# Chapter 9 Principles of Solidification¶

## Example 9_1 pgno:333¶

In [1]:
from math import pi
# Initialisation of Variables
deltaT=236;#Typical Undercooling for HomogeneousNucleation from the table 9-1 for cooper
Tm=1358;#Freezing Temperature from the table 9-1 for cooper in degree celsius
deltaH=1628;# Latent Heat of Fusion from the table 9-1 for cooper in J/cm**3
sigma1=177*10**-7;#Solid-Liquid Interfacial Energyfrom the table 9-1 for cooper in J/cm**2
a0=3.615*10**-8;#The lattice parameter for FCC copper in cm
#CALCULATIONS
r=(2*sigma1*Tm)/(deltaH*deltaT);#Critical Radius of copper in cm
V=a0**3;#Volume of FCC unit cell of copper in cm**3
V2=(4./3.)*pi*r**3;#Critical volume of FCC copper
N=V2/V#The number of unit cells in the critical nucleus
Nc=4*round(N);#Since there are four atoms in each unit cell of FCC metals
print "Critical Radius of copper in cm:",r*10**8
print "Volume of FCC unit cell of copper in cm**3:",V
print "Critical volume of FCC copper :",V2
print "The number of unit cells in the critical nucleus :",round(N)
print "Since there are four atoms in each unit cell of FCC metals:",Nc

Critical Radius of copper in cm: 12.5122850123
Volume of FCC unit cell of copper in cm**3: 4.7241633375e-23
Critical volume of FCC copper : 8.2053761484e-21
The number of unit cells in the critical nucleus : 174.0
Since there are four atoms in each unit cell of FCC metals: 696.0


## Example 9_2 pgno:339¶

In [2]:
from math import pi
#page 255
# Initialisation of Variables
d=18.;#Diameter of the casting in in
x=2.;#Thickness  of the casting in in
B=22.#Mold constant of casting
V=(pi/4.)*d**2;#Volume of the casting in in**3
A=2*(pi/4.)*d**2+pi*d*x;#The surface area of the casting in contact with the mold
x=(0.708*A)/V
print "The thickness in inches=",round(x,2),
#the diffrence in asnwer is due to round foo error

The thickness in inches= 1.73


## Example 9_4 pgno:342¶

In [10]:
#page 250

# Initialisation of Variables
d=5#inches
t1=5#mins
t2=20#mins
d2=1.5#inches
#CALCULATIONS
ksol=0.447#inches/mts**0.5
c1=0.5;
t=((d-3+c1)/ksol)**2
print "the total solidification time is ",round(t,2)#mins

the total solidification time is  31.28


## Example 9_5 pgno:342¶

In [1]:
#page 250
from math import pi

# Initialisation of Variables
l=12 #inchs
w=8#inchs
ts=40000#psi
mc=45#mins/in**2
x=1
v=8*12*x
a=2*8*12+2*x*12
#by solving the two equations we get
x=64/82.67;
print round(x,2)

0.77