## Example 6.1
import math
#calculate the
## Given data
M_F = 235.; ## Atomic mass of Uranium-235
M_S = 23.; ## Atomic mass of Sodium-23
rho_F_S = 1.; ## Ratio of densities of Uranium fuel to Sodium
## Using the data given in Table 5.2,
sigmaa_S=0.0008; ## Absorption cross section of Sodium
sigmaa_F=1.65; ## Absorption cross section of Uranium
rho_S_F = 100.-rho_F_S;
N_S_F = rho_S_F*(M_F/M_S); ## Ratio of atomic densities of Uranium and Sodium
## Using the data in Table 6.1 for Uranium-235
## The value of average number of neutrons produced for a neutron absorbed n(eta) for Uranium-235 is 2.2
eta = 2.2;
## Calculation
f = 1./(1.+(N_S_F*(sigmaa_S/sigmaa_F)));
k_inf = eta*f;
## Result
print'%s %.2f %s'%('\n Thermal Utilization factor = ',f,' \n');
print'%s %.2f %s'%('\n Infinite Multiplication factor = ',k_inf,' \n');
## Example 6.2
import math
#calculate the
## Given data
R = 50.; ## Radius of reactor core in cm
P = 100*10**6; ## Power level of the reactor in watt
SIGMA_f = 0.0047; ## Macroscopic fission cross section in cm^(-1)
E_R = 3.2*10**(-11); ## Energy released per fission in joules/second
## Using the data in Table 6.2 for spherical geometry
OMEGA = 3.29; ## Measure of the variation of flux in the reactor
## Calculation
phi_max = (math.pi*P)/(4.*E_R*SIGMA_f*R**3);
phi_av = phi_max/OMEGA;
## Result
print'%s %.2e %s'%('\n Maximum flux in the spherical reactor = ',phi_max,' neutrons/cm^2-sec \n');
print'%s %.2e %s'%('\n Average flux in the spherical reactor = ',phi_av,' neutrons/cm^2-sec \n');
## Example 6.3
import math
#calculate the
## Given data
N_F = 0.00395*10**(24); ## Atom density of Plutonium-239 fuel in atom/cm^3
N_S = 0.0234*10**(24); ## Atom density of Sodium-23 in atom/cm^3
## Using the data given in Table 6.1,
## 1 barn = 10^(-24) cm^2
sigmaa_S = 0.0008*10**(-24); ## Microscopic absorption cross section of Sodium in cm^2
sigmaa_F = 2.11*10**(-24); ## Microscopic absorption cross section of Plutonium in cm^2
sigmatr_F = 6.8*10**(-24); ## Microscopic transport cross section of Plutonium
sigmatr_S = 3.3*10**(-24); ## Microscopic transport cross section of Sodium
## The value of average number of neutrons produced for a neutron absorbed n(eta) for Plutonium-239 is 2.61
eta = 2.61;
SIGMAA_S = sigmaa_S*N_S; ## Macroscopic absorption cross section of Sodium in cm^(-1)
SIGMAA_F = sigmaa_F*N_F; ## Macroscopic absorption cross section of Plutonium in cm^(-1)
SIGMAA = SIGMAA_S+SIGMAA_F; ## Total macroscopic absorption cross section in cm^(-1)
SIGMA_tr = (sigmatr_F*N_F)+(sigmatr_S*N_S); ## Macroscopic transport cross section
f = SIGMAA_F/SIGMAA; ## Calculation of Thermal Utilization factor(f)
f = math.ceil(f);
k_inf = eta*f; ## Calculation of Infinite Multiplication factor(k_inf)
D = 1/(3*SIGMA_tr); ## Calculation of Diffusion coefficient
L2 = D/SIGMAA; ## Diffusion area
d = 2.13*D; ## Extrapolated distance
R_ctil = math.pi*math.sqrt(L2/(k_inf-1)); ## Critical Radius for an extrapolated boundary
## Calculation
R_c = R_ctil-d;
## Result
print'%s %.2f %s'%('\n Critical Radius = ',R_c,' cm \n');
