# Phase Transformation Questions and Answers – Solidification – Growth of a Pure Solid

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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

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(ΔTi/Tm)
b) ΔG= L(Tm/ΔTi)
c) ΔG= ΔTi/(Tm*L)
d) ΔG= Tm/(L*ΔTi)

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(ΔTi/Tm).

3. The equation related to the net rate of solidification is given as_______ (K has the properties of boundary mobility)
a) R = K*ΔTi
b) R = K*ΔTi/Tm
c) R = K*Tm
d) R = K*(Tm/ΔTi)

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 (ΔTi) 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 (k/ΔTi)
c) 1/ΔTi
d) Exp (-k/ΔTi)

Explanation: The surface nucleation rate is proportional to Exp (-k/ΔTi). 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

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? (Assume the driving force for solidification as 45kJ/kg?
a) 150K
b) 600K
c) 900K
d) 450K

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 (ΔG/L)*(Tm). Substituting the values we get 600*45/30 = 900K.

7. Calculate the extent of interfacial undercooling if the value of k (Mobility) is given as 0.05(m/ (sec*Kelvin)) and the rate of solidification(R) is given as 5m/sec?
a) 100K
b) 1000K
c) 200K
d) 2000K

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_______ (K material constant)
a) R = K(ΔTi)
b) R = K(ΔTi)3
c) R = K(ΔTi)2
d) R = K/(ΔTi)2

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

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? (Kl, Ks are respective thermal conductivities of liquids and solids, L the latent heat of fusion per unit volume, v growth rate).
a) KsTs = Kl*Tl /(v*L)
b) KsTs = Kl*Tl – v*L
c) KsTs = Kl/(Tl +v*L)
d) KsTs = Kl*Tl +v*L

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______ (Latent heat of fusion per unit volume)
a) ΔTr = 2γTm/(L*r)
b) ΔTr = 2Tm/(L*r)
c) ΔTr = 2Tm/(L*γ*r)
d) ΔTr = 2γTm/L

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/(L*r). The minimum possible radius of curvature of the tip is when ΔTr equals the total undercooling ΔTo = Tm – T(infinity). This is just the critical nucleus radius r* given by (2γTm/ L*ΔTo).

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______

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 (cooler than melt), 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

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? (Assume the rate of growth as 5m/s)
a) 0
b) 5
c) 25
d) 100

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 ______ (r* is the critical radius)
a) R = 2r*
b) R = 3r*
c) R = r*
d) R = 4r*

Explanation: V= K/L((1/R)-(r*/R2)), 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.

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