) Annealing small void inside a crystals
Answer: d
Explanation: Equilibrium shapes can be determined experimentally by annealing small single crystals at high temperatures in an inert atmosphere, or it can be done by annealing small voids present inside a crystal.
13. When the macroscopic surface plane has a high or irrational {h k l} index the surface will appear as a stepped layer structure where each layer is a close packed plane.
a) True
b) False
Answer: a
Explanation: When the macroscopic surface plane has a high or irrational {h k l} index the surface will appear as a stepped layer structure where each layer is a close packed plane. This can be illustrated for a simple cubic crystal. A crystal plane at an angle φ to the close-packed plane will contain broken bonds in excess of the close-packed plane due to the atoms at the steps.
14. For the given notation {222}, the spacing of equivalent atom plane is given by________
a) 1/√3
b) 1/√2
c) √3
d) √2
Answer: a
Explanation: For a given notation {h k l} the spacing of the equivalent atom plane is given by the formula a/√ (h 2 +k 2 +l 2 ), where ‘a’ is the lattice parameter. Here substituting the value of h, k, l as 2 we get 1/√3 as the answer.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Boundaries in Single Phase Solids – 1”.
1. The lattices of any two grains can be made to coincide by ________
a) Twisting across the plane
b) Twirling about multiple axis
c) Turning in the direction of grain boundary
d) Rotating one of them through a suitable angle about a single axis
Answer: d
Explanation: The nature of any given boundary depends on the disorientation of the two adjoining grains and the orientation of the boundary plane relative to them. The lattices of any two grains can be made to coincide by rotating one of them through a suitable angle about a single axis.
2. Which of the following boundary occurs when the axis of rotation is parallel to the plane of boundary?
a) Roll in boundary
b) Twist boundary
c) Tilt boundary
d) Roll out boundary
Answer: c
Explanation: There are two special types of boundary that are relatively simple. A tilt boundary occurs when the axis of rotation is parallel to the plane of the boundary, whereas a twist boundary is formed when the rotation axis is perpendicular to the boundary.
3. Which among the following boundary is an array of parallel edge dislocations?
a) Low angle twist boundary
b) High angle tilt boundary
c) High angle twist boundary
d) Low angle tilt boundary
Answer: d
Explanation: The low-angle tilt boundary is an array of parallel edge dislocations. These is an idealized boundary and are symmetric in nature. In this case the atoms in the regions between the dislocations fit almost perfectly into both adjoining crystals whereas the dislocation cores are regions of poor fit in which the crystal structure is highly distorted.
4. Which of the following boundary is related to screw dislocation?
a) Low angle twist boundary
b) High angle tilt boundary
c) High angle cross boundary
d) Low angle tilt boundary
Answer: a
Explanation: The twist boundary is a cross-grid of two sets of screw dislocations. Here the disorientation occurs around the axis that is perpendicular to the boundary plane. And these dislocations in the boundary remain isolated and distinct.
5. Calculate the magnitude of Burgers vector for the BCC lattice expressed as ½<148>, where 1, 4, 8 are their respective components of Burgers vector and the value of “a’’ is given as 1?
a) 6.5
b) 4
c) 5.5
d) 4.5
Answer: d
Explanation: The magnitude is given by the equation ||b||= *√(1 2 +4 2 +8 2 ) = 4.5.The direction of the vector depends on the dislocation plane and in most cases it is usually found on one of the closest packed crystallographic planes.
6. In the edge dislocations, the Burger vectors and dislocation line are _________
a) Parallel to each other
b) Perpendicular to each other
c) Close to each other
d) Far away from each other
Answer: b
Explanation: In the edge dislocations, the Burger vectors and dislocation line are perpendicular to each other whereas in case of the screw dislocations they are parallel to each other.
7. “α” is the angular misorientation across the boundary. At very small values of α, the dislocation spacing is very large and the grain boundary energy ꙋ is approximately proportional to the____
a) Position of dislocation
b) Density of dislocation
c) Type of dislocation
d) Volume of dislocation
Answer: b
Explanation: At very small values of α, the dislocation spacing is very large and the grain boundary energy ꙋ is approximately proportional to the density of dislocations in the boundary . However, as α increases the strain fields of the dislocations progressively cancel out so that ꙋ increases at a decreasing rate.
8. For simple arrays, the spacing of dislocation is given by the formula________
a) D = b/sinα
b) D = b/cosα
c) D = b*sinα
d) D = b*cosα
Answer: a
Explanation: The energy of a low-angle grain boundary is simply the total energy of the dislocations within unit area of boundary and this depends on the spacing of the dislocations which, for the simple arrays is given by D = b/sinα.
9. If the tilt boundary is unsymmetrical, dislocations with different Burgers vectors are required to accommodate the misfit.
a) True
b) False
Answer: a
Explanation: Here we require different Burgers vector to accommodate the misfit as the boundary is having unsymmetrical dislocation and in some cases the tilt boundary can be symmetrical with respect to the two adjoining crystals at that instance however we don’t need different Burgers vector to accommodate the misfit.
10. When α > 10-15° the boundary is known as a_______
a) Random high-angle grain boundary
b) Random low-angle grain boundary
c) Super angle grain boundary
d) Axial grain boundary
Answer: a
Explanation: When α > 10-15° the boundary is known as a Random high-angle grain boundary. The difference in structure between low-angle and high-angle grain boundaries can be lucidly illustrated by the bubble-raft model. High angle boundaries contain large areas of poor fit and have a relatively open structure.
11. Which among the following is the property of a low angle boundary?
a) Areas of poor fit
b) Relatively open structure
c) Highly disordered
d) Areas of perfect fit
Answer: d
Explanation: The reason behind the perfect fit in the low-angle boundaries is that most of the atoms correctly fit into both lattices so that there is very less free volume and the interatomic bonds are only slightly distorted. The regions of poor fits are also observed but rarely and it is restricted to the dislocation cores.
12. What is most likely to happen to the grain boundary energy of a pure metal on alloying?
a) It increases
b) It reduces
c) It increases first and finally becomes constant
d) No effect
Answer:
Explanation: Generally, the grain boundary energy of a pure metal changes on alloying. Often it is reduced. Under these circumstances the concentration of alloying element in the boundary is higher than that in the matrix.
13. A recrystallization texture is sometimes an advantage.
a) True
b) False
Answer: a
Explanation: Yes, this statement is true. For example, the proper texture in Fe-3wt% Si alloys makes them much better soft magnets for use in transformers. Another application is in the production of textured sheet for the deep drawing of such materials as low-carbon steel.
14. When the boundary moves the solute atoms migrate along with the boundary and exert a drag that reduces the boundary velocity. The magnitude of the drag will depend on ______
a) Shape of the boundary
b) Texture
c) Concentration in the boundary
d) It is an independent property
Answer: c
Explanation: The magnitude of the drag will depend on the binding energy and the concentration in the boundary. The higher mobility of special boundaries can, therefore, possibly be attributed to a low solute drag on account of the relatively more c1ose-packed structure of the special boundaries.
This set of Phase Transformation Questions and Answers for Experienced people focuses on “Boundaries in Single Phase Solids – 2”.
1. When the two grains are related by a rotation about a <100> axis, it can be seen that most high-angle boundaries have _____
a) Almost same energy
b) Different energy
c) Highly ordered structure
d) Nothing can be said about their energies
Answer: a
Explanation: When the two grains are related by a rotation about a <100> axis it can be seen that most high-angle boundaries have about the same energy and should therefore have a relatively disordered structure characteristic of random boundaries.
2. When the two grains are related by a rotation about a <110> axis, there are several large-angle orientations which have significantly lower energies than the random boundaries.
a) True
b) False
Answer: a
Explanation: When the two grains are related by a rotation about a <110> axis, there are several large-angle orientations which have significantly lower energies than the random boundaries. When Θ = 70.5°, it corresponds to the coherent twin boundary, but low-energy boundaries are also found for several other values of Θ.
3. When two grains meet in a plane it is known as _____
a) Grain corner
b) Grain boundary
c) Grain edge
d) Grain center
Answer: b
Explanation: When two grains meet in a plane it is known as a grain boundary, when three grains meet in a line then it is called a grain edge and when four grains meet at a point then it is called a grain corner.
4. What is the need for a grain boundary in an annealed material?
a) To control the grain size
b) To maintain its shape
c) It can produce a stable equilibrium at the grain edges
d) During annealing it produce a metastable equilibrium at the grain boundary intersections
Answer: d
Explanation: The boundaries are all high-energy regions that increase the free energy of a polycrystal relative to a single crystal. It is the reason that a polycrystalline material is never a true equilibrium structure but it has the property to adjust themselves during annealing to produce a metastable equilibrium and this happens at the intersections of the grain boundary.
5. What will be the torque acting on the boundary if the boundary happens to be at the orientation of a cusp in the free energy?
a) 500 Nm
b) Greater than 500 Nm
c) Less than 500 Nm
d) Zero
Answer: d
Explanation: If the boundary happens to be at the orientation of a cusp in the free energy, there will be no torque acting on the boundary since the energy is a minimum in that orientation. However, the boundary will be able to resist a pulling force Fy of up to cusp without rotating.
6. Under which of the following circumstances the grain boundary behaves as a thin soap film?
a) When the boundary energy is dependent on the orientation
b) When boundary energy is 0
c) When the torque is greater than zero
d) When the boundary energy is independent on the orientation
Answer: d
Explanation: If the boundary energy is independent of orientation the torque term is zero and the grain boundary behaves like a soap film. Under these conditions the requirement for metastable equilibrium at a junction between three grains.
7. Calculate the pulling force F, if the chemical potential and the molar volume is given as respectively?
a) ΔG-V
b) ΔG+V
c) ΔG/V
d) ΔG*V
Answer: c
Explanation: F = ΔG/V. In other words the force on the boundary is simply the free energy difference per unit volume of material. In case of grain growth ΔG arises from the boundary curvature, but this equation applies equally to any boundary whose migration causes a decrease in free energy.
8. Calculate the chemical potential if the molar volume is given as 5 mm 3 /mol? (Given the value of radius as 4mm and the interfacial energy γ is 8 kJ/mm 2 ?
a) 10
b) 20
c) 24
d) 22
Answer: b
Explanation: The effect of the pressure difference caused by a curved boundary is to create a difference in free energy or chemical potential that drives the atoms across the boundary. In a pure metal ΔG and Δμ are identical and are given by ΔG = 2γV/r. This free energy difference can be thought of as a force pulling the grain boundary towards the grain with the higher free energy.
9. During recrystallization, the boundaries between the new strain-free grains and the original deformed grains are acted on by a force ΔG /V where, in this case, ΔG is due to the difference in dislocation strain energy between the two grains.
a) False
b) True
Answer: b
Explanation: Consider a dislocation- free recrystallized grain expanding into the heavily deformed surroundings. In this case the total grain-boundary area is increasing, therefore the driving force for recrystallization must be greater than the opposing boundary tension forces. Such forces are greatest when the new grain is smallest, and the effect is therefore important in the early stages of recrystallization.
10. For low mole fractions of solute in the matrix 0.5 , the boundary solute concentration Xb (in 10 -1 2 ) is given by_____
a) 6.94
b) 7.94
c) 6.14
d) 9.80
Answer: a
Explanation: For low mole fractions of solute in the matrix , the boundary solute concentration Xb is given by the equation Xb= Xo*exp. Substituting the respective value we get 6.94*10 -1 2 as the value of Xb.
11. It is possible that the higher mobility of special grain boundaries plays a role in the development of recrystallization textures.
a) True
b) False
Answer: a
Explanation: It is possible that the higher mobility of special grain boundaries plays a role in the development of recrystallization textures. If a polycrystalline metal is heavily deformed, by say rolling to a 90% reduction, a deformation texture develops such that the rolled material resembles a deformed single crystal.
12. On heating to a sufficiently high temperature, what happens to the new grains?
a) They fall off
b) They disperse
c) Velocity decreases
d) They grow and nucleate
Answer: d
Explanation: On heating to a sufficiently high temperature new grains nucleate and begin to grow. However, not all grains will grow at the same rate: those grains which are specially oriented with respect to the matrix should have higher mobility boundaries and should overgrow the boundaries of the randomly oriented grains.