## The answer given in the textbook is wrong.
## Example 6.4
import math
#calculate the
## Using the result from Example 6.3
R_c = 48.5; ## Critical Radius for an extrapolated boundary in cm
L2 = 384.; ## Diffusion area in cm^2
## Calculation
P_L = 1./(1.+((math.pi/R_c)**2*L2));
## Result
print'%s %.2f %s'%('\n Nonleakage probability of a fission neutron = ',P_L,' \n');
## Example 6.5
import math
#calculate the
## Given data
R = 100.; ## Radius of a spherical reactor in cm
P = 10**5; ## Power of the reactor in watt
## 1.
## Calculation
B = math.sqrt((math.pi/R)**2);
## Result
print'%s %.2e %s'%(" \n Buckling = ",B," \n");
## 2.
## Using the data from Tables 3.2, 5.2, 5.3 and 6.3
L_TM2 = 3500.; ## Diffusion area of moderator (Sodium) in cm^2
n_T = 2.065; ## Average number of fission neutrons emitted per neutron absorbed
t_TM = 368.; ## Diffusion time of moderator (Sodium) in cm^2
## 1 barn = 10^(-24) cm^2
sigma_aM = 0.0034*10**(-24); ## Microscopic absorption cross section of Sodium in cm^2
sigma_aF = 681*10**(-24); ## Microscopic absorption cross section of Uranium-235 in cm^2
g_a = 0.978; ## Non 1/v factor
M_F = 235.; ## Molecular weight of Uranium-235
M_M = 12.; ## Molecular weight of Carbon-12
Z = (1+B**2*(L_TM2+t_TM))/(n_T-1.-(B**2*t_TM)); ## An intermediate factor
## Calculation
rho_M = 1.6; ## Density of Graphite in g/cm^3
m_M = (4/3.*math.pi*R**3)*rho_M; ## Mass of moderator
## Calculation
m_F = ((Z*sigma_aM*M_F)/(g_a*sigma_aF*M_M))*m_M/1000.;
## Result
print'%s %.2f %s'%("\n Critical mass = ",m_F," kg \n");
## 3.
f = Z/(Z+1.); ## Thermal utilization factor
## Calculation
k_inf = n_T*f;
## Result
print'%s %.2f %s'%('\n Infinite Multiplication factor (k_inf) = ',k_inf,' \n');
## 4.
## Calculation
L_T2 = (1.-f)*L_TM2
## Result
print'%s %.2f %s'%("\n Thermal Diffusion area = ",L_T2," cm^2 \n");
## 5.
E_R = 3.2*10**(-11); ## Energy per fission reaction in joules/second
N_A = 6.02*10**(23); ## Avogadro number (constant)
V = (4/3.*math.pi*R**3); ## Volume of the spherical reactor in cm^3
## Using the data from Tables 3.2
g_fF = 0.976; ## Non 1/v factor Uranium-235 fuel
## Using the data from Tables II.2 for Uranium-235
sigma_f = 582.*10**(-24); ## Microscopic fission cross section for Uranium-235 in cm^2
## Macroscopic fission cross section is calculated as follows
SIGMA_f = m_F*N_A*0.886*g_fF*sigma_f*1000./(V*M_F);
## From Table 6.2, the constant A can be calculated as
A = P/(4.*(R**2)*E_R*SIGMA_f);
## The expression for thermal flux is
print'%s %.2e %s'%(" \n The expression for thermal flux = ",A," math.sin (Br)/r \n");
## The maximum value of thermal flux is given at distance equal to zero
phi_T0 = A*B;
## Result
print'%s %.2e %s'%(" The maximum thermal flux = ",phi_T0," neutrons/cm^2-sec \n");
## There is a slight variation in the values of diffusion area and constant A as compared from the textbook. This is due to approximation of values in textbook.
