Chapter 7 : Diffusion in Solids

Example 7.1 Page No : 207

In [1]:
import math 
from scipy.special import erfinv

# Variables
D = 1.28*10**(-11);			#diffusion coefficient of carbon in given steel in m2/s
c_s = 0.9;			#Surface concentration of diffusion element in the surface
c_o = 0.2;			#Initial uniform concentration of the element in the solid
c_x = 0.4;			#Concentration of the diffusing element at a distance x from thesurface
x = 0.5*10**(-3);			#depth from the surface in m

# Calculation
#(c_s-c_x)/(c_s-c_o) = erf(x/(2*(D*t)**(1/2)))
t = (x/(2*erfinv((c_s-c_x)/(c_s-c_o))*D**(1./2)))**2;			#time required for carburization(in sec)

# Results
print 'time required for carburization in %.1f min'%(t/60)

time required for carburization in 142.8 min

Example 7.2 Page No : 208

In [1]:
			
import math 
from scipy.special import erfinv

# Variables
D = 4*10**(-17);			#diffusion coefficient of carbon in given steel in m2/s
c_s = 3*10**26;			#Surface concentration of boron atoms in the surface
c_1 = 0;			#Initial uniform concentration of the element in the solid
c_x = 10**23;			#Concentration of the diffusing element at a distance x from thesurface
x = 2*10**(-6);			#depth from the surface in m

# Calculation and Results
#(c_s-c_x)/(c_s-c_1) = erf(x/(2*(D*t)**(1/2)))
a = (erfinv((c_s-c_x)/(c_s-c_1)));
print ' C1  = ',a
t = (x**2/(D*4*(2.55)**2));			#time required to get a boron content of 1023 atoms per m3 at a depth of 2 micro meter
print 'time required to get a boron content of 1023 atoms per m3 at a depth of 2 micro meter is = %.0f sec'%t

 C1  =  0.0
time required to get a boron content of 1023 atoms per m3 at a depth of 2 micro meter is = 3845 sec

Example 7.3 Page No : 208

In [2]:
import math 

# Variables
t_1 = 736.;			#Temperature in °C
t_2 = 782.;			#Temperature in °C
T_1 = t_1+273;			#Temperature in K
T_2 = t_2+273;			#Temperature in K
D_1 = 2.*10**(-13);			#Coefficient of diffusion at T_1 (in m2/s)
D_2 = 5.*10**(-13);			#Coefficient of diffusion at T_2 (in m2/s)
k = 1.38*10**(-23);			#in J/K

# Calculation and Results
#math.log(d_1) = math.log(d_o)-E/(k*T_1)
#math.log(d_2) = math.log(d_o)-E/(k*T_2)
E = (math.log(D_1)-math.log(D_2))/((1/(k*T_1))-(1/(k*T_2)));			#
print 'activation energy = %.2e J'%-E
D_o = 2.*10**(-13)/math.exp(E/(k*T_1));
print 'constant of the equation = %.2e m2/s'%D_o
t_4 = 500.;			#Temperature in °C
T_4 = t_4+273;			#Temperature in °K
D_4 = D_o*math.exp(E/(k*T_4));			#diffusion coefficient at 500°C
print 'diffusion coefficient at 500°C = %.2e m2/s'%D_4

# rounding off error

activation energy = 2.93e-19 J
constant of the equation = 2.68e-04 m2/s
diffusion coefficient at 500°C = 3.27e-16 m2/s

Example 7.4 Page No : 210

In [5]:
			
import math 

# Variables
D_500 = 4.8*10**(-14);			#Diffusion coefficient for copper in aluminimum at 500*C(in m**2/s)
D_600 = 5.3*10**(-13);			#Diffusion coefficient for copper in aluminimum at 600*C(in m**2/s)
t_600 = 10;			#time of diffussion at 600*C(in Hours)

# Calculation
#D_500*t_500 = D_600*t_600
t_500 = D_600*t_600/D_500;			#time of diffussion at 500*C

# Results
print 'Time at 500*C that will produce the same diffusion as in 600*C in %.1f Hours'%t_500

Time at 500*C that will produce the same diffusion as in 600*C in 110.4 Hours