## Exa15.1 : : Page-652 (2011)
#find Total leakage factor
import math
N_0_235 = 1.; ## Number of uranium atom
N_0_c = 10**5; ## Number of graphite atoms per uranium atom
sigma_a_235 = 698.; ## Absorption cross section for uranium, barns
sigma_a_c = 0.003; ## Absorption cross section for graphite, barns
f = N_0_235*sigma_a_235/(N_0_235*sigma_a_235+N_0_c*sigma_a_c ); ## Thermal utilization factor
eta = 2.08; ## Number of fast fission neutron produced
k_inf = eta*f; ## Multiplication factor
L_m = 0.54; ## Material length, metre
L_sqr = ((L_m)**2.*(1.-f)); ## diffusion length, metre
tau = 0.0364; ## Age of the neutron
B_sqr = 3.27; ## Geometrical buckling
k_eff = round (k_inf*math.exp(-tau*B_sqr)/(1+L_sqr*B_sqr)); ## Effective multiplication factor
N_lf = k_eff/k_inf; ## Non leakage factor
lf = (1-N_lf)*100.; ## Leakage factor, percent
print'%s %.2f %s'%("\n Total leakage factor = ",lf," percent")
## Result
## Total leakage factor = 31.3 percent
## Exa15.2 : : Page-652 (2011)
#find Neutron multiplication factor
import math
N_m = 50.; ## Number of molecules of heavy water per uranium molecule
N_u = 1.; ## Number of uranium molecules
sigma_a_u = 7.68; ## Absorption cross section for uranium, barns
sigma_s_u = 8.3; ## Scattered cross section for uranium, barns
sigma_a_D = 0.00092; ## Absorption cross section for heavy water, barns
sigma_s_D = 10.6; ## Scattered cross section for uranium, barns
f = N_u*sigma_a_u/(N_u*sigma_a_u+N_m*sigma_a_D ); ## Thermal utilization factor
zeta = 0.570; ## Average number of collisions
N_0 = N_u*139./140.; ## Number of U-238 atoms per unit volume
sigma_s = N_m/N_0*sigma_s_D; ## Scattered cross section, barns
sigma_a_eff = 3.85*(sigma_s/N_0)**0.415; ## Effective absorption cross section, barns
p = math.exp(-sigma_a_eff/sigma_s); ## Resonance escape probablity
eps = 1.; ## Fast fission factor
eta = 1.34; ## Number of fast fission neutron produced
k_inf = eps*eta*p*f; ## Effective multiplication factor
print'%s %.2f %s'%("\nNeutron multiplication factor = ", k_inf,"");
## Result
## Neutron multiplication factor = 1.2
## Exa15.3 : : Page-652 (2011)
#find The required multiplication factor
import math
## For graphite
sigma_a_g = 0.0032; ## Absorption cross section for graphite, barns
sigma_s_g = 4.8; ## Scattered cross section for graphite, barns
zeta = 0.158; ## Average number of collisions
N_m = 50.; ## Number of molecules of graphite per uranium molecule
## For uranium
sigma_f = 590.; ## Fissioning cross section, barns
sigma_a_u = 698.; ## Absorption cross section for U-235, barns
sigma_a_238 = 2.75; ## Absorption cross section for U-238, barns
v = 2.46; ## Number of fast neutrons emitted
N_u = 1 ## Number of uranium atoms
f = N_u*sigma_a_u/(N_u*sigma_a_u+N_m*sigma_a_g ); ## Thermal utilization factor
N_0 = N_u*(75./76.); ## Number of U-238 atoms per unit volume
sigma_s = N_m*76./75.*sigma_s_g/N_u; ## Scattered cross section, barns
sigma_eff = 3.85*(sigma_s/N_0)**0.415; ## Effective cross section, barns
p = math.exp(-sigma_eff/sigma_s); ## Resonance escape probability, barns
eps = 1.; ## Fast fission factor
eta = 1.34; ## Number of fast fission neutron produced
k_inf = eps*eta*p*f; ## Multiplication factor
print'%s %.2f %s'%("\nThe required multiplication factor = ", k_inf,"");
## Result
## The required multiplication factor = 1.1
## Exa15.4 : : Page-653 (2011)
#find The ratio of number of uranium atoms to graphite atoms
import math
eta = 2.07; ## Number of fast fission neutron produced
x = 1./(eta-1.);