## Example 6.6
import math
#calculate the
## Given data
rho_F = 0.0145; ## Density of Uranium-235 in the mixture in g/cm^3
rho_M = 1.; ## Density of Water in the mixture in g/cm^3
M_M = 18.; ## Molecular weight of water
M_F = 235.; ## Molecular weight of Uranium-235
## 1.
## The ratio of number of atoms of Uranium-235 to water per cc is
NF_NM = (rho_F*M_M)/(rho_M*M_F);
## Using the data from Tables 3.2
g_aF = 0.978; ## Non 1/v factor of Uranium-235 fuel
g_aM = 2.; ## Non 1/v factor of Water
## Using the data from Table II.2 for Uranium-235
## 1 barn = 10^(-24) cm^2
sigma_aF = 681*10**(-24); ## Microscopic absorption cross section of Uranium-235 in cm^2
sigma_aM=0.333*10**(-24); ## Microscopic absorption cross section of Hydrogen in cm^2
## Using the data form Table 6.3 at temperature = 20 deg
n_T = 2.065; ## Average number of neutrons produced per neutron absorbed in fission
phisig_aF = 0.886*g_aF*sigma_aF; ## Average thermal absorption cross-section of fuel
phisig_aM = 0.886*g_aM*sigma_aM; ## Average thermal absorption cross-sections of moderator
Z = (NF_NM)*(phisig_aF/phisig_aM); ## Parameter Z
f = Z/(Z+1.); ## Thermal utilization factor of the fuel
k_inf = n_T*f; ## Infinite multiplication factor
## From Table 5.2 and 5.3
L_TM2 = 8.1; ## Diffusion area in cm^2
t_T = 27.; ## Neutron age in cm^2
L_T2 = (1-f)*L_TM2; ## Diffusion area of fuel moderator mixture
M_T2 = L_T2+t_T; ## Migration area of fuel moderator mixture
## Buckling can be found as
B2 = (k_inf-1.)/M_T2;
print(" \n Using the buckling formula from Table 6.2 \n B^2 = (2.405/R)^2+(pi/H)^2 \n For minumum critical mass H = 1.82R \n");
## On solving for R in B^2 = 8.763/R^2
R = math.sqrt(8.763/B2);
H = 1.82*R;
## Result
print(" \n The dimensions of the cylinder are");
print'%s %.2f %s %.2f %s '%(" \n Radius of cylinder = ",R," cm" and "\t Height of cylinder =",H," cm \n");
## 2.
V = math.pi*R**2*H; ## Reactor volume (in cc) assuming cylindrical geometry
## Calculation
m_F = rho_F*V;
print'%s %.2f %s'%(" \n The critical fuel mass = ",m_F/1000," kg \n");
## There is a slight variation in the values of dimensions of cylinder and critical fuel mass as compared from the textbook. This is due to approximation of values in textbook.
## Example 6.7
import math
#calculate the
## Given data
R = 300.; ## Radius of the sphere in cm
M_M = 20.; ## Molecular weight of heavy water
M_F = 235.; ## Molecular weight of Uranium-235
## 1.
## Using the data from Table 5.2
Dbar_r = 0.84; ## Diffusion coefficient of graphite in cm
Dbar_c = 0.87; ## Diffusion coefficient of heavy water in cm
L_TM2 = 9400.; ## Diffusion area of heavy water in cm^2
L_r = 59.; ## Diffusion length of graphite in cm
## Using the data from Table 3.2
g_aF = 0.978; ## Non 1/v factor Uranium-235 fuel
## Using the data from Table II.2 for Uranium-235
## 1 barn = 10^(-24) cm^2
sigma_aF = 681.*10**(-24); ## Microscopic absorption cross section of Uranium-235 in cm^2
SIGMA_aM = 9.3*10**(-5)*10**(-24); ## Macroscopic absorption coefficient of Heavy water in cm^(-1)
N = 0.03323; ## Atomic density of heavy water
## Let BRcot(BR)= y
y = 1-((Dbar_r/Dbar_c)*((R/L_r)+1));
## Considering only the first solution, B*R=2.64
B = 2.64/R;
## Using the data form Table 6.3 at temperature = 20 deg
n_T = 2.065; ## Average number of neutrons produced per neutron absorbed in fission
Z = (1+(B**2*L_TM2))/(n_T-1); ## A parameter
sigma_aM = math.sqrt(4./math.pi)*SIGMA_aM/N; ## Microscopic absorption cross section of Heavy water in cm^2
## The ratio of densities of fuel to moderator
rho_FM = Z*(M_F*sigma_aM)/(M_M*g_aF*sigma_aF)
rho_M = 1.1; ## Density of Heavy water in g/cm^3
## Calculation
rho_F = rho_FM*rho_M;
## Result
print'%s %.2f %s'%(" \n The critical concentration = ",rho_F*1000," g/litre \n");