13. If unit area of grain boundary advances a distance 4mm, the number of moles of material that enter in the grain B is 4* (1/ 8(mm 3 /mol)) and the free energy released is given by_______
a) 5 kJ/mol*mm 2
b) 20 kJ/mol*mm 2
c) 10 kJ/mol*mm 2
d) 40 kJ/mol*mm 2
Answer: a
Explanation: ΔG, Free energy difference can be thought of as a force pulling the grain boundary towards the grain with the higher free energy. If unit area of grain boundary advances a distance dx, the number of moles of material that enter grain B is dx* and the free energy released is given by ΔG*.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Kinetics of Grain Growth”.
1. In a single-phase metal the rate at which the mean grain diameter D’ increases with time will depend on the ____________
a) Flow rate
b) Texture
c) Grain boundary size
d) Grain boundary mobility
Answer: d
Explanation: At sufficiently high temperatures the grain boundaries in a recrystallized specimen will migrate so as to reduce the total number of grains and thereby increase the mean grain diameter. In a single-phase metal the rate at which the mean grain diameter D’ increases with time will depend on the grain boundary mobility and the driving force for boundary migration.
2. The driving force or the rate of grain growth is given by_________
a) v’ = k*M*
b) v’ = k + M*
c) v’ = k*M*
d) v’ = k*M*
Answer: a
Explanation: We assume that the mean radius of curvature of all the grain boundaries is proportional to the mean grain diameter D, the mean driving force for grain growth will be proportional to 2γ/D. Therefore v’ = k*M*, where k is a proportionality constant of the order unity. This equation implies that the rate of grain growth is inversely proportional to D and increases rapidly with increasing temperature due to increased boundary mobility M.
3. Calculate the mean diameter of the grain at t=5sec, Assume the grain boundary energy and mobility are given as 5 Nmm -1 and 10mm 3 J -1 s -1 respectively?
a) 109mm
b) 1009mm
c) 10.9m
d) 9mm
Answer: b
Explanation: The mean diameter at time t is given by the equation D = D 2 ’ + 4Mγt, where M represents the mobility, ‘’t’’ the time at which it happens, γ the grain boundary energy and D’ the initial mean diameter.
4. Experimentally one can find the grain size as function of time. If the K value is given as 3mm/sec and n is 0.2. Find the grain diameter at t =2sec?
a) 8.76mm
b) 2.67mm
c) 3.45mm
d) 4.56mm
Answer: c
Explanation: The equation linked with the above mentioned question is given as D = Ktᶯ. Here K is the temperature dependent proportionality constant and n is a number less than 0.5. Substituting the respective values we get the D value as 3.45mm.
5. Coarsening is a process that can be considered quite identical to grain growth.
a) True
b) False
Answer: b
Explanation: The above statement is true. Coarsening is actually curvature driven and in that the driving force is provided by the interphase interfacial energy. During coarsening the small sized precipitates expands or grows at the cost of the larger ones.
6. Surface grooving where grain boundaries intersect free surfaces leads to_______
a) Surface roughness
b) Smoothening
c) Texture
d) Increase in grain size
Answer: a
Explanation: Actually, this leads to the hardening of the surface or removing the smooth part, possibly shattering up of thin films. The energy found extra in the interface under all circumstances implies a driving force for the deduction in the total surface area. This happens for grain growth not for the recrystallization.
7. Experimentally for finding the grain size we use the equation, D = Ktᶯ. Here K is the temperature dependent proportionality constant and n is a number less than 0.5 and this n can be taken as 0.5 under which circumstances?
a) Low temperature
b) High pressure
c) Impure metals
d) High temperature
Answer: d
Explanation: However, the experimentally determined values of n are usually much less than 0.5 and only approach 0.5 in very pure metals or at very high temperatures. The reason for this are not fully understood, but the most likely explanation is that the velocity of grain boundary migration, v, is not a linear function of the driving force, ΔG, that is the mobility in the equation M*ΔG/Vm is not a constant but varies with ΔG and therefore also with D.
8. There can be this growth of just a few grains to very large diameters. Such situation is known as_____
a) Unusual grain growth
b) Abnormal grain growth
c) Altered grain growth
d) Acceptable grain growth
Answer: b
Explanation: Occasionally so-called abnormal grain growth can occur. This situation is characterized by the growth of just a few grains to very large diameters. These grains then expand consuming the surrounding grains, until the fine grains are entirely replaced by a coarse-grained array.
9. The nature of normal grain growth in the presence of a second phase deserves special consideration. The moving boundaries will be attached to the particles exert a pulling force on the boundary restricting its motion. Therefore if the boundary intersects the particle surface at 90° the particle will feel a pull of *sinΘ. This will be counterbalanced by an equal and opposite force acting on the boundary. As the boundary moves over the particle surface Θ changes and the drag reaches a maximum value, this happens when Θ becomes______
a) 90
b) 45
c) 60
d) 30
Answer: d
Explanation: As the boundary moves over the particle surface Θ changes and the drag reaches a maximum value when sinΘ*cosΘ is a maximum, i.e. at Θ = 45°. The maximum force exerted by a single particle is therefore given by πrγ.
10. If there is a volume fraction f of particles all with a radius r the mean number of particles intersecting unit area of a random plane is_________
a) 2f/πr 2
b) 5f/πr 2
c) 3f/πr 2
d) 3f/2πr 2
Answer: d
Explanation: If there is a volume fraction f of particles all with a radius r the mean number of particles intersecting unit area of a random plane is so that the restraining force per unit area of boundary is approximately the product of this and the maximum force exerted by a single particle.
11. Calculate the maximum grain size D’max possible, if the radius of the spherical particle is 5mm and the volume fraction is given as 0.2?
a) 33.33mm
b) 66.66mm
c) 10mm
d) 38.33mm
Answer: a
Explanation: The driving force will be insufficient to overcome the drag of the particles and grain growth stagnates. A maximum grain size will be given by 4r/3f. Substituting the values of f and r as 0.2 and 5mm we get the required answer.
12. Stabilization of a fine grain size during heating at high temperatures requires__________
a) Large fraction of small particles
b) Small fraction of large particles
c) Large fraction of large particles
d) Small fraction of small particles
Answer: d
Explanation: It can be seen that the stabilization of a fine grain size during heating at high temperatures requires a large volume fraction of very small particles. Unfortunately, if the temperature is too high, the particles tend to coarsen or dissolve.
13. Aluminium-killed steels contain aluminium nitride precipitates which stabilize the austenite grain size during heating.
a) False
b) True
Answer: b
Explanation: Aluminium-killed steels contain aluminium nitride precipitates which stabilize the austenite grain size during heating. However, their effectiveness disappears above about 1000 degree Celsius when the aluminium nitride precipitates start to dissolve. And this an example where the transformation of the fine-grain array into a very coarse-grain structure takes place.
14. Assume the volume fraction 0.4 of particles all with a radius 3mm, the grain boundary energy is given as 6kJ/mm. Calculate the restraining force per unit area of boundary?
a) 1.2
b) 1.8
c) 2.4
d) 3.6
Answer: a
Explanation: This can be calculated using the formula 3fγ/2r, where the f is the volume fraction, r the radius and γ the boundary energy. This is actually the product of the mean number of particles intersecting unit area of a random plane and the maximum force exerted by a single particle.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Interphase Interfaces in Solids – 1”.
1. HCP silicon-rich k phase and the FCC copper-rich α-matrix in Cu-Si alloys forms ___
a) A coherent interface
b) An incoherent interface
c) Mixed interface
d) A semi coherent interface
Answer: d
Explanation: Here it forms a fully coherent interface. This requires the two crystals to be oriented relative to each other in a special way such that the interfacial plane has the same atomic configuration in both phases, disregarding chemical species.
2. Within the bulk of each phase every atom has an optimum arrangement of nearest neighbours that produces a low energy. At the interface, however, there is usually a change in composition so that each atom is partly bonded to wrong neighbours across the interface.
a) True
b) False
Answer: a
Explanation: At the interface there is usually a change in composition so that each atom is partly bonded to wrong neighbours across the interface. This increases the energy of the interfacial atoms and leads to a chemical contribution to the interfacial energy.
3. The resultant lattice distortion to maintain the coherency is known as__
a) Strain rupture
b) Coherency strain
c) Rupture plane
d) Maintenance plane
Answer: b
Explanation: It is possible to maintain the coherency even when the distance between the atoms in the interface is not identical and this can be done by straining one or both of the two lattices and the lattice distortions which results from this is known as coherency strain.
4. In which among the following case there is only one plane that can form a coherent interface?
a) Simple cubic
b) BCC
c) Edge centered lattice
d) HCP
Answer: d
Explanation: Only a single plane can form a coherent interface in case of a HCP/FCC interface, no other plane is identical in both crystal lattices. However all the lattice planes are identical apart from the differences in the composition if the two adjoining phases have the same crystal structure and lattice parameter.
5. In a semi coherent interface, the disregistry is periodically taken up by _________
a) Coarsened structure
b) Proper fit
c) Misfit dislocations
d) Cross dislocation
Answer: c
Explanation: The misfit dislocations periodically take up the disregistry and it becomes energetically more favorable to replace the coherent interface with a semi coherent interface.The strains associated with a coherent interface raise the total energy of the system.
6. Assume that 5mm and 5mm are the unstressed interplanar spacing’s of matching planes in the α and ß phases respectively, the disregistry, or misfit between the two lattices is defined by__
a) 0
b) 0.5
c) 1
d) 2
Answer: a
Explanation: If dα and dβ are the unstressed interplanar spacing’s of matching planes in the α and β phases respectively, the disregistry, or misfit between the two lattices is defined by Δ = /dα, here since the interplanar spacing’s are same, the misfit will be 0.
7. A spacing ‘’D’’ can be used to completely accommodate the lattice misfit in the one direction without any long-range strain field by a set of edge location. Calculate the value of D if the interplanar spacing’s of matching planes are given as 20mm, 19mm respectively?
a) 390mm
b) 240mm
c) 795mm
d) 800mm
Answer: a
Explanation: The value of D is given as b/Δ. Where the value of b = /2, in this case 20mm and 19mm and this is actually the Burgers vector for dislocation. And the value of Δ is given as /20, that is 0.05 and hence substituting this we get the required solution.
8. The interfacial energy of a semi coherent interface can be approximately considered as the sum of two parts. What are they?
a) Chemical and structural contribution
b) Chemical and bulk contribution
c) Magnetic contribution and structural
d) Physical and bulk contribution
Answer: d
Explanation: The interfacial energy of a semi coherent interface can be approximately considered as the sum of two parts: a chemical contribution, same as for a fully coherent interface, and a structural term, which is the extra energy due to the structural distortions caused by the misfit dislocations.
9. For small values of misfit ∆, the structural contribution to the interfacial energy is approximately proportional to the____ .
a) Size and shape of dislocation
b) Density of dislocation
c) Texture
d) Position of dislocation
Answer: d
Explanation: As the misfit Δ increases the dislocation spacing diminishes. For small values of Δ the structural contribution to the interfacial energy is approximately proportional to the density of dislocations in the interface. However, the interfacial energy increases less rapidly as Δ becomes larger and it levels out at a particular value of Δ.
10. When the value of misfit is greater than 0.25, the kind of interface is known as _________
a) Coherent interface
b) Mixed interface
c) Semi coherent interface
d) Incoherent interface
Answer: d
Explanation: When Δ > 0.25, that is one dislocation every four interplanar spacing’s, the regions of poor fit around the dislocation cores overlap and the interface cannot be considered as coherent, and it is known as the incoherent interface.
11. Incoherent interface can also exist between crystals with an orientation relationship if the interface has a different structure in the two crystals.
a) True
b) False
Answer: a
Explanation: In general, incoherent interfaces result when two randomly oriented crystals are joined across any interfacial plane. They may, however, also exist between crystals with an orientation relationship if the interface has a different structure in the two crystals and they have many features in common with high angle grain boundaries.
