This set of Engineering Materials Science "Numerical Problems Questions & Answers" are based on "PHASE TRANSFORMATION"
PHASE TRANSFORMATION
Numerical Problems
Hello reader, here you can find some Numerical Problems with answers. But before reading this I will suggest you take a practice test on these questions given below. so you can understand your strength on this topic.
Click here to go Practice Test section on this topic namely Phase Transformation.
But if you want to read this topic first and then practice it then read it carefully. The Numerical Problems based on Phase Transformation are given below:
1) Calculate the ratio of critical free energy of heterogeneous nucleation to homogeneous nucleation.
Given,
Solid-liquid interfacial energy = 0.5 J.m-2
Interface-liquid interfacial energy = 0.3 J.m-2
Interface - solid interfacial energy = 0.3 J.m-2
Correct Answer is "C"
We know that, for heterogeneous nucleation,
γSL cosθ + γIL = γIL
Or, cosθ = (γIL-γIL )/( γSL ) = (0.3-0.3 )/0.5 = 0
Again, we know critical free energy of heterogeneous nucleation = [(4πγIL3)/(3(∆g)2 )] (2 - 3cosθ + cos3θ )
Critical free energy of heterogeneous nucleation = (16πγ3)/(3(∆g)2 )
(critical free energy of heterogeneous nucleation )/(critical free energy of homogeneous nucleation )
= 1/4 × (2 - 3cosθ + cos3θ )
= 1/4 × (2 – 0 + 0 ) = 0.5
γSL cosθ + γIL = γIL
Or, cosθ = (γIL-γIL )/( γSL ) = (0.3-0.3 )/0.5 = 0
Again, we know critical free energy of heterogeneous nucleation = [(4πγIL3)/(3(∆g)2 )] (2 - 3cosθ + cos3θ )
Critical free energy of heterogeneous nucleation = (16πγ3)/(3(∆g)2 )
(critical free energy of heterogeneous nucleation )/(critical free energy of homogeneous nucleation )
= 1/4 × (2 - 3cosθ + cos3θ )
= 1/4 × (2 – 0 + 0 ) = 0.5
2) The critical free energy for homogeneous nucleation is equal to that of heterogeneous nucleation if the contact angle is
Correct Answer is "D"
(critical free energy of heterogeneous nucleation )/(critical free energy of homogeneous nucleation )
= 1/4 × (2 - 3cosθ + cos3θ ) = 1
Or, -3cosθ + cos3θ = 2
θ = 180°
= 1/4 × (2 - 3cosθ + cos3θ ) = 1
Or, -3cosθ + cos3θ = 2
θ = 180°
3) Calculate the undercooling that is required for a liquid to crystal transformation in copper. The enthalpy of fusion for copper is 1.88 G Jm-3. Appreciable nucleation occurs when the free energy of the critical nucleus is 1.5 × 10-19 J. The liquid-crystal interfacial energy is 0.144 J.m-2. The melting point of copper is 1085°C
Correct Answer is "A"
We know that,
∆G* = [(16πγ3Tm2)/(3∆H2 )] [1/(Tm- T)]2
Or, (Tm- T)2 = [16 × 3.14 × (0.144)3 × (1085+273)2 ]/[3 ×(1.88 ×109)2 ×(1.5 ×10-19)] = 173943.59
Or, Tm- T = √(173943.59) = 417K
∆G* = [(16πγ3Tm2)/(3∆H2 )] [1/(Tm- T)]2
Or, (Tm- T)2 = [16 × 3.14 × (0.144)3 × (1085+273)2 ]/[3 ×(1.88 ×109)2 ×(1.5 ×10-19)] = 173943.59
Or, Tm- T = √(173943.59) = 417K
4) Calculate the critical radius of the tin nucleus in nm during solidification of liquid tin at 169°C. The enthalpy of fusion of tin is 0.42 GJm-3. The liquid-crystal interfacial energy is 0.055 Jm-2. The melting point of tin is 231.9°C.