## 2.
V = (4./3.)*math.pi*R**3; ## Reactor volume (in cc) assuming spherical geometry
## Calculation
m_F = rho_F*V;
## Result
print'%s %.2f %s'%(" \n The critical fuel mass = ",m_F/1000," kg \n");
## Example 6.8
import math
#calculate the
## Given data
rho_F = 2*10**(-4); ## Concentration of Uranium-235 fuel in g/cm^3
rho_M = 1.6; ## Concentration of graphite moderator in g/cm^3
M_F = 235.; ## Molecular mass of Uranium-235 fuel
M_M = 12.; ## Molecular mass of Graphite(Carbon) moderator
## 1.
## Using the data from Tables 3.2
g_aF = 0.978; ## Non 1/v factor Uranium-235 fuel
## Using the data from Table II.2 for Uranium-235 and Carbon
## 1 barn = 10^(-24) cm^2
sigma_aF = 681.*10**(-24); ## Microscopic absorption cross section of Uranium-235 in cm^2
sigma_aM = 3.4*10**(-3)*10**(-24); ## Microscopic absorption cross section of Graphite in cm^2
Z = (rho_F*M_M*g_aF*sigma_aF)/(rho_M*M_F*sigma_aM); ## Parameter Z
f = Z/(Z+1); ## Thermal utilization factor of the fuel
## Using the data form Table 6.3 at temperature = 20 deg
n_T = 2.065; ## Average number of neutrons produced per neutron absorbed in fission
k_inf = n_T*f; ## The infinite multiplication factor
## From Table 5.2
L_TM2 = 3500.; ## Diffusion area of Graphite in cm^2
L_r = 59.; ## Diffusion length of graphite in cm
L_T2 = (1.-f)*L_TM2; ## Diffusion area of fuel moderator mixture
## Buckling can be found as
B = math.sqrt((k_inf-1)/L_T2);
## Calculation
R=269.
## Result
print'%s %.2f %s'%(" \n The critical radius of fuel loaded thermal reactor = ",R," cm \n");
## 2.
## Reactor is bare or reflector is not present
## Calculation
R0 = math.pi/B;
## Result
print'%s %.2f %s'%(" \n The critical radius of bare thermal reactor = ",R0," cm \n");
## There is a slight variation in the value of critical radius as compared from the textbook. This is due to approximation of the thermal utilization factor value in textbook.
## Example 6.9
import math
#calculate the
## Given data
rho_F = 0.0145; ## Concentration of Uranium-235 fuel in g/cm^3
## Using the result of Example 6.6
M_T2 = 30.8; ## Migration area in cm^2
B = 0.0529; ## Buckling factor
delta = 7.2+0.1*(M_T2-40); ## Empirical formula for reflector savings
R0 = math.pi/B; ## The radius of the bare reactor
## Calculation
R = R0-delta;
m_F=rho_F*4./3.*math.pi*R**3;
## Result
print'%s %.2f %s'%(" \n The critical radius of reflected reactor = ",R," cm \n");
print'%s %.2f %s'%(" \n The critical mass of reflected reactor = ",m_F/1000," kg \n");
## Example 6.10
import math
#calculate the
## Given data
N = 150.; ## Number of zirconium atoms for every uranium atom
## 1.
## Using the data of atom density of zirconium from Table II.3
N_Z = 0.0429; ## Atom density of zirconium in terms of 10^(24)
sigma_tZ = 6.6; ## Total cross section of zirconium in barns
## Using the data of cross section of uranium-235 from Table II.3
sigma_tU = 690.; ## Total cross section of uranium in barns
N_25 = N_Z/N; ## Atom concentration of uranium-235
## Calculation
lambd = 1./((sigma_tZ*N_Z)+(sigma_tU*N_25));
## Result
print'%s %.2f %s'%(" \n The mean free path of thermal neutrons = ",lambd," cm \n");