12. The degree of coherency can be greatly increased if_______
a) Macroscopically irrational interface is formed
b) Microscopically rational interface is formed
c) Macroscopically rational interface is formed
d) Microscopically irrational interface is formed
Answer: a
Explanation: The degree of coherency can, however, be greatly increased if a macroscopically irrational interface is formed, that means the indices of the interfacial plane in either crystal structure are not small integers and the detailed structure is very complex in nature.
13. The interfacial energies of semi coherent interfaces are generally in the range of_______
a) 0-200 mJm -2
b) 200-500 mJm -2
c) 500-1000 mJm -2
d) 10000 mJm -2
Answer: c
Explanation: The energies of semi coherent interfaces are generally in the range 200-500 mJm -2 . In general, coherent interfacial energies range up to about 200 mJm -2 and the incoherent interfaces are characterized by a high energy (500-1000 mJm -2 ).
This set of Phase Transformation Interview Questions and Answers for Experienced people focuses on “Interphase Interfaces in Solids – 2”.
1. Fully coherent precipitate is also known as GP zone, GP stands for _______
a) Gaff and Pearl
b) Gas and Pressure
c) Gatsby and Prince
d) Guinier and Preston
Answer: d
Explanation: GP for Guinier and Preston who first discovered their existence. This discovery was made independently by Preston in the USA and Guinier in France, both employing X-ray diffraction techniques.
2. Ag-rich zones in an AI-4 atomic % Ag alloy is an example of GP zone .
a) True
b) False
Answer: a
Explanation: The above mentioned statement is true and these zones are silver rich FCC region within the aluminum-rich FCC matrix. And one more major thing to note is that the diameters of aluminum and silver differ only by 0.7% hence the coherency strains doesn’t make much contribution to the total free energy of the alloy.
3. From an interfacial energy standpoint it is favourable for a precipitate to be surrounded by____
a) High-energy coherent interfaces
b) High-energy incoherent interfaces
c) Low-energy incoherent interfaces
d) Low-energy coherent interfaces
Answer: d
Explanation: If we consider the interfacial energy standpoint it always favorable to be surrounded by low-energy coherent interfaces but when the crystal structures of the of the precipitates and matrix are different it is usually hard to find a plane in the lattice that is common to both planes.
4. Which among the following are most unlikely to arise when a second-phase particle is located on a grain boundary?
a) Precipitate can have incoherent interfaces with both grains
b) Precipitate can have coherent or semi coherent interface with one grain and an incoherent interface with the other
c) Precipitate can have a coherent or semi coherent interface with both grains.
d) Nothing is predictable
Answer: c
Explanation: The first two options are commonly encountered but the third possibility is unlikely since the very restrictive crystallographic conditions imposed by coherency with one grain are unlikely to yield a favourable orientation relationship towards the other grain.
5. If μ is the shear modulus of the matrix and V is the volume of the unconstrained hole in the matrix and the elastic energy does not depend on the shape of the precipitate, if so, calculate the elastic strain energy?
a) ΔG=4μΔ 2 /V
b) ΔG=4μ/Δ 2 *V
c) ΔG=4μΔ 2 *V
d) ΔG=4μΔ 2 – V
Answer: c
Explanation: In general, the total elastic energy depends on the shape and elastic properties of both matrix and inclusion. However, if the matrix is elastically isotropic and both precipitate and matrix have equal elastic moduli, the total elastic strain energy ΔG is independent of the shape of the precipitate and is given as ΔG=4μΔ 2 *V.
6. If the precipitate and inclusion have different elastic moduli the elastic strain energy is no longer shape independent.
a) False
b) True
Answer: b
Explanation: If the precipitate and inclusion have different elastic moduli the elastic strain energy is no longer shape independent but is a minimum for a sphere if the inclusion is hard and a disc if the inclusion is soft. The above statements are applicable to isotropic matrices.
7. What kind of misfit arises if the inclusion is the wrong size for the hole it is located?
a) Volume misfit
b) Lattice misfit
c) Vertical misfit
d) Lateral misfit
Answer: a
Explanation: When the inc1usion is incoherent with the matrix, there is no attempt at matching the two lattices and lattice sites are not conserved across the interface. Under these circumstances there are no coherency strains. Misfit strains can, however, still arise if the inclusion is the wrong size for the hole it is located in. In this case the lattice misfit has no significance and it is better to consider the volume misfit.
8. Calculate the punching stress P, if the constrained misfit is given as 1/3 and the shear modulus of the matrix is given as 5Pa?
a) 5Pa
b) 3Pa
c) 2Pa
d) 7Pa
Answer: a
Explanation: The punching stress P is independent of the precipitate size and depends only on the constrained misfit Ɛ. If the shear modulus of the matrix is μ, the punching stress is given as P=3μƐ. Substitute the respective value we get the required solution.
9. Under which circumstances does a glissile semi coherent interface gets formed?
a) If the dislocations do not have a Burgers vector that can glide on matching planes in the adjacent lattices
b) Depends on the extent of gliding
c) When the orientation of the plane is unmatching
d) If the dislocations have a Burgers vector that can glide on matching planes in the adjacent lattices
Answer: d
Explanation: It is however possible, under certain circumstances, to have glissile semi coherent interfaces which can advance by the coordinated glide of the interfacial dislocations. This is possible if the dislocations have a Burgers vector that can glide on matching planes in the adjacent lattices.
10. The formation of martensite in steel and other alloy systems occurs by the motion of_______
a) Incoherent-dislocation interfaces
b) Cropped-dislocation interfaces
c) Mixed-dislocation interfaces
d) Glissile-dislocation interfaces
Answer: d
Explanation: The formation of martensite in steel and other alloy systems occurs by the motion of glissile-dislocation interfaces. These transformations are characterized by a macroscopic shape change and no change in composition.
11. The point of intersection of the curves in the graph represents?
a) Max radius
b) Cross radius
c) Critical radius
d) Inter-common point
Answer: c
Explanation: For a given lattice misfit, ΔG and ΔG vary with r as shown in above. When small, therefore, the coherent state gives the lowest total energy, while it is more favourable for large precipitates to be semi coherent or incoherent . At the critical radius ΔG = ΔG.
12. If the dislocation network glides into the FCC crystal it results in a transformation of_____
a) FCC->HCP
b) HCP->FCC
c) HCP->BCC
d) BCC->HCP
Answer: a
Explanation: The glide planes of the interfacial dislocations are continuous from the FCC to the HCP lattice and the Burgers vectors of the dislocation’s, which necessarily lie in the glide plane, are at an angle to the macroscopic interfacial. If the dislocation network glides into the FCC crystal it results in a transformation of FCC->HCP.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Interface Migration”.
1. In a way we can say that an interface is created during the____
a) Nucleation
b) Destruction
c) Vibration
d) Growth
Answer: a
Explanation: In a way we can say that an interface creation occurs during the nucleation stage and then the next step is that it migrates into the surrounding parent phase during the growth stage. Essentially this type of transformation is heterogeneous in nature.
2. Most of the transformation products are formed during the growth stage by the_____
a) Transfer of atoms across the moving parent/product interface
b) Movement of atoms in the formed nuclei
c) Nucleation
d) Random oscillation of parent particle
Answer: a
Explanation: Even though the nucleation stage plays the key role in determining many of the features of transformation but most of the transformation products are formed or created during the growth stage and this takes place by the transfer of atoms across the moving parent interface.
3. Athermal migration is related to which of the following interface?
a) Mixed
b) Corrupted
c) Non-Glissile
d) Glissile
Answer: d
Explanation: There are basically two different types of interface: glissile and non-glissile. Glissile interfaces migrate by dislocation glide that results in the shearing of the parent lattice into the product. The motion of glissile interfaces is relatively insensitive to temperature and is therefore known as athermal migration.
4. The migration of which among the following interfaces is extremely sensitive to temperature?
a) Mixed
b) Corrupted
c) Glissile
d) Non-Glissile
Answer: d
Explanation: In many cases it is found that the interfaces are non-glissile in nature and migrate by the more or less random jumps of individual atoms across the interface. The thermal activation is the process by which we get the extra energy that the atom needs to break free of one phase and attach itself to the other. The migration of non-glissile interfaces is therefore extremely sensitive to temperature.
5. Civilian transformation is a phenomenon that occurs when the uncoordinated transfer of atoms occurs across a____________
a) Glissile interface
b) Mixed interface
c) Partially coherent interface
d) Non-Glissile interface
Answer: d
Explanation: Transformations that are created by the migration of a glissile interface are referred to as military transformations. This gives a clarity and of the analogy between the coordinated motion of atoms crossing the interface and that of soldiers moving in ranks on the parade ground. In contrast the uncoordinated transfer of atoms across a non-glissile interface results in what is known as a civilian transformation.
6. During civilian transformations the parent and product will have _____ .
a) Same composition
b) Different composition
c) Time dependent composition
d) Nothing can be predicted
Answer: d
Explanation: Here in the case of a civilian transformation nothing can predicted about its composition, it can be same or may be different and if there is no change in composition, the new phase growth will be almost at the speed at which the atoms cross the interface.
7. Martensitic transformations belongs to____
a) Base transformation
b) Civilian transformation
c) Military transformation
d) Mixed transformation
Answer: c
Explanation: During a military transformation the nearest neighbours of any atom are essentially unchanged. Therefore, the parent and product phases must have the same composition and no diffusion is involved in the transformation. Martensitic transformations belong to this group
8. Just for a simplicity consider a pure substance B. By analogy with the migration of a high-angle grain boundary the net flux of B across the interface will produce an interface velocity v given by____
a) MΔμ/V
b) MΔμ +V
c) M/Δμ*V
d) MV/Δμ
Answer: a
Explanation: The interface velocity is the product of the interface mobility and the driving force per unit molar volume, so in this case the interface mobility is given as M, Molar volume V, and the driving force is given as Δμ so the required velocity is MΔμ/V.
9. If we consider a β precipitate of an almost pure metal B with Interface mobility given as M, Molar volume V of the β phase, and the driving force is given as Δμ. Calculate the flux across the interface?
a) J = MΔμ/V²
b) J = -MΔμ/V
c) J = -MΔμ/V²
d) J = MΔμ/V
Answer: c
Explanation: Actually, the interface velocity per molar volume is the flux across the surface. The interface energy is given as the product of the interface mobility and the driving force per unit molar volume hence the flux is given as J = -MΔμ/V², here the negative sign indicates that flux is in the negative direction along the oriented axis.
10. In the limit of a very low mobility it is possible that ____
a) interface = 0
b) interface > 0
c) interface < 0
d) interface = 1
Answer: a
Explanation: In the limit of very low mobility the value of approach 0. Under this situation the growth is described as interface controlled and here the chance of having a maximum possible driving force is very high.
11. What happens to the driving force Δμ, when Xi approaches Xe ?
a) Δμ=0
b) Δμ>0
c) Δμ<0
d) Δμ cannot be predicted
Answer: a
Explanation: The driving force Δμ approaches 0 when Xe=Xi, this happens because the driving force is directly proportional to and it can be calculated using the given formula Δμ= *, this is only applicable for dilute or ideal solution.
12. The accommodation factor at the incoherent interfaces and diffuse solid/liquid interfaces can be close to________
a) 0
b) 1
c) 0.5
d) 6
Answer: b
Explanation: The accommodation factor , i.e. the probability that an atom crossing the boundary will be accommodated on arrival at the new phase. It is likely that incoherent interfaces and diffuse solid/liquid interfaces, as high angle grain boundaries, will have values of close to unity.
13. If a single atom attaches itself to a flat close-packed interface it will raise the interfacial free energy.
a) False
b) True
Answer: b
Explanation: If a single atom attaches itself to a flat close-packed interface it will raise the interfacial free energy and will therefore tend to detach itself again. It can thus be seen that continuous growth at the above type of interfaces will be very difficult which means the mobility and the accommodation factor will be very low.
14. Growth ledges are by no means restricted to solid/solid systems.
a) True
b) False
Answer: a
Explanation: Growth ledges are by no means restricted to solid/solid systems. The first evidence for the existence of growth ledges came from studies of solid/vapour interfaces. They are also found on faceted solid/liquid interfaces.
