## Example 5.2
import math
#calculate the
## Given data
## 1 barn = 10^(-24) cm^2
sigma_s = 4.8*10**(-24) ## Scattering cross section of carbon in cm^2
A_C = 12.; ## Atomic Mass number for carbon-12
E = 1.; ## Energy of carbon-12 atom in eV
## Using the data given in Table II.3, for carbon (graphite) at energy 1 eV
N = 0.08023*10**(24); ## Atom density in terms of atom/cm^3
mu_bar = 2./(3.*A_C); ## Average value of the cosine of the angle at which neutrons are scattered in the med/ium
SIGMA_s = N*sigma_s ## Macroscopic scattering cross section of carbon-12
## Calculation
D = 1/(3.*SIGMA_s*(1.-mu_bar));
## Result
print'%s %.2f %s'%('\n Diffusion coefficeint of graphite at 1 eV = ',D,' cm \n');
## Example 5.5
import math
#calculate the
## Given data
phi1 = 6*10**(14); ## Neutron flux of Group 1
phi2 = 1*10**(15); ## Neutron flux of Group 2
phi3 = 3*10**(15); ## Neutron flux of Group 3
## 1.
## Using the data given in Table II.3, for atom density of sodium
N = 0.02541*10**(24); ## Atom density in terms of atom/cm^3
## Using the data given for sigmay (Microscopic radiative capture cross section) in Table II.3,
## 1 barn = 10^(-24) cm^2
sigmay1 = 0.0005*10**(-24); ## Microscopic gamma cross section of Group 1
sigmay2 = 0.001*10**(-24); ## Microscopic gamma cross section of Group 2
sigmay3 = 0.001*10**(-24); ## Microscopic gamma cross section of Group 3
## Calculation
F_a = N*((sigmay1*phi1)+(sigmay2*phi2)+(sigmay3*phi3));
## Result
print'%s %.2e %s'%('\n Total absorption rate for three groups = ',F_a,' neutrons/cm^3-sec \n');
## 2.
## Calculation
sigmag_12 = 0.24*10**(-24); ## Microscopic scattereing cross section of neutrons from Group 1 to Group 2
F_12 = N*sigmag_12*phi1;
## Result
print'%s %.2e %s'%('\n Neutron scattering rate from the first to second group = ',F_12,' neutrons/cm^3-sec \n');
## Example 5.6
import math
#calculate the
## Given data
S = 10**7; ## Strength of neutron source in neutrons/sec
r = 15.; ## Distance over which neutron flux is to be calculated in cm
## Using the data given in Table 5.2,
L_T = 2.85; ## Thermal diffusion length in cm
D_bar = 0.16; ## Diffusion coefficient in cm
## Calculation
phi_T = S*math.exp(-r/L_T)/(4.*math.pi*D_bar*r);
## Result
print'%s %.2f %s'%('\n Neutron flux = ',phi_T,' neutrons/cm^2-sec \n');
## Example 5.7
import math
#calculate the
## Given data
T_F = 500.; ## Temeperature in Fahrenheit
P = 2000.; ## Pressure in psi
rho = 49.6; ## Density in terms of lb/ft^3
## Converting the given temperature from Fahrenheit to Celsius
T_C = (5./9.)*(T_F-32.);
## Converting the temperature from Celsius to Kelvin scale
T_K = 273.+T_C;
## Using the data given in Table 5.2,
D_bar_0 = 0.16; ## Diffusion coefficient at 293 K
rho_0 = 62.4; ## Density at 293 K in terms of lb/ft^3
L_T2_0 = 8.1; ## Diffusion area at 293 K in cm^2
T_0 = 293.; ## Standard Temperature in kelvin
m = 0.47; ## Material specific constant
## Calculation
D_bar = D_bar_0*(rho_0/rho)*(T_K/T_0)**m;
L_T2 = L_T2_0*(rho_0/rho)**2*(T_K/T_0)**(m+1./2.);
## Result
print'%s %.2f %s'%('\n Diffusion coefficient of ordinary water = ',D_bar,' cm \n');
print'%s %.2f %s'%('\n Diffusion area of ordinary water =',L_T2,' cm^2 \n');