Chapter10-Radiation Shielding¶
Ex1-pg553¶
In [1]:
## Example 10.1
import math
## Given data
E0 = 2.; ## Energy of gamma rays in MeV
a = 10.; ## Thickeness of lead shield in cm
phi0 = 10**6; ## Intensity of gamma rays in gamma-rays/cm^2-sec
## 1.
## Using the data from Table II.4 for E0 = 2 MeV
mu_rho = 0.0457; ## The ratio of total attenuation coefficient to density in cm^2/g
## From standard data tables for lead
rho = 11.34; ## Density of lead in g/cm^3
## Calculation
phi_u = phi0*math.exp(-(mu_rho*rho*a));
## Result
print'%s %.2f %s'%("\n Uncollided flux at the rear side of lead shield = ",phi_u," gamma-rays/cm^2-sec \n")
## 2.
## Using the data from Table 10.1 for 2 MeV of lead material
mua = mu_rho*rho*a;
B_4 = 2.41; ## Buildup factor if mu*a = 4
B_7 = 3.36; ## Buildup factor if mu*a = 7
## Using two point method of straight line for calculating buildup factor at mu*a
B_m = B_4+((mua-4.)*((B_7-B_4)/(7.-4.)));
## Calculation
phi_b = phi_u*B_m;
## Result
print'%s %.2f %s'%("\n Buildup flux at the rear side of lead shield = ",phi_b," gamma-rays/cm^2-sec \n");
## 3.
## Using the data from Table II.5 for 2 MeV
mua_rho_air = 0.0238; ## The ratio of total attenuation coefficient to density of air in cm^2/g
## Calculation
X_dot = 0.0659*E0*mua_rho_air*phi_b;
## Result
print'%s %.2f %s'%("\n Exposure rate at the rear side of lead shield = ",X_dot," mR/hour \n");
Ex2-pg555¶
In [38]:
## Example 10.2
import math
import warnings
warnings.filterwarnings('ignore')
import numpy
%matplotlib inline
import matplotlib
from matplotlib import pyplot
## Given data
E = 1.; ## Energy of gamma rays in MeV
X_dot = 1.; ## Exposure rate in mR/hour
phi0 = 10**8; ## Intensity of gamma rays in gamma-rays/cm^2-sec from isotropic point source
## Using the data from Table II.5 for 1 MeV
mua_rho_air = 0.028; ## The ratio of total attenuation coefficient to density of air in cm^2/g
phi_b = X_dot/(0.0659*E*mua_rho_air); ## Buildup flux in gamma-rays/cm^2-sec
## Using Eq 10.14
print'%s %.2f %s'%(" \n The equation to calculate radius is \n %.2E = %E * Bp*exp(-mu*R)/(4*pi*R^2) \n",phi_b,phi0);
## Using the data from Table II.4 for E = 1 MeV for Iron
mu_rho = 0.0595; ## The ratio of total attenuation coefficient to density in cm^2/g
## From standard data tables for iron
rho = 7.864; ## Density of iron in g/cm^3
mu = mu_rho*rho;
## On solving the right hand side of equation
## RHS = 3.22*10^3*Bp*exp(-mu*R)/(mu*R)^2
## Let mu*R = x
## Using the data from Table 10.2 for isotropic point source of 1 MeV incident on iron material
Bp = numpy.array([1.87, 2.89, 5