15. Calculate the driving force at temperature 300K if the value of Xi and Xe are given as 4mols and 2mols respectively?
a) 300Rln3
b) 300Rln2
c) 150Rln2
d) 150Rln3
Answer: b
Explanation: Here the value of driving force can be calculate using the equation Δμ = RTln . So at a temperature of 300K the value of the driving force is give as 300Rln2 but in the case of an ideal solution the equation can be re-written as Δμ= *.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Solidification – Nucleation in Pure Metal”.
1. Consider a given volume of liquid at a temperature T below Tm with a free energy G’. If some of the atoms of the liquid cluster together to form a small sphere of solid, the free energy of the system will change to G, given by________ .
a) V1G1 + V2G2 + Aγ
b) V1G2 + V2G1 + Aγ
c) V1G1 + V2G2 – Aγ
d) V1G2 + V2G1 – Aγ
Answer: a
Explanation: If some of the atoms of the liquid cluster together to form a small sphere of solid, the free energy of the system will change to G2 is given by the formula V1G1 + V2G2 + Aγ and this is a part of homogeneous nucleation and the formation of the solid results in the change in the free energy which is given by ΔG = G1-G2.
2. If the volume of the solid sphere and the volume of the liquid is given as V1 and V2 respectively, G1 and G2 are the free energies per unit volume of solid and liquid respectively. The free energy of the system without any solid present is given by_______
a) G = G2
b) G = V1G1 + V2G2
c) G = V1G2 + V2G1
d) G = G1
Answer: a
Explanation: The free energy of the system without any solid present is given by the formula G = *G2 and hence the formation of solid will directly lead to a change in free energy, G*-G where the value of G* is given as V1G1 + V2G2 + Aγ.
3. If the latent heat of fusion is given as L and the melting temperature of that substance is given as x. For an undercooling ΔT, the change in free energy is given as _________
a) ΔG = /x
b) ΔG = L*ΔT * x
c) ΔG = L*x/ΔT
d) ΔG = L/
Answer: a
Explanation: The change in free energy is given as ΔG = /x. Below Tm, ΔGv is positive so that the free energy change associated with the formation of a small volume of solid has a negative contribution due to the lower free energy of a bulk solid, but there is also a positive contribution due to the creation of a solid/liquid interface.
4. Let’s take the equilibrium melting temperature of a fluid ‘A’ as Tm and which is cooled below its equilibrium melting temperature then there is always possibility of a driving force F for solidification and it might be expected that the liquid phase of ‘A’ would spontaneously solidify.
a) True
b) False
Answer: b
Explanation: The above mentioned statement is not always true there are some exceptions. Let’s take the case of nickel, under suitable conditions nickel can be undercooled to 250 K below Tm and held there indefinitely without any transformation occurring.
5. The free energy change associated with homogeneous nucleation of a sphere has an interfacial energy term and a volume free energy. Calculate the value of the interfacial term if the interfacial energy is given as п kJ/m 2 Kmol?
a) 1000 kJ/Kmol
b) 10 kJ/Kmol
c) 1 kJ/Kmol
d) 100 kJ/Kmol
Answer: d
Explanation: The interfacial energy term is directly proportional to the square of the radius and the related equation is the product of the surface area and the interfacial energy and is given as 4пr 2 γ.
6. Calculate the volume free energy associated with homogenous nucleation of sphere if the radius of the sphere is given as 5m and the ΔGv is given as п kJ/m 3 Kmol?
a) 500 kJ/Kmol
b) 50 kJ/Kmol
c) 5 kJ/Kmol
d) 1000 kJ/Kmol
Answer: a
Explanation: Here when it comes to the interface term is directly proportional to r 2 and the volume free energy is directly proportional to r 3 here in order to calculate the volume free energy we require the value of r and ΔGv and the related equation is given as 4пr 3 ΔGv.
7. For a given undercooling there is a certain radius r*, which is associated with a maximum excess free energy. If r< r* the system can lower its free energy by_____
a) Creating a void
b) Maintaining its volume
c) Transferring liquid
d) Dissolution of solid
Answer: d
Explanation: Actually, the interface term is directly proportional to r 2 and the volume free energy is directly proportional to r 3 and hence the creation of small particles of solid always leads to a free energy increase. It is this increase that is able to maintain the liquid phase in a metastable state almost indefinitely at temperatures below Tm. If r< r* the system can lower its free energy by dissolution of solid.
8. Calculate the critical radius if the ΔGv and γ values are given as 5 kJ/m 3 Kmol and 2.5 kJ/m 2 Kmol respectively?
a) 4m
b) 2m
c) 1m
d) 0.5m
Answer: c
Explanation: Critical radius can be calculate if we know the values of interfacial energy and ΔGv and the related equation is as 2γ/ΔGv and when we substitute the respective values we get r* as 2*2.5/5=1m.
9. Calculate the ΔG’ if the ΔGv and γ values are given as √Π kJ/m 3 Kmol and 3kJ/m 2 Kmol respectively?
a) 144 kJ/Kmol
b) 81 kJ/Kmol
c) 125 kJ/Kmol
d) 16 kJ/Kmol
Answer: a
Explanation: Here in this case the ΔG’ can be calculated using the formula ΔG’ = 16Πγ 3 /3 2 . Here if we substitute the respective value we get the required solution and its vlue is given as 144 kJ/Kmol.
10. Calculate the critical radius, if the interface energy is given as 3kJ/m 2 Kmol and the undercooling is given as 300K ? as 150K).
a) 3/L
b) 6/L
c) 6L
d) 3L
Answer: a
Explanation: Here in this case the critical radius using the formula *, where γ denotes the interface so if the substitute the respective values in the formula we get, * =3/L.
11. If the liquid contains x atoms per unit volume, calculate the number of clusters that have reached the critical size , if the temperature is taken at T and the critical value of change in free energy at homogenous nucleation is given as ΔG’?
a) Y = x*exp
b) Y = x*exp
c) Y = x*exp
d) Y = x*exp
Answer: b
Explanation: Here the number of clusters the reach the critical size is dependent on the temperature and the critical value of free energy and the related formula is given as Y =x*exp where k represents the Boltzmann constant.
12. Calculate the value of ΔGv if the interfacial energy is given as 3kJ/mm 2 Kmol and the critical radius is given as 1.5mm?
a) 2 kJ/mm 3 Kmol
b) 4 kJ/mm 3 Kmol
c) 5 kJ/mm 3 Kmol
d) 3 kJ/mm 3 Kmol
Answer: b
Explanation: The value of ΔGv can be calculated using the critical radius and the interfacial energy and the related formula is given by twice the interfacial energy divided by the critical radius gives the value of ΔGv and here when we substitute the required value, we get b as the answer.
13. Value of shape factor of heterogeneous nucleation if Θ is given as 60°?
a) 5/32
b) 5/16
c) 5/4
d) 5/8
Answer: a
Explanation: The value of shape factor is given by s = * 2 /4 and hence when we substitute the value of Θ as 60, we get the value of cos as 0.5 and further when we substitute the value in the above mentioned equation we get the required solution.
14. In some metallic systems the liquid can be rapidly cooled to temperatures below the so-called glass transition temperature without the formation of crystalline solid.
a) False
b) True
Answer: b
Explanation: Since atomic jumps from the liquid on to the cluster are thermally activated, fₒ will in fact diminish with decreasing temperature. In some metallic systems the liquid can be rapidly cooled to temperatures below the so-called glass transition temperature without the formation of crystalline solid. fₒ is very small at these temperatures and the supercooled liquid is a relatively stable metallic glass or amorphous metal.
15. Which among the following are true about the activation barrier of homogeneous and heterogeneous nucleation?
a) ΔG > ΔG
b) ΔG << ΔG
c) ΔG is always a constant but ΔG varies with temperature
d) Nothing can be predicted
Answer: d
Explanation: This happens because the activation barrier of heterogeneous nucleation is given as ΔG = ΔG*s and here the extra term is known as the shape factor and is dependent on the wetting angle hence this statement is proved.
This set of Phase Transformation Questions & Answers for Exams focuses on “Solidification – Growth of a Pure Solid”.
1. Name the process by which the migration of rough interfaces takes place?
a) Lateral growth
b) Vertical wipe
c) Batch growth
d) Continuous growth
Answer: d
Explanation: Here the rough interface actually migrate using the continuous growth process and in case of a flat interface the migration takes place by the process called lateral growth process and in this case it involves ledges.
2. In the continuous growth process the driving force for solidification ΔG is given as ______
a) ΔG= L
b) ΔG= L
c) ΔG= ΔTi/
d) ΔG= Tm/
Answer: a
Explanation: Driving force for solidification in the case of a continuous growth process is dependent on the latent heat of melting and on the undercooling of the interface below the equilibrium melting temperature and the related equation is given as ΔG= L.
3. The equation related to the net rate of solidification is given as_______
a) R = K*ΔTi
b) R = K*ΔTi/Tm
c) R = K*Tm
d) R = K*
Answer: a
Explanation: The equation related to the net rate of solidification is given as R = K*ΔTi. A full theoretical treatment indicates that K has such a high value that normal rates of solidification can be achieved with interfacial undercooling of only a fraction of a degree Kelvin.
4. The surface nucleation rate governs the rate of growth normal to the interface .A theoretical treatment shows that this is proportional to_______
a) ΔTi
b) Exp
c) 1/ΔTi
d) Exp
Answer: d
Explanation: The surface nucleation rate is proportional to Exp . Once nucleated the disc will spread rapidly over the surface and the rate of growth normal to the interface will be governed by the surface nucleation rate.
5. Materials with a high entropy of melting prefer to form atomically smooth, close-packed interfaces. For this type of interface the minimum free energy also corresponds to the minimum internal energy.
a) True
b) False
Answer: a
Explanation: This statement directly implies a minimum number of broken solid bonds. Here in this case the minimum internal energy corresponds to the minimum free energy internal energy and is valid for this type of interface. The interfacial energy gets increased once the single atom leaves the liquid and gets attached itself to the flat solid surface.
6. Calculate the interfacial undercooling if the melting temperature is 600K and the latent heat of melting is given as 30kJ/kg? 150K
b) 600K
c) 900K
d) 450K
Answer: c
Explanation: The migration of a diffuse solid/liquid interface can be treated in a similar way to the migration of a random high-angle grain boundary. The free energy of an atom crossing the S/L interface will vary. And in this case the interfacial cooling can be calculated using the formula *. Substituting the values we get 600*45/30 = 900K.
7. Calculate the extent of interfacial undercooling if the value of k is given as 0.05) and the rate of solidification is given as 5m/sec?
a) 100K
b) 1000K
c) 200K
d) 2000K
Answer: a
Explanation: Here the extent of undercooling can be determined using the formula R/k, substituting the respective values we get 100K as the extent of interfacial undercooling. The above treatment is applicable to diffuse interfaces where it can be that atoms can be received at any site on the solid surface, that is, the accommodation factor A is approximately unity.So substituting the values we get 5/0.05 = 100K.
8. If the solid contains dislocations that intersect the S/L interface the problem of creating new interfacial steps can be circumvented. A complete theoretical treatment of this situation shows that for spiral growth the normal growth rate v and the undercooling of the interface ΔTi are related by an expression given as_______
a) R = K
b) R = K 3
c) R = K 2
d) R = K/ 2
Answer: c
Explanation: If we consider the growth of rough interfaces, the necessary undercooling will be the least for it at the given solid growth rate and for a particular or given undercooling the mobility of faceted interface are much less and it follows the spiral growth mechanism.
9. In solidification it is quite common for materials showing faceting to solidify as two crystals in twin orientations.
a) False
b) True
Answer: b
Explanation: The pattern or the orientation normally found once after the materials showing faceting solidifies is the twin orientation where the 2 crystals gets involved and because of this there is an intersection at the twin boundaries for the interfacial facets which finally act as new step providing n simple growth mechanism.
10. In pure metals solidification is controlled by the rate at which the latent heat of solidification can be conducted away from the solid/liquid interface. Which among the following equation satisfies the heat flow and the interface stability? .
a) KsTs = Kl*Tl /
b) KsTs = Kl*Tl – v*L
c) KsTs = Kl/
d) KsTs = Kl*Tl +v*L
Answer: d
Explanation: KsTs = Kl*Tl +v*L, the heat flow away from the interface through the solid must balance that from the liquid plus the latent heat generated at the interface. This equation is quite general for a planar interface and even holds when heat is conducted into the liquid.
