Ch-15, Phase transformation

example-15.1 page no-461

In [6]:
from __future__ import division
from math import log, exp, pi
#given
#reaction comlete in 500 mins at 10 degree celcius and in 1 min at 80 degrees celcius
#so
t1=1  #min
t2=500  #min
T1=273+80  #kelvin
T2=273+10  #Kelvin
#gas constant 
R=8.314
#as we know that phie=1/t*
#so
#log(phie1/phie2)=log(t2/t1)
#so we get
#log(t2/t1)=Ea/2.303*(1/T1-1/T2)
#so Ea is given by
Ea=2.303*log(t2/t1)*(1/(1/T1-1/T2))*R
#to find the reaction completion time when temperature is 40 degrees celcius
#so
T3=273+40   #Kelvin
t3=t1*exp(Ea/2.303/R*(1/T1-1/T3))  #min
print " the reaction completion tym at 40 degree celcius is %0.2f min"%(t3)
 the reaction completion tym at 40 degree celcius is 24.80 min

example-15.2 page no-463

In [7]:
#given
#temp
T=983+273  #K
Tm=1083+273  #K
deltaT=Tm-T  #K
#given that
#latent heat of fusion of copper
deltaHm=1.88*10**9  #J/m**3
#interface energy/unit area
gammasL=0.144  #J/m**2
#change in free energy of vapour
deltaGv=deltaHm*deltaT/Tm    #J/m**3
#critical radius of copper during solidification
#for notation only we are using r*=R
R=2*gammasL/deltaGv*10**10  #A
print " the critical radius of copper during solidification is %0.3f A"%(R)
 the critical radius of copper during solidification is 20.773 A

example-15.3 page no-463

In [9]:
#given
#latent heat of fusion of pure gold
deltaHf=-1.16*10**9  #J/m**2
#surface free energy
gama=0.132  #J/m**2
#melting point of gold
Tm=1064+273  #K
#supercooling value
deltaT=230  #K
#critical radius is given by R intead of r* just for notation
R=(-2*gama*Tm/deltaHf)*(1/deltaT)*10**9  #nm
print " critical radius of pure gold at 230 degree celcius is %0.3f nm"%(R)
#change in free energy of vapours is
deltaGv=deltaHf*deltaT/Tm  #J/m**3
#activation free energy
#for notation deltaG* we  use deltaG
deltaG=16*(pi)*(gama)**3/3/(deltaGv)**2  #J
print " the activation free energy is %0.3e J"%(deltaG)
 critical radius of pure gold at 230 degree celcius is 1.323 nm
 the activation free energy is 9.677e-19 J

example-15.5 page no-472

In [10]:
#given
#diameter of hard inert particle
D=2  #m
#average centre to centre distance between the particles
l=20*10**-6   #m becaus 1 micro m=10**-6 m
#shear modulus of copper
G=41*10**9  #Pa
#burgers vector
b=0.64*10**-9  #m
#contribution of these particles on the yeild strength of material
tau=G*b/l/10**6  #MPa
print " the contribution of these particles on the yeild strength of material is %0.3f MPa"%(tau)
 the contribution of these particles on the yeild strength of material is 1.312 MPa

example-15.7 page no-474

In [12]:
#given
#heat for recovery of yeild point in zinc
Q=83.14*10**3  #J/mole
#gas constant
R=8.314*10**3  #J/mol K
#temperature
T1=0+273  #K
T2=27+273  #K
#recovery time at 0 degree celcius
t2=5  #min
#the recovery time for the two different temp is given by
#1/t1=A*exp(-Q/R/T1)   --------------(1)
#1/t2=A*exp(-Q/R/T2)  -----------(2)
#taking ratio of (1) to (2)
#we get
t1=t2*exp(Q/R*(1/T1-1/T2))  #min (there is calculation mistake I myself have checked from calculator too)
print " the recovery time at 27 degree celcius is %0.2f min"%(t2)
 the recovery time at 27 degree celcius is 5.00 min