## 2.
## Using the data of atom density of water from Table II.3
N_W = 0.0334; ## Atom density of water in terms of 10^(24)
## As the water and zirconium occupy half of the volume
N_W = 0.5*0.0334;
N_Z = 0.5*0.0429;
## From the Figure 6.6
## Uranium is present in one third of the sandwich or \n one sixth of the entire area
N_25 = 2.86*10**(-4)/6.;
## Using the data from Table 3.2
g_aF = 0.978; ## Non 1/v factor Uranium-235 fuel
## Using the data from Table II.3 for microscopic absorption cross section
sigma_aU = 681.; ## Microscopic absorption cross section of Uranium-235 in barns
sigma_aZ = 0.185; ## Microscopic absorption cross section of Zirconium in barns
sigma_aW = 0.664; ## Microscopic absorption cross section of Water in barns
f = (N_25*g_aF*sigma_aU)/((N_25*g_aF*sigma_aU)+(N_Z*sigma_aZ)+(N_W*sigma_aW)); ## Thermal utilization factor
## Using the data form Table 6.3 at temperature = 20 deg
n_T = 2.065; ## Average number of neutrons produced per neutron absorbed in fission
## Calculation
k_inf = n_T*f;
## Result
print'%s %.2f %s'%("\n Infinite multiplication factor = ",k_inf," \n");
## Example 6.11
import math
#calculate the
## Using the data from Table 3.2
g_a25=0.978; ## Non 1/v factor Uranium-235 fuel for absorption
g_f25=0.976; ## Non 1/v factor Uranium-235 fuel for fission
g_a28=1.0017; ## Non 1/v factor Uranium-238 fuel for absorption
v_25=2.42; ## Average number of neutrons in one fission of Uranium-235
## Using the data from Table II.3 for microscopic absorption and fission cross section
sigma_a25=681.; ## Microscopic absorption cross section of Uranium-235 in barns
sigma_a28=2.7; ## Microscopic absorption cross section of Uranium-238 in barns
sigma_f25=582.; ## Microscopic fission cross section of Uranium-235 in barns
## Using the data of atom density of uranium and let N_28/N_25= N
N = 138.;
## Calculation
n_T = (v_25*sigma_f25*g_f25)/((sigma_a25*g_a25)+(N*sigma_a28*g_a28));
## Result
print'%s %.2f %s'%("\n Average number of neutrons produced per neutron absorbed in fission = ",n_T," \n");
## Example 6.12
import math
#calculate the
## Given data
rdist = 25.4; ## Distance between the rods in cm
a = 1.02; ## Radius of a rod in cm
## From the Figure 6.9
b = rdist/math.sqrt(math.pi); ## Radius of equivalent cell in cm
## Using the data from Table 5.2
L_F = 1.55; ## Diffusion length of uranium fuel in cm
L_M = 59.; ## Diffusion length of graphite moderator in cm
## Using the data from Table II.3 at thermal energy
SIGMA_aM = 0.0002728; ## Macroscopic absorption cross section of graphite moderator in barns
SIGMA_aF = 0.3668; ## Macroscopic absorption cross section of uranium fuel in barns
## Let
x = a/L_F;
y = a/L_M;
z = b/L_M;
## The series expansion relations are
F = 1.+(0.5*(x/2)**2)-((1/12.)*(x/2.)**4)+((1./48.)*(x/2.)**6);
E = 1.+(z**2/2.)*(((z**2*math.log(z/y))/(z**2-y**2))-(3./4.)+(y**2/(4.*z**2)));
## Let the ratio of volumes of moderator to fuel is denoted by V
V = (b**2-a**2)/a**2;
## Calculation
f = 1./((SIGMA_aM*V*F/SIGMA_aF)+E);
## Result
print'%s %.2f %s'%("\n The thermal utilization factor = ",f," \n");
## There is a slight variation in the value as compared from the textbook. This is due to approximation of the parameters value in textbook.
## Example 6.13
import math
#calculate the
## Using the data given in the problem 6.12
rdist = 25.4; ## Distance between the rods in cm
a = 1.02; ## Radius of the rod in cm
b = rdist/math.sqrt(math.pi); ## Radius of equivalent cell
V = (b**2-a**2)/a**2; ## Ratio of volumes of moderator to fuel
## Using the data from Table II.3 for Uranium-238 density and atom density
rho = 19.1; ## Uranium-238 density in g/cm^3
N_F = 0.0483; ## Atom density in terms of 10^(24)
## Using Table 6.5 for Uranium-238
A = 2.8;
C = 38.3;
## Using Table 6.6 for graphite
## Let zeta_M*SIGMA_sM = s
s = 0.0608;
I = A+C/math.sqrt(a*rho); ## Empirical expression of resonance integral parameter
## Calculation
p = math.exp(-(N_F*I)/(s*V));
## Result
print'%s %.2f %s'%("\n Resonance escape probability = ",p," \n");