11. Let us now take a closer look at the tip of a growing dendrite. The situation is different from that of a planar interface because heat can be conducted away from the tip in three dimensions. As a result of the Gibbs-Thomson effect equilibrium across a curved interface occurs at an undercooling ΔTr below Tm given by______
a) ΔTr = 2γTm/
b) ΔTr = 2Tm/
c) ΔTr = 2Tm/
d) ΔTr = 2γTm/L
Answer: a
Explanation: As a result of the Gibbs-Thomson effect equilibrium across a curved interface occurs at an undercooling ΔTr below Tm given by ΔTr = 2γTm/. The minimum possible radius of curvature of the tip is when ΔTr equals the total undercooling ΔTo = Tm – T. This is just the critical nucleus radius r* given by .
12. Let’s take the tip of a growing dendrite. Here in this case it can be seen that the tip velocity tends to zero for a particular value of r known as______
a) Critical radius
b) Open radius
c) Closed radius
d) Extended radius
Answer: a
Explanation: The above mentioned condition takes place when the radius tends to the critical radius and this is due to the Gibbs-Thomson effect and the r tends to infinity due to slower heat conduction. The maximum velocity can be acquired when the value of r = 2r*.
13. When the solidification takes place from the mould wall , this leads to the heat conduction through the solids. However the heat flow into the liquid arises if a certain condition is satisfied. Which among the following corresponds to the same?
a) If the liquid is supercooled below Tm
b) If the liquid is brought in contact with mould wall
c) If the liquid is supercooled at any temperature
d) Superheated above Tm
Answer: a
Explanation: The above mentioned condition arises if the liquid is supercooled below Tm. This type of situation is common at the beginning of the solidification but it should satisfy the condition that the nucleation should occur at impurity particles in the bulk of the liquid.
14. Calculate the value of latent heat of fusion per unit volume if the thermal conductivities of solid and liquid are given as 15 and 20 kW/mK respectively and the temperature gradient of liquid and solid are given as 2K/m and 1.5K/m respectively?
a) 0
b) 5
c) 25
d) 100
Answer: a
Explanation: Here when we balance the heat flow away from the interface through the solid and that from the liquid plus the latent heat generated at the interface we get the value of latent heat of fusion as 0 because when we substitute the respective values in the equation KsTs = Kl*Tl +v*L, we get KsTs = Kl*Tl, so hence L=0.
15. The maximum velocity at the tip of growing dendrite occurs when ______
a) R = 2r*
b) R = 3r*
c) R = r*
d) R = 4r*
Answer: a
Explanation: V= K/L-(r*/R 2 )), differentiate this equation with respect to radius and equate it to 0, so that we get the value of R for which the velocity at the tip of the dendrite is maximum. Finally, we obtain that at R = 2r* this happens.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Alloy Solidification”.
1. The solidification of pure metals is rarely encountered in practice. Which among the following is a reason for this?
a) Pure metals contain sufficient impurities
b) Pure metals have high value of Tm
c) Pure metals have low melting temperature
d) Pure metals can withstand high pressure
Answer: a
Explanation: There are a lot of reasons behind this but the major one is that even though we say something is pure or commercially pure, it contains an amount of impurities that we can consider sufficient to change the characteristics from pure to alloy behavior.
2. If we consider an idealized phase diagram by assuming that the solidus and liquidus are straight lines. Then the partition coefficient can be taken as _____
a) K = /Xs
b) K = /Xl
c) K = Xl/
d) K = Xs/Xl
Answer: d
Explanation: It is useful to take the partition coefficient as K = Xs/Xl, where Xs and Xl are the mole fractions of solid and liquid respectively. The way in which such alloys solidify in practice depends in rather a complex way on temperature gradients, cooling rates and growth rates.
3. “Non-equilibrium lever rule” is also known by the name _________
a) Matano analysis
b) Cape theory
c) Dobrienier theory
d) Scheil’s equation
Answer: d
Explanation: The equation XL =X o fl , represents the Scheil’s equation. For j < 1 these equations predict that when there is no diffusion in the solid there will always be some eutectic in the last drop of liquid to solidify, no matter how little solute is present.
4. When there will be a rapid build-up of solute ahead of the solid and a correspondingly rapid increase in the composition of the solid formed. This is known as ___
a) Final transient
b) Cover up
c) Random order
d) Initial transient
Answer: d
Explanation: Diffusion is the only mode of transportation for the solute rejected from the solid when there is no stirring or convection in the liquid phase. As an after effect there will be a rapid build-up of the solute much before the solid hence there will be a rapid increase in the composition of the solid formed. This is known as the initial transient.
5. Which among the following equation is related to diffusional mixing in liquid ?
a) –DCl = v
b) DCl = /v
c) DCl = Cs-Cl/v
d) DCl = v
Answer: a
Explanation: Steady state itself gives an idea that the particular state should be balanced. Here in this case the rate at which the solute is rejected from the solidifying liquid should be balanced by the rate at which solute diffuses down the concentration gradient away from the interface. That is –DCl = v.
6. Under the steady state growth of the critical gradient, if we consider T1 and T3 to be the liquidus and solidus temperatures for the bulk composition Xo. The equilibrium freezing range of the alloy solidus can be taken as _______
a) T3-T1
b) Xo*
c) T3/T1
d) T1-T3
Answer: d
Explanation: T1 – T3, is known as the equilibrium freezing range of the alloy. Clearly planar front solidification is most difficult for alloys with a large solidification range and high rates of solidification. Except under well-controlled experimental conditions alloys rarely solidify with planar solid/liquid interfaces.
7. In practice alloy solidification will usually possess certain features. Actually there is some stirring caused by the turbulence of the liquid which can happen due to many reasons. Which among the following is not a reason for that?
a) Pouring
b) Convection currents
c) Gravity currents
d) Composition of alloy
Answer: d
Explanation: There are a lot of reasons for this stirring to take place and this can happen because of liquid turbulence which may be caused by pouring or it can be caused by the convection currents or the gravity effects may also play a role in this.
8. As the A-rich α phase solidifies excess B diffuses a short distance laterally where it is incorporated in the B-rich β phase. For an interlamellar spacing K, there is a total of X m 2 of α/β interface per m 3 of eutectic. Find the value of X?
a) 1/K
b) 3/K
c) 2/K
d) 2/K 2
Answer: b
Explanation: For an interlamellar spacing K, there is a total of 2/K m 2 of α/β interface per m 3 of eutectic. However, there is a lower limit to K determined by the need to supply the α/β interfacial energy. Thus small interlamellar space should lead to rapid growth.
9. Unidirectional solidification has commercial importance.
a) False
b) True
Answer: b
Explanation: The production of creep resistant aligned microstructure for gas turbine blades is actually a commercial example of the unidirectional solidification which highlights its importance. Not only this but this is used during the zone refining for the production of extremely pure metals.
10. Which among the following can be categorized as a peritectic reaction?
a) Liquid + α->β
b) Liquid + α->β
c) Liquid + α->β
d) Liquid + α->β
Answer: d
Explanation: Liquid + α->β, this kind of reactions are known as peritectic reaction. In peritectic reaction a second solid phase is created as a result of the reaction between a liquid phase and a solid phase.
11. In general the tendency to form dendrites increases as the solidification range______
a) Becomes constant
b) Increases
c) Decreases
d) Nothing can be predicted
Answer: b
Explanation: In general the tendency to form dendrites increases as the solidification range increases. Therefore the effectiveness of different solutes can vary widely. For solutes with a very small partition coefficient cellular or dendritic growth can be caused by the addition of a very small fraction of a per cent solute.
12. In the case of the Al₆Fe-Al rod-like eutectic, the impurity causing the cellular structure is mainly___
a) Aluminum
b) Ferrous
c) Nickel
d) Copper
Answer: d
Explanation: A planar eutectic front is not always stable. If for example the binary eutectic alloy contains impurities, or if other alloying elements are present, the interface tends to break up to form a cellular morphology. In this case the impurity that leads to the cellular structure is copper.
13. Various different types of eutectic solidification are possible and these are usually classified as____
a) Normal and anomalous
b) Perfect and Imperfect
c) Proper and Dismantled
d) Even and uneven
Answer: a
Explanation: They are classified as normal and anomalous. Normal structures occur when both phases have low entropies of fusion. Anomalous structures, on the other hand, occur in systems when one of the solid phases is capable of faceting.
14. Under what condition is it possible to solidify an off-eutectic alloy without permitting the formation of the primary dendritic phase?
a) Controlled solidification
b) Rapid solidification
c) Multi-directional solidification
d) No such condition exist
Answer: a
Explanation: Under controlled solidification conditions, e.g. in unidirectional solidification experiments, it is possible to solidify an off-eutectic alloy without permitting the formation of the primary dendritic phase.
15. The total undercooling at the eutectic front has two contributions.
a) True
b) False
Answer: a
Explanation: ΔTo =ΔTr+ ΔTd, ΔTr is the undercooling required to overcome the interfacial curvature effects and ΔTd is the undercooling required to give a sufficient composition difference to drive the diffusion.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Solidification of Ingots and Castings”.
1. Which among the following is not a zone in solidified alloy ingots?
a) Columnar zone
b) Outer chill zone
c) Central equiaxed zone
d) Coverage zone
Answer: d
Explanation: An outer chill zone of equiaxed crystals, A columnar zone of elongated or column-like grains, and A central equiaxed zone. If you take in general these are the three different zones that can be distinguished in solidified alloy ingots.
2. Most engineering alloys begin by being poured or cast into a fireproof container or mould. If the as-cast pieces are permitted to retain their shape afterwards, or are reshaped by machining, they are called as _______
a) Metallurgy
b) Extrusion
c) Casting
d) Ingots
Answer: c
Explanation: If the as-cast pieces are permitted to retain their shape afterwards, or are reshaped by machining, they are called castings. If they are later to be worked, e.g. by rolling, extrusion or forging, the pieces are called ingots, or blanks if they are relatively small.
3. In the chill zone of ingots, if the pouring temperature is low the whole of the liquid will be rapidly cooled below the liquidus temperature and the crystals swept into the melt may be able to continue to grow. This is known as______
a) Big Bang nucleation
b) Rapid quenching
c) Big Boss nucleation
d) Casting prep
Answer: a
Explanation: The above mentioned is known as ‘big-bang’ nucleation since the liquid is immediately filled with a myriad of crystals. This produces an entirely equiaxed ingot structure, i.e. no columnar zone forms.
4. The chill zone grow dendritically in certain crystallographic directions. In the case of cubic metals it is given as______
a) <111>
b) <100>
c) <222>
d) <101>
Answer: b
Explanation: Very soon after pouring the temperature gradient at the mould walls decreases and the crystals in the chill zone grow dendritically in certain crystallographic directions, e .g. <100> in the case of cubic metals. Those crystals with a <100> direction close to the direction of heat flows grow fastest and are able to outgrow less favourably oriented neighbours.
5. What happens to most metals on solidification?
a) Shrink
b) Expand
c) Elongate
d) No change
Answer: a
Explanation: In general most of the metals shrink on solidification and these affects final ingot structure. There is a decrease in the level of the liquid remaining and this is a consequence of thicker outer shell of solid and with a narrow freezing range the mushy zone is also narrow.
6. The equiaxed zone consists of equiaxed grains. They are oriented randomly in which part the ingot?
a) Bottom edge
b) Centre
c) Right side
d) Top corner
Answer: b
Explanation: If we take the equiaxed zone, the center of the ingot is found to contain the equiaxed grains which are just randomly oriented. The melted-off dendrite side-arms are assumed to be the origin of these grains.
7. There is approximately no temperature gradient perpendicular to the growth direction for a ______
a) Multidirectional solidification
b) Unidirectional solidification
c) Mixed solidification
d) Partial solidification
Answer: b
Explanation: If we take the temperature gradient perpendicular to the growth direction it turn out to be zero in case of unidirectional solidification. The spacing in the dendrite arm is probably the one which reduces constitutional super cooling in the intervening liquid to a very low level.