Correct Answer is "B"
We know that,
r* = - [2γTm/∆H ] [1/(Tm- T)]
= - [2 ×0.055 ×(231.9+273)/(-0.42 ×109 ] × [1/(231.9-169)]
= 2.1 × 10-9 m = 2.1 nm
r* = - [2γTm/∆H ] [1/(Tm- T)]
= - [2 ×0.055 ×(231.9+273)/(-0.42 ×109 ] × [1/(231.9-169)]
= 2.1 × 10-9 m = 2.1 nm
5) If the interfacial energy increases by 15%, the homogeneous nucleation barrier for a spherical particle increases by
Correct Answer is "C"
The nucleation barrier for a spherical particle is the activation energy of the nucleation
We know that, ∆G* ∝ γ3
So, ∆G*2/∆G*1 = [γ23/γ13] = (100+15)3/1003 = 1.52
We know that, ∆G* ∝ γ3
So, ∆G*2/∆G*1 = [γ23/γ13] = (100+15)3/1003 = 1.52
6) If the product phase completely wets a nucleating agent, the nucleation barrier as a fraction of the homogeneous barrier is
Correct Answer is "A"
If the product phase completely wets a nucleating agent then the contact angle is 0°
So, (critical free energy of heterogeneous nucleation )/(critical free energy of homogeneous nucleation )
= 1/4 × (2 - 3cosθ + cos3θ )
= 1/4 × [2 – 3(1) + (1) ]
= 0
So, (critical free energy of heterogeneous nucleation )/(critical free energy of homogeneous nucleation )
= 1/4 × (2 - 3cosθ + cos3θ )
= 1/4 × [2 – 3(1) + (1) ]
= 0
7) When the contact angle is 60°, the heterogeneous nucleation barrier expressed as a fraction of the homogeneous barrier is
Correct Answer is "D"
(critical free energy of heterogeneous nucleation )/(critical free energy of homogeneous nucleation )
= 1/4 × (2 - 3cosθ + cos3θ )
= 1/4 × [2 - 3cos60° + cos360° ] = 5/64
= 1/4 × (2 - 3cosθ + cos3θ )
= 1/4 × [2 - 3cos60° + cos360° ] = 5/64
8) For some transformation having kinetics that obeys the Avrami equation, the parameter n is known to have a value of 2.5. If the reaction is 25% complete after 125 s, how long (total time) will it take the transformation to go to 70% completion?
Correct Answer is "B"
From Avrami equation,
y = 1 – exp (-ktn)
According to the given problem,
0.25 = 1 – exp (-k × 1252.5)
Or, ln 0.75 = -k × 174692.8
Or, k = 1.646 × 10-6
Again,
0.7 = 1 – exp (-1.646 × 10-6 × t2.5)
731453.7 = t2.5
731453.7(1/2.5) = 221.6s
y = 1 – exp (-ktn)
According to the given problem,
0.25 = 1 – exp (-k × 1252.5)
Or, ln 0.75 = -k × 174692.8
Or, k = 1.646 × 10-6
Again,
0.7 = 1 – exp (-1.646 × 10-6 × t2.5)
731453.7 = t2.5
731453.7(1/2.5) = 221.6s
9) For the solidification of iron, values for the latent heat of fusion and surface free energy are –1.85 × 109 J/m3 and 0.204 J/m2, respectively. supercooling value 295 K. Now calculate the number of atoms found in a nucleus of critical size. Assume a lattice parameter of 0.292 nm for solid iron at its melting temperature. The melting point of iron is 1538°C.
Correct Answer is "B"
We know that,
Critical radius, r* = - [2γTm/∆H][ 1/(Tm- T)]
= - (2 ×0.204 ×(1538+273))/(-1.85 ×109 ) × 1/295
= 1.35 × 10-9 m
=1.35 nm
Now, Unit cell per particle
= Critical nucleus volume / unit cell volume
= (4/3) π r3 / a3
= (4/3) × 3.14 × 1.353 /0.2923
=414
At 1538°C iron has a BCC structure which contains 2 equivalence atoms per critical nucleus
So the total no of atom per critical nucleus = 414 × 2 = 828
Critical radius, r* = - [2γTm/∆H][ 1/(Tm- T)]
= - (2 ×0.204 ×(1538+273))/(-1.85 ×109 ) × 1/295
= 1.35 × 10-9 m
=1.35 nm
Now, Unit cell per particle
= Critical nucleus volume / unit cell volume
= (4/3) π r3 / a3
= (4/3) × 3.14 × 1.353 /0.2923
=414
At 1538°C iron has a BCC structure which contains 2 equivalence atoms per critical nucleus
So the total no of atom per critical nucleus = 414 × 2 = 828
10) Compute the rate of some reaction in min-1 that obeys Avrami kinetics, assuming that the constants n and k have values of 4.0 and 7 × 10-4, respectively, for time expressed in seconds.
Correct Answer is "C"
From Avrami equation,
y = 1 – exp (-ktn)
According to the given problem,
0.5 = 1 – exp (-7 × 10-4 × t0.54)
t0.5 = 5.6
Rate = 1/t0.5 1/5.6 = 0.178 s-1 = 10.68 min-1
y = 1 – exp (-ktn)
According to the given problem,
0.5 = 1 – exp (-7 × 10-4 × t0.54)
t0.5 = 5.6
Rate = 1/t0.5 1/5.6 = 0.178 s-1 = 10.68 min-1
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