# Phase Transformation Questions and Answers – Kinetics of Grain Growth

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This set of Phase Transformation Multiple Choice Questions & Answers (MCQs) 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

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_________ (γ-surface energy, D- mean grain diameter, M-mobility)
a) v’ = k*M*(2γ/D)
b) v’ = k + M*(2γ/D)
c) v’ = k*M*(2D/γ)
d) v’ = k*M*(γ/D)

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*(2γ/D), 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 10mm3J-1s-1 respectively? (Assume the mean size of the grain at t=0 as 3mm)
a) 109mm
b) 1009mm
c) 10.9m
d) 9mm

Explanation: The mean diameter at time t is given by the equation D = D2’ + 4Mγt, where M represents the mobility, ‘’t’’ the time at which it happens, γ the grain boundary energy and D’ the initial mean diameter.
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4. Experimentally one can find the grain size as function of time. If the K value is given as 3mm/sec (temperature dependent proportionality constant) and n is 0.2. Find the grain diameter at t =2sec?
a) 8.76mm
b) 2.67mm
c) 3.45mm
d) 4.56mm

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

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

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 (Virtually) 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

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

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 (2πrγcosΘ)*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

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/πr2
b) 5f/πr2
c) 3f/πr2
d) 3f/2πr2

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

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

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

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? (Approximately)
a) 1.2
b) 1.8
c) 2.4
d) 3.6

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.

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