8. If the lamellar distance k tends to infinity, the free energy change associated with the solidification of 1 mol of liquid is given by_____
a) ΔG = ΔG + 2Vm/k
b) ΔG = ΔG
c) ΔG = ΔG + c
d) ΔG = ΔG – 2Vm/k
Answer: b
Explanation: The free energy change associated with the solidification of 1 mol of liquid is given by, ΔG = ΔG + 2Vmγ/k. where Vm is the molar volume of the eutectic and ΔG is the free energy decrease for very large values of k. Since here k->infinity, the second term disappears.
9. Which among the following can be the reason for the disappearance of the primary dendrites?
a) Growth rate is raised above critical level
b) Under slow growth rate
c) Very minute temperature gradient
d) Forced Growth rate
Answer: a
Explanation: If the growth rate is raised above a critical level this happens. The reason for the disappearance of the primary dendrites is that for a given growth velocity the eutectic is able to grow at a higher temperature than the dendrite tips.
10. In practice the maximum casting speed and billet cross-section are less for steel than for aluminium or copper.
a) True
b) False
Answer: a
Explanation: In case of continuous casting, the isotherm distributions are affected by the conductivity of the solidifying metal and its speed of withdrawal from the mould. This means, for example, that the depth of the liquid pool in continuous casting is much greater for steel than for aluminium alloys under comparable conditions. This implies that in practice the maximum casting speed and billet cross-section are less for steel than for aluminium or copper.
11. In alloys with a wide freezing range the mushy zone can occupy the whole of the ingot. In this case no central pipe is formed.
a) True
b) False
Answer: a
Explanation: It is commonly found in the alloys with a wide freezing range that ingot is completely occupied by the mushy zone hence there is no chance for the formation of central pipe. As a result in order to compensate for the shrinkage of the dendrites the liquid level gradually falls across the width of the ingot as liquid flows down.
12. What are the two different types of segregation that can be distinguished in solidified structures?
a) Competent and non-competent
b) Semi and complete
c) Macro and micro
d) Full and partial
Answer: c
Explanation: Generally, there are two types of segregations which can be distinguished in the solidified structures. The first one is micro, that occurs on the scale of the secondary dendrite arm spacing and the second done is the macro which changes composition over comparable distance.
13. Which among the following is not an important variable in weld solidification or continuous casting?
a) Thickness of plate being welded
b) The rate of heat input
c) Speed of arc movement
d) External condition
Answer: d
Explanation: Except the external conditions rest everything plays an important role in the weld solidification and continuous casting even the thermal conductivity of the metal being welded also plays a crucial role in the weld solidification.
14. Which among the following is not a factor that leads to macro-segregation in ingots?
a) Shrinkage due to solidification and thermal contraction
b) Density differences in the interdendritic liquid
c) Density differences between the solid and liquid
d) Expansion due to mild heating
Answer: d
Explanation: Shrinkage due to solidification and thermal contraction; density differences in the interdendritic liquid; density differences between the solid and liquid; and convection currents driven by temperature-induced density differences in the liquid. These are four factors that lead to macro segregation in ingots.
15. Shrinkage effects can give rise to what is known as inverse segregation. The effect is particularly marked in alloys with a_______
a) Wide freezing range
b) Short freezing range
c) Wide melting range
d) Short melting range
Answer: a
Explanation: The effect is particularly marked in alloys with a wide freezing range. As the columnar dendrites thicken solute-rich liquid ,k < 1) must flow back between the dendrites to compensate for shrinkage and this raises the solute content of the outer parts of the ingot relative to the centre. Example Cu-Sn alloy.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Solidification of Fusion Welds”.
1. Which among the following is a property of fusion welding not of solid state welding?
a) Filler metal is added
b) Heat is applied
c) Used to join metals
d) Nothing can differentiate them
Answer: a
Explanation: In case of a fusion welding coalescence is accomplished by melting the two parts to be joined, in some cases adding filler metal to the joint, but in case of solid state welding filler metal is not added.
2. A fusion-welding operation in which no filler metal is added is referred to as a/an _____
a) Compile weld
b) Autogenous weld
c) Filler Weld
d) Damper weld
Answer: b
Explanation: When the filler metals are not added in a fusion-welding operation then it is referred as an autogenous weld Fusion-welding processes and it uses heat to melt the base metals just like the other fusion welding process. Mainly these fillers are added to facilitate the process and to give a strength or toughness to the joints. .
3. Which among the following is not a fusion welding technique?
a) Arc welding
b) Oxy-fuel welding
c) Resistance welding
d) Diffusion welding
Answer: d
Explanation: Here if we check the first three options they belong to the category of fusion welding technique and they use heat mainly to join the metals but in the diffusion welding two surfaces are held together by using the pressure at a pretty high temperature and the parts coalesce by solid-state diffusion.
4. In which joint type, the parts lie in the same plane and are joined at their edges?
a) Butt
b) Corner
c) Lap
d) Tee
Answer: a
Explanation: This happens in the case of butt joints. Whereas in corner joints the parts in a corner joint form a right angle and are joined at the corner of the angle and in the case of lap joint the joints consist of two overlapping parts.
5. The heat to transform the metal from solid to liquid phase at the melting point, which depends on the metal’s heat of fusion. Calculate the unit energy for melting if the melting temperature is 1000K? (K = 3.33 * 10 -6 ).
a) 3.33
b) 6.66
c) 9.99
d) 3.33*10 -3
Answer: a
Explanation: If we consider a fair approximation the total amount of heat or the quantity of the heat can be calculated using the following equation Um = K 2 , where Um = the unit energy for melting and Tm is the melting point of the metal. Um is the total amount or quantity of heat required to melt a unit volume of metal taking the initial temperature as room temperature.
6. Typical fusion-weld joint in which filler metal has been added consists of several zones.
a) True
b) False
Answer: a
Explanation: The several zones which include the heat affected zone, fusion zone, the heat unaffected zone that is present away from the fusion zone and weld interface zone.
7. Calculate the net heat available for welding if the heat transfer factor f1, melting factor f2 and the total heat supplied are given as 0.8, 0.9, 1000J respectively?
a) 720J
b) 100J
c) 1000J
d) 110J
Answer: a
Explanation: The net heat available for welding is actually the product of the heat transfer factor, melting factor and the total heat supplied. In this case it, H = 0.9*0.8*1000J =720J.
8. In the case of arc welding find the arc time, if the time for which the arc is switched on is given as 3hrs and the hours worked with that is given as 1hr?
a) 3hrs
b) 1.5hrs
c) 6hrs
d) 9hrs
Answer: a
Explanation: Arc time = Time arc is on/ hours worked. Productivity is also an issue. It is often measured as arc time , that is the proportion of hours worked that arc welding is being accomplished.
9. A gas tungsten arc-welding operation is performed at a current of 300 A and voltage of 20V. Calculate the total power in the operation?
a) 6000W
b) 150W
c) 600W
d) 1500W
Answer: a
Explanation: The total power generated is given as P = IV, so in this case it is 300*20 = 6000W. Power Source in Arc Welding, both direct current and alternating current are used in arc welding.
10. If the total power generated is given as 3000W. Calculate the rate of heat generation at the weld if the melting factor is given as 0.6 and the heat transfer factor as 0.8?
a) 1444J/s
b) 1434J/s
c) 1240J/s
d) 1440J/s
Answer: d
Explanation: The rate of heat generation at the weld is given as the product of the total input power, the heat transfer factor and the melting factor. So in this case 0.8*0.6*3000 which gives the required answer 1440J/s.
11. A gas tungsten arc-welding operation is performed at a current of 500 A and voltage of 20 V. The melting factor f2 = 0.5, and the unit melting energy for the metal Um = 10 Jmm -3 . Calculate the volume of metal welded?
a) 350mm/sec
b) 950mm/sec
c) 450mm/sec
d) 550mm/sec
Answer: a
Explanation: Here the volume of the metal welded can be calculated if we know the total heat generation at the weld. This can be calculated if we know the total power required. So here the total power is given as 500*20 = 10000W, hence the rate is given as 0.7*0.5*10000 = 3500J/sec and the volume is given as 3500/10=350mm/sec.
12. The power source in a particular welding setup generates 4000 W that can be transferred to the work surface with a heat transfer factor = 0.9. The metal to be welded is low carbon steel, whose melting temperature, is 1800 K. The melting factor in the operation is 0.6. A continuous fillet weld is to be made with a cross-sectional area = 20 mm 2 . Determine the travel speed at which the welding operation can be accomplished? (K = 3.33 *10 -6 )
a) 59.5mm/sec
b) 10.49mm/se
c) 3.65mm/sec
d) 4.56mm/sec
Answer: b
Explanation: Let us first find the unit energy required to melt the metal Um, which can be calculated as Um = 3.33*10 -6 *1800 2 = 10.78Jmm -3 .To calculate the travel velocity we take the equation f1*f2*R=Um*A*v, hence from this we get the value of v the required velocity, v = /10.3*20 = 10.49mm/sec.
13. The quantity of heat required to melt a given volume of metal does not depend on______
a) The melting point of the metal
b) Composition of elements in the melt
c) The heat to transform the metal from solid to liquid phase at the melting point
d) The heat to raise the temperature of the solid metal to its melting point
Answer: b
Explanation: The quantity of heat required to melt a given volume of metal depends on the melting point of the metal, the heat to transform the metal from solid to liquid phase at the melting point which depends on metals heat of fusion and the heat to raise the temperature of the solid metal to its melting point.
14. A flux is a substance used to prevent the formation of oxides and other unwanted contaminants, or to dissolve them and facilitate removal.
a) False
b) True
Answer: b
Explanation: A flux is a substance used to prevent the formation of oxides and other unwanted contaminants, or to dissolve them and facilitate removal. During welding, the flux melts and becomes a liquid slag, covering the operation and protecting the molten weld metal. The slag hardens upon cooling and must be removed later by chipping or brushing.
15. Arc shielding is accomplished by covering the electrode tip, arc, and molten weld pool with a blanket of gas or flux, or both, which inhibit exposure of the weld metal to air.
a) True
b) False
Answer: a
Explanation: At the high temperatures in arc welding, the metals being joined are chemically reactive to oxygen, nitrogen, and hydrogen in the air. The mechanical properties of the weld joint can be seriously degraded by these reactions. Thus, some means to shield the arc from the surrounding air is provided in nearly all AW processes. Arc shielding is accomplished by covering the electrode tip, arc, and molten weld pool with a blanket of gas or flux, or both, which inhibit exposure of the weld metal to air.
This set of Phase Transformation Quiz focuses on “Solidification During Quenching from the Melt”.
1. Solidification can also occur at much higher rates of 10 4 -10 7 K/s. In which among the following it is not applicable?
a) Liquid metal atomization
b) Melt spinning
c) Roller quenching
d) Solidification of weld
Answer: d
Explanation: Here except the fourth option rest everything can occur at much higher rates. Processes such as liquid metal atomization, melt spinning, roller-quenching or plasma spraying, as well as laser or electron beam surface treatment occur at this rate. By quenching melts, it is possible to achieve various metastable solid states not predicted by equilibrium phase diagrams.
2. Which among the following solidification can occur without microsegregation?
a) Crystalline solidification
b) Mixed solidification
c) Partial solidification
d) Revert solidification
Answer: a
Explanation: Crystalline solidification can occur without microsegregation or with cells or secondary dendrites spaced much more finely than in conventional solidification processes. Whether the solid is crystalline or amorphous, rapid solidification processing offers a way of producing new materials with improved magnetic or mechanical properties.
3. The breakdown of solid/liquid interface at the local equilibrium can be due to___
a) Slow cooling
b) Slow heating
c) Rapid cooling
d) Rapid heating
Answer: c
Explanation: One consequence of rapid cooling can be that local equilibrium at the solid/liquid interface breaks down. Melts can solidify with no change in composition, i.e. partition less solidification or solute trapping can occur.
4. It is seen that for crystal growth rate, R, to keep pace with the welding speed, 3m/s, the condition must be met that for this to happen is____
a) R = 1.5m/s
b) R = 3m/s
c) R = 0
d) R = 1.33m/s
Answer: a
Explanation: For this to happen it should satisfy the condition that R the crystal growth rate should be equal to V*cosφ , so here in this case φ = 60, hence cosφ equal to 0.5 hence the rate is given as 3*0.5 = 1.5 m/s.
5. Welding can be considered as a ______
a) Dynamic process
b) Static process
c) Rapid process
d) Explosive process
Answer: a
Explanation: Welding is essentially a dynamic process in which the heat source is continuously moving. This means that the maximum temperature gradients are constantly changing direction as the heat source moves away.
6. An important effect of increasing the welding speed is that the shape of the weld pool changes from an elliptical shape to a ___________
a) Circular shape
b) Pear shape
c) Rectangular shape
d) Broad shape
Answer: b
Explanation: Here the main change observed is in the shape of the weld pool, it changes from an elliptical shape to a narrower, pear shape. Corresponding to the angular geometry of the melt, the pear-shaped weld pool keep fairly constant thermal gradients up to the weld centre-line.
7. TIG welding of thin plates will give steeper thermal gradients than submerged arc welding of thick plates, the latter process will have the higher heat input.
a) True
b) False
Answer: a
Explanation: The above mentioned statement is true. Here the base metal is an efficient heat sink and the temperature noticed directly underneath the arc is so high, hence as a consequence the degree of super cooling turns be so low.
8. Typical shielding gases in TIG welding includes ______
a) Helium
b) Carbon
c) Oxygen
d) Methane
Answer: a
Explanation: Here the melting point of tungsten is fairly high and it is about 3140 degree Celsius hence it act as a good electrode. Argon, helium or the mixture of these gas elements act as the typical shielding gases.
9. Plasma arc welding is a special form of ____
a) TIG welding
b) Electron beam welding
c) Resistance welding
d) Submerged arc welding
Answer: a
Explanation: Plasma arc welding is a special type or form of gas tungsten arc welding in which the metal to metal joint takes place in molten state and in the weld area we have the a constricted plasma arc. It is extensively used in the electronic industries or for coating on the turbine blade we use this type of welding.
10. Which among the following is the first automated arc welding process?
a) TIG
b) Submerged arc welding
c) Plasma arc welding
d) Resistance arc process
Answer: b
Explanation: This process, developed during the 1930s, was one of the first AW processes to be automated. Submerged arc welding is an arc-welding process that uses a continuous, consumable bare wire electrode, and arc shielding is provided by a cover of granular flux.
11. Calculate the heat density if the power transferred is 1000W and the corresponding surface area is 10m 2 ?
a) 100
b) 10
c) 1
d) 1000
Answer: a
Explanation: Heat density can be calculated using the formula P/A, where P refers to the total power and A the corresponding area. So, in this case 1000/10 = 100W/m 2 . High-density heat energy is supplied to the faying surfaces, and the resulting temperatures are sufficient to cause localized melting of the base metals
12. In general dendritic and cellular substructures in welds tend to be on a finer scale than in casting.
a) False
b) True
Answer: b
Explanation: The statement is true and this is mainly due to the comparatively high solidification rates of weld metal. Here the greater rates of solidification can be achieved by the higher welding speeds or thicker base metals, hence the finest and the smoothest substructures are associated with these welds.
13. How much impact does a nucleation barrier can create to the solidification when melt has approximately the same composition as the base metal, ‘wetting’ of the base metal is very efficient?
a) It can increase the undercooling
b) It can decrease the rate of undercooling
c) It can melt the base metal
d) No major effect
Answer: d
Explanation: There is almost no nucleation barrier to solidification and hence very little undercooling occurs. Solidification is thus predicted to occur epitaxially, that means the nuclei will have the same lattice structure and orientation as the grains at the solid-liquid surface of the base metal.
14. Wire diameters in the Gas metal arc welding ranges from _______
a) 0.5-0.8cm
b) 1-3m
c) 1-4mm
d) 0.65-0.8mm
Answer: d
Explanation: Wire diameters ranging from 0.8 to 6.5 mm are used in GMAW, the size depending on the thickness of the parts being joined and the desired deposition rate. Gases used for shielding include inert gases such as argon and helium, and active gases such as carbon dioxide.
15. General coarseness of the microstructure is largely determined by the _______
a) Grain size of base metal
b) Composition of base metal
c) Texture
d) Nothing can be said
Answer: a
Explanation: the general coarseness of the microstructure is largely determined at the melt by the grain size of the base metal. Unfortunately, the base metal at the transition zone receives the most severe thermal cycle and after high-energy welding in particular the grains in this zone tend to grow and become relatively coarse.
This set of Phase Transformation MCQs focuses on “Solidification – Case Studies of Practical Casting and Weld”.
1. The process in which molten metal flows by gravity or other force into a mold where it solidifies in the shape of the mold cavity is known as_____
a) Rolling
b) Ingots
c) Extrusion
d) Casting
Answer: d
Explanation: The above mentioned process is known as casting. Part made from casting process is also termed as CASTING this process is carried out in foundry. It can be used to create complex parts, very large parts can be manufactured.
2. Upper half of a mold is known as ______
a) Drag
b) Gating
c) Cope
d) Riser
Answer: a
Explanation: Upper and lower half of a mold is known as Drag and Cope respectively. Gating system in a casting mold is the channel or network of channels by which molten metal flows into the cavity from outside the mold and the riser is a reservoir.
3. Which among the following is not a challenge in determining heat input?
a) Specific may be different for solid and liquid
b) Alloy melt over a range of temperature
c) Significant heat loss to environment
d) Manpower
Answer: d
Explanation: The general problems encountered are as follows. Specific heat and other thermal properties of a solid metal vary with temperature, especially if the metal undergoes a change of phase during heating. Specific heat may be different for solid and liquid state. Alloys melt over range of temperatures. Values in eq. for typical alloys are not easily available. There are significant heat losses to the environment during heating.
4. One cubic meter of some alloy is heated in a crucible from room temperature to 100 degree Celsius above its melting point for casting. The alloy density = 7.5 g/cm 3 , melting point = 800 0C, specific heat = 0.33 J/gC in the solid state and 0.29 J/gC in liquid state and heat of fusion = 160 J/g. How much heat energy must be added to accomplish the heating, assuming no losses? Ambient temp = 25 °C. Density of solid and liquid are same.
a) 4535 * 10 6 J
b) 8735 * 10 6 J
c) 9135 * 10 6 J
d) 7835 * 10 6 J
Answer: d
Explanation: Here the heat energy that should be added is given as = density*V*[Cs(Tm-T] + Hf +Cl*[Tp – Tm]. Cs represents weight specific heat for solid metal, Tm represents the melting temperature of metal, T represents the ambient temperature, and Hf represents the heat of fusion. Tp represents the pouring temperature.
5. A mold sprue is 20 cm long and the cross sectional area at its base is 2.5cm 2 . The velocity of the flowing metal is given as 198cm/sec. Volume rate of flow?
a) 495cm 2 /sec
b) 895cm 2 /sec
c) 585cm 2 /sec
d) 695cm 2 /sec
Answer: a
Explanation: According to the continuity rule the volume flow rate is given as the product of area and velocity, thus the increase in area will result in a decrease in the velocity so in this case substituting the values will give the required answer.
6. A mold sprue is 20 cm long and the cross sectional area at its base is 2.5cm 2 . Calculate the velocity of the flowing metal at the base of the sprue?
a) 109.1cm/s
b) 198.1cm/s
c) 186.1cm/s
d) 208.1cm/s
Answer: b
Explanation: The velocity of the flowing metal at the base of the sprue is given as square root of 2*g*H and substituting the respective values will give √2gh = 198.1cm/s.
7. The sprue feeds a horizontal runner leading into mold cavity whose volume is 1560cm 3 .Find the time required to fill the mold if the volumetric flow rate is given as 495cm 2 /sec?
a) 7.2sec
b) 6.4sec
c) 3.2sec
d) 4.8sec
Answer: c
Explanation: Here we assume the runner from the sprue base to the mold cavity to be horizontal so that the volume flow rate through the gate into the mold cavity remains equal to v*A at the base. So in this case the time required is given as Volume / Volumetric rate = 1560/495= 3.2sec.
8. In the casting of steel under certain mold condition the mold constant in Chvorinov’s rule is known to be 4 min/cm 2 .The volume and the surface area of the casting plate is known to be 600cm 3 and 760cm 2 respectively. Calculate the solidification time?
a) 2.49Hr
b) 2.49min
c) 3.64min
d) 3.45min
Answer: b
Explanation: Higher Volume to surface area ratio will cool and solidify slowly than with lower ration. Solidification time for riser must be more than the solidification time for casting. Riser with higher volume to area ratio can lead to solidification of casting prior to riser. Here in this case the solidification time is given as C 2 = 4 2 .
9. Metal forming can be done in the cold working process. Which among the following is a disadvantage of this process compared to hot working?
a) Surface oxidation
b) Shorter tool life
c) Poor finish
d) Higher forces or power required
Answer: d
Explanation: Here in this case the cold working process gives good surface finish, better tool life and the chance of surface oxidation is very less but the power required or the force required for this process is bit high compared to the hot working process.
10. Which among the following is a property of warm working process?
a) Enhanced plastic deformation
b) Working temperature is more than the melting temperature
c) Higher forces are required
d) Work piece is not ductile
Answer: a
Explanation: In this case, forming is performed at temperatures just above room temperature but below the recrystallization temperature. The working temperature is taken to be 0.3 Tm where Tm is the melting point of the work piece. Enhanced plastic deformation properties lower forces required are some of the major advantages of this form of working.
11. Which among the process is used to create objects of a fixed cross sectional profile?
a) Extrusion
b) Rolling
c) Casting
d) Ingot
Answer: a
Explanation: Extrusion is the process that is used to create objects of a fixed cross sectional profile. Extrusion is a compression process in which the work metal is forced to flow through a die opening to produce a desired cross sectional shape.
12. In the case of arc welding find the arc time, if the time for which the arc is switched on is given as 4hrs and the hours worked with that is given as 2hr?
a) 4hrs
b) 2hrs
c) 6hrs
d) 8hrs
Answer: b
Explanation: It is often measured as arc time , that is the proportion of hours worked that arc welding is being accomplished. Arc time = Time arc is on/ hours worked. Productivity is also an issue. So in this case 4/2 = 2hrs.
13. Which among the following is not a part of electrode composition of stainless steel?
a) Cr
b) Ni
c) C
d) Ca
Answer: d
Explanation: The electrode composition of stainless steel contains nickel, chromium, carbon, silicon and magnesium, even the phosphorus and Sulphur is present in minor traces.
14. Permanent Mold casting uses two molds that are designed for easy precise opening and closing. The materials behind the making the mold are_____________
a) Nickel and calcium
b) Cast iron or steel
c) Tungsten and steel
d) Plastic material
Answer: b
Explanation: Permanent Mold casting uses two molds made up of steel or cast iron that are designed for easy precise opening and closing. Cavity and gating system is machined in two halves for good accuracy and surface finish.
15. Which among the following is not an example of permanent mold casting?
a) Gravity
b) Slush
c) Vacuum
d) Sand casting
Answer: d
Explanation: Option d is an example of expandable mold casting. Expendable mold casting requires new mold for every casting and permanent mold casting mold can be reused. Permanent Mold casting uses two mold made up of steel or cast iron that are designed for easy precise opening and closing.
This set of Phase Transformation Multiple Choice Questions & Answers focuses on “Homogeneous Nucleation in Solids”.
1 For both coherent and incoherent inclusion the misfit strain energy is proportional to____
a) Shape of the inclusion
b) Mass of the inclusion
c) Density of the inclusion
d) Volume of the inclusion
Answer: d
Explanation: In general the transformed volume will not fit perfectly into the space originally occupied by the matrix and this gives rise to a misfit strain energy ΔGs per unit volume of β. For both coherent and incoherent inclusions, ΔGs is proportional to the volume of the inclusion.
