Chapter 10 :Phase Transformation In Metals

Example 10.1 Page no 318

In [11]:
Hf=-1.16*10**9     # in J/m**3  latent heat of fusion
Y=0.132            # in J/m**2  Surface energy
Tm=1064.0+273.0        # in K  Melting point of gold

import math
T=Tm-230.0           # in K  230 is supercooling value
r=-2*Y*Tm/(Hf*(Tm-T))
G=16*math.pi*Y**3*Tm**2/(3*Hf**2*(Tm-T)**2)

a=0.413*10**-9  # in m  Unit Cell edge length
u_c=4*math.pi*r**3/(3*a**3)
n=4            #In FCC . no of atoms in per unit cell
U_c=int(u_c)*n

print"(a)Critical Radius is ",round(r/10**-9,2),"nm"
print"Activation free energy is ",round(G,21),"J"
print"(b)Total no. of atoms per critical nucleus are ",U_c

(a)Critical Radius is  1.32 nm
Activation free energy is  9.68e-19 J
(b)Total no. of atoms per critical nucleus are  548

Example 10.2 Page no 335

In [13]:
n=5.0
y=0.3
t=100.0          #in min

k=-math.log(1-y)/t**n
thalf=(-math.log(1-0.5)/k)**(1/n)
rate=1/thalf

print"Rate is ",rate,"min**-1"

Rate is  0.00875567087531 min**-1
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