# Chapter6-Neutron Reactor Theory¶

## Ex1-pg269¶

In [1]:
## 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');
Thermal Utilization factor =  0.67

Infinite Multiplication factor =  1.48

## Ex2-pg282¶

In [2]:
## 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');
Maximum flux in the spherical reactor =  4.18e+15  neutrons/cm^2-sec

Average flux in the spherical reactor =  1.27e+15  neutrons/cm^2-sec

## Ex3-pg283¶

In [3]:
## 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.

## Ex4-pg285¶

In [4]:
## 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');
Nonleakage probability of a fission neutron =  0.38

## Ex5-pg294¶

In [5]:
## 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.

Buckling =  3.14e-02

Critical mass =  4.60  kg

Infinite Multiplication factor (k_inf) =  1.80

Thermal Diffusion area =  445.03  cm^2

The expression for thermal flux =  5.52e+13  math.sin (Br)/r

The maximum thermal flux =  1.73e+12  neutrons/cm^2-sec

## Ex6-pg296¶

In [6]:
## 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.

Using the buckling formula from Table 6.2
B^2 = (2.405/R)^2+(pi/H)^2
For minumum critical mass H = 1.82R

The dimensions of the cylinder are

Radius of cylinder =  55.85 	 Height of cylinder = 101.65  cm

The critical fuel mass =  14.44  kg

## Ex7-pg302¶

In [7]:
## 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");

The critical concentration =  0.10  g/litre

The critical fuel mass =  11.25  kg

## Ex8-pg303¶

In [8]:
## 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.

The critical radius of bare thermal reactor =  322.75  cm

## Ex9-pg307¶

In [9]:
## 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");

The critical radius of reflected reactor =  53.11  cm

The critical mass of reflected reactor =  9.10  kg

## Ex10-pg310¶

In [10]:
## 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");

The mean free path of thermal neutrons =  2.08  cm

Infinite multiplication factor =  1.40

## Ex11-pg312¶

In [11]:
## 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");
Average number of neutrons produced per neutron absorbed in fission =  1.32

## Ex12-pg315¶

In [12]:
## 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.
The thermal utilization factor =  0.83

## Ex13-pg317¶

In [13]:
## 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");
Resonance escape probability =  0.95