sigma_a_u = 687.; ## Absorption cross section for uranium, barns
sigma_a_g = 0.0045; ## Absorption cross section for graphite, barns
N_ratio = x*sigma_a_g/sigma_a_u; ## Ratio of number of uranium atoms to graphite atoms
print'%s %.2e %s'%("\nThe ratio of number of uranium atoms to graphite atoms = ", N_ratio,"");
## Result
## The ratio of number of uranium atoms to graphite atoms = 6.12e-006
## Exa15.5 : : Page-653 (2011)
import math
#find The multiplication factor for LOPO reactor
f = 0.754; ## Thermal utilization factor
sigma_s_o = 4.2; ## Scattered cross section for oxygen, barns
sigma_s_H = 20.; ## Scattered cross section for hydrogen, barns
N_O = 879.25; ## Number of oxygen atoms
N_238 = 14.19; ## Number of uranium atoms
N_H = 1573.; ## Number of hydrogen atoms
sigma_s = N_O/N_238*sigma_s_o+N_H/N_238*sigma_s_H; ## Scattered cross section, barns
N_0 = 14.19; ## Number of U-238 per unit volume
zeta_o = 0.120; ## Number of collision for oxygen
zeta_H = 1.; ## Number of collision for hydrogen
sigma_eff = (N_0/(zeta_o*sigma_s_o*N_O+zeta_H*sigma_s_H*N_H )); ## Effective cross section, barns
p = math.exp(-sigma_eff/sigma_s); ## Resonance escape probablity
eta = 2.08; ## Number of fission neutron produced.
eps = 1; ## Fission factor
K_inf = eps*eta*p*f; ## Multiplication factor
print'%s %.2f %s'%("\nThe multiplication factor for LOPO reactor = ", K_inf,"");
## Result
## The multiplication factor for LOPO reactor = 1.6
## Exa15.6 : : Page-654 (2011)
#find The required controlled cross section
import math
r = 35; ## Radius of the reactor, centi metre
B_sqr = (math.pi/r)**2; ## Geometrical buckling, per square centi metre
D = 0.220; ## Diffusion coefficient, centi metre
sigma_a_f = 0.057; ## Rate of absorption of thermal neutrons
v = 2.5; ## Number of fast neutrons emitted
tau = 50.; ## Age of the neutron
sigma_f = 0.048; ## Rate of fission
sigma_a_c = -1/(1+tau*B_sqr)*(-v*sigma_f+sigma_a_f+B_sqr*D+tau*B_sqr*sigma_a_f); ## Controlled cross section
print'%s %.2f %s'%("\nThe required controlled cross section = ", sigma_a_c,"");
## Result
## The required controlled cross section = 0.0273
## Exa15.7 : : Page-655 (2011)
#find side of the cubical reactor nd critical radius of the reactor
import math
B_sqr = 65.; ## Geometrical buckling
a = math.sqrt(3*math.pi**2/B_sqr)*100.; ## Side of the cubical reactor, centi metre
R = round(math.pi/math.sqrt(B_sqr)*100.); ## Radius of the cubical reactor,centi metre
print'%s %.2f %s %.2f %s '%("\nThe side of the cubical reactor =",a," cm"and" \nThe critical radius of the reactor =",R," cm");
## Result
## The side of the cubical reactor = 67.5 cm
## The critical radius of the reactor = 39 cm
## Exa15.8 : : Page-655 (2011)
#find The critical volume of the reactor
import math
sigma_a_u = 698.; ## Absorption cross section for uranium, barns
sigma_a_M = 0.00092; ## Absorption cross section for heavy water, barns
N_m = 10**5; ## Number of atoms of heavy water
N_u = 1.; ## Number of atoms of uranium
f = sigma_a_u/(sigma_a_u+sigma_a_M*N_m/N_u); ## Thermal utilization factor
eta = 2.08; ## Number of fast fission neutron produced
k_inf = eta*f; ## Multiplication factor
L_m_sqr = 1.70; ## Material length, metre
L_sqr = L_m_sqr*(1-f); ## Diffusion length, metre
B_sqr = 1.819/0.30381*math.exp(-1/12.)-1./0.3038; ## Geometrical buckling, per square metre
V_c = 120./(B_sqr*math.sqrt(B_sqr)); ## Volume of the reactor, cubic metre
print'%s %.2f %s'%("\nThe critical volume of the reactor = ",V_c," cubic metre");
## Result
## The critical volume of the reactor = 36.4 cubic metre