2. If values of the misfit energy term and reduction of volume free energy are the same then the total free energy during a homogeneous transformation is given as______
a) 0
b) VΔGs + VΔGv
c) Aγ
d) Cannot be determined
Answer: c
Explanation: -VΔGv + VΔGs + Aγ is the total free energy equation of a homogeneous transformation and from the above given condition it is already given that -VΔGv + VΔGs = 0, as their values are same and hence the total free energy is just Aγ.
3. What will happen to critical radius if the value of ΔGs, ΔGv becomes same?
a) Tends to infinity
b) Tends to 0
c) Unity
d) No relation
Answer: a
Explanation: Here in this case when the value of ΔGs, ΔGv becomes equal the critical radius of the nuclei will tend to infinity as they are linked by the equation R* = 2γ/ , so when the denominator becomes zero the entire relation will tend to infinity.
4. Calculate the value of critical free energy if the value of -ΔGv) is given as 22/7? γ 3
b) γ 3
c) γ 3
d) γ 3
Answer: c
Explanation: When we differentiate the equation related to the total free energy with respect to free energy we end up with the equation G* = πγ 3 / 2 , substituting the respective values in the equation will end up giving the result as γ 3 .
5. Consider the precipitation of B-rich β from a supersaturated A-rich α solid solution. Which among the following will lead to the creation of activation energy barrier?
a) Creation of α/β interface
b) Diffusion of α matrix
c) Diffusion of β matrix
d) Nothing can be predicted
Answer: a
Explanation: Here the first step in the for the nucleation of β is that the B-atoms within the α matrix must first diffuse together and they try to form a small volume with the β composition, and then, when its necessary, the atoms must reorder or rearrange themselves into the crystal structure β . Then the α/β interface must be created during the process with the liquid->solid transformation, this leads to an activation energy barrier.
6. Calculate the concentration of critical sized nuclei if the critical free energy tends to 0?
a) Cx
b) Cx*exp
c) Cx*exp
d) Nothing can be predicted
Answer: a
Explanation: Since the value of critical free energy is 0, the concentration of critical sized nuclei formed is Cx. This can be understood easily when we substitute the respective values in the related equation that is C* = Cx*exp , since the value of ΔG*->0 we get C*= Cx.
7. If we ignore the variation of γ with interface orientation and assume the nucleus is spherical with a radius of curvature r, determine the value of γ if the value of tends to zero?
a) ΔG/4πr 2
b) ΔG/4πr 3
c) ΔG/*πr 2 )
d) ΔG/4πr
Answer: a
Explanation: If we consider the value of ΔG as a function of r we finally end up in the equation, ΔG = *πr 3 * + 4πr 2 γ, so when this value tends to zero we can directly equate the interfacial energy term with the total energy.
8. In which among the following system does the equilibrium phase nucleate homogeneously?
a) Cu-Co
b) S-P
c) Cu-Ag
d) Au-Cu
Answer: a
Explanation: In the Cu-Co system Cu alloys containing 1-3% Co can be solution treated and quenched to a temperature where Co precipitates. Both Cu and Co are fee with only a 2% difference in lattice parameter. Therefore, very little coherency strain is associated with the formation of coherent Co particles.
9. If each nucleus can be made supercritical at a rate of f per second the homogeneous nucleation rate will be given by________
a) N = f*C
b) N = f/C
c) N = f+C
d) N = f-C
Answer: a
Explanation: The value of homogeneous nucleation rate is given as N = f*C. f depends on how frequently a critical nucleus can receive an atom from the ‘α’ matrix. This will depend on the surface area of the nucleus and the rate at which diffusion can occur.
10. If each nucleus can be made supercritical at a rate of f per second the homogeneous nucleation rate is given as f*C . The value of f depends on ______
a) Vibrational frequency of atoms
b) Radius of atoms
c) Texture of the phase
d) Nothing can be predicted
Answer: a
Explanation: If the activation energy for atomic migration is ΔGm per atom, f can be written as w exp where w is a factor that includes the vibration frequency of the atoms and the area of the critical nucleus.
11. Which among the following can make the homogeneous reaction feasible?
a) Contraction relationship
b) Circuit orientation
c) Orientation relationship
d) It is not possible
Answer: c
Explanation: Incoherent nuclei have such a high value of γ that incoherent homogeneous nucleation is virtually impossible. If, however, the nucleus has an orientation relationship with the matrix, and coherent interfaces are formed, ΔG* is greatly reduced and homogeneous nucleation becomes feasible.
12. Equilibrium phase is probably formed homogeneously at a few tens of degrees undercooling is at the precipitation of Ni3Al in many Ni-rich alloys.
a) False
b) True
Answer: b
Explanation: This is an example where the equilibrium phase nucleate. Depending on the system the misfit varies up to a maximum of 2%, and interfacial energy is probably less than 30 mJmˉ 2 . Most other examples of homogeneous nucleation, however, are limited to metastable phases, usually GP zones
13. Which among the following curve represents the surface energy term?
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: The curve 1 represents the surface energy term and it is directly proportional to the square of the radius. Curve 3 depicts the volume terms and is directly proportional to the cube of the radius and it is preceded by a negative sign.
14. Driving force for precipitation is the main factor in controlling_________
a) The energy barrier or critical free energy
b) Size of the nuclei
c) Surface energy
d) Rate of fusion
Answer: a
Explanation: The main factor that controls ΔG* is the driving force for precipitation. And this ΔG* is strongly dependent on the temperature. This critical free energy is obtained by differentiating the total energy equation and equating it to 0.
15. In most systems the α and β phases have such different crystal structures that it is impossible to form coherent low-energy interfaces and homogeneous nucleation of the equilibrium β phase is then impossible.
a) True
b) False
Answer: a
Explanation: Even though this scenario exists, it is often possible to form a coherent nucleus of some other, metastable phase which is not present in the equilibrium phase diagram. The most common example of this is the formation of GP zones.
This set of Phase Transformation Question Bank focuses on “Solids Diffusional Transformations – Heterogeneous Nucleation”.
1. Nucleation in solids, as in liquids, in most cases is always __________
a) Homogeneous
b) Heterogeneous
c) Mixed
d) Semi heterogeneous
Answer: b
Explanation: This is the scenario found in most cases because they prefer the non-equilibrium defects as there nucleation sites, in many cases it occurs in the grain boundaries, excess vacancies and in the stacking faults.
2. If the creation of a nucleus results in the destruction of a defect this will result in the_______
a) Destruction of new nuclei
b) Mass production of new nuclei
c) Change of phase
d) Reduction of activation energy barrier
Answer: d
Explanation: The creation of a nucleus results in the destruction of a defect this will result in the production of some energy, let it be ΔGd and this will be released hence making an impact in the activation energy barrier and that too in the reduction of activation energy barrier.
3. Which among the following equation satisfies heterogeneous nucleation?
a) ΔGhet = – V* + Aγ – ΔGd
b) ΔGhet = – V* – ΔGd
c) ΔGhet = – V* + Aγ + ΔGd
d) ΔGhet = – V* + Aγ
Answer: a
Explanation: Equation, ΔGhet = – V* + Aγ – ΔGd corresponds to heterogeneous nucleation. Here the only difference between the homogeneous and the heterogeneous nucleation comes in the last term ΔGd and this the energy released during the destruction of defects.
4. If we ignore the effect of misfit strain energy, which among the following could minimize the interfacial free energy?
a) Optimum embryo texture
b) Optimum embryo shape
c) Optimum embryo volume
d) Optimum embryo length
Answer: b
Explanation: Ignoring any misfit strain energy, the optimum embryo shape should be that which minimizes the total interfacial free energy. The optimum shape for an incoherent grain-boundary nucleus will consequently be two abutted spherical caps.
5. The ability of the grain boundary to reduce the value of critical free energy ΔG* depends on
a) CosΦ
b) SinΦ
c) TanΦ
d) CotΦ
Answer: a
Explanation: The ability of a grain boundary to reduce ΔG , i.e. its potency as a nucleation site, depends on cosΦ. The cosΦ is actually the ratio of γαα/γαβ .
6. Calculate the shape factor if the angle tends to 90 degree?
a) 0
b) 1
c) 2
d) 4
Answer: b
Explanation: The shape factor will tend to unity when the angle tends to 90 degree, this can be proved easily using the following equation 0.5* 2 . Substitute the value 90 in the place of Φ then we get the required answer as 1.
7. If the ΔG* heterogeneous is given as 5kJ/mol and ΔG* homogeneous is given as 10kJ/mol. Calculate the shape factor?
a) 0.5
b) 1
c) 2
d) 2.5
Answer: a
Explanation: Here in this case the shape factor can be determined by the ratio of ΔG*heterogeneous/ΔG*homogeneous. So here for this particular case the shape factor value is 5/10 =0.5. Actually the grain boundary has the ability to reduce ΔG* heterogeneous, it’s the power it has as a nucleation site.
8. Excess vacancies are retained during the quench if the age hardening alloys are retained from______
a) Dark texture
b) Room temperature
c) Low temperature
d) High temperature
Answer: d
Explanation: Excess vacancies are retained during the quench if the age hardening alloys are retained from high temperature. This can actually increase the diffusion rate or it can relieve the misfit strain energies hence it assist the nucleation.
9. What is special about the dark-field electron microscope micrograph?
a) Precipitates can be imaged bright and matrix dark
b) 360 degree rotation is possible
c) High speed imaging is possible
d) Anti roller scheme available
Answer: a
Explanation: The so-called dark-field electron microscope micrograph in which the precipitates are imaged bright and the matrix dark. The precipitates lie in rows along dislocations. And this can be used in case of niobium carbonitride precipitates on dislocations in a ferritic iron matrix
10. The relative magnitudes of the heterogeneous and homogeneous volume nucleation rate is given as 0.5. Calculate the factor C/C1 ? – ΔG*/kT) to be 1)
a) 0.5
b) 1.2
c) 0.25
d) 1.25
Answer: a
Explanation: The relative magnitudes of the heterogeneous and homogeneous volume nucleation rate is given by the equation * – ΔG*/kT). So in this case the nucleation rate is 0.5 which itself is the factor C/C1.
11. At very small driving forces, when activation energy barriers for nucleation are high, the highest nucleation rates will be produced by grain corner nucleation.
a) False
b) True
Answer: b
Explanation: The driving force is the factor which determines whether the type of site gives the highest volume nucleation rate or not. At very small driving forces, when activation energy barriers for nucleation are high, the highest nucleation rates will be produced by grain corner nucleation.
12. A coherent nucleus with a negative misfit can reduce the critical volume free energy by forming a region of _________
a) Compressive strain
b) Compressive stress
c) Tensile strain
d) Tensile stress
Answer: a
Explanation: This directly implies that the smaller volume than the matrix can reduce its ΔG* by forming a region of compressive strain above an edge dislocation, whereas if the misfit is positive it is energetically favourable for it to form below the dislocation.
13. In FCC crystals the a/2*[110] unit dislocations can dissociate to produce a ribbon of stacking fault. Which among the following is as an example of the same?
a) a/2*[110]->a/6*[121] + a/6[211]
b) a/2*[110]->a/6*[101] + a/6[201]
c) a/2*[110]->a/6*[821] + a/6[911]
d) a/2*[110]->a/6*[161] + a/6[221]
Answer: a
Explanation: a/2*[110]->a/6*[121] + a/6*[211] is an example and this gives a stacking fault on separated by two Shockley partials. Since the stacking fault is in effect four close-packed layers of hcp crystal, it can act as a very potent nucleation site for an hcp precipitate.
14. Which among the following curve represent the grain boundary?
phase-transformation-questions-answers-heterogeneous-nucleation-q14
a) 1
b) 2
c) 3
d) Cannot be predicted
Answer: a
Explanation: Curve one represents the grain boundary and this can be determined from the effect of angle on the activation energy for grain boundary nucleation relative to homogeneous nucleation. Curve 2 and 3 represents the grain edges and grain corners respectively.
15. Nucleation on dislocations may also be assisted by solute segregation which can raise the composition of the matrix to nearer that of the precipitate.
a) True
b) False
Answer: a
Explanation: Nucleation on dislocations may also be assisted by solute segregation which can raise the composition of the matrix to nearer that of the precipitate. The dislocation can also assist in growth of an embryo beyond the critical size by providing a diffusion pipe with a lower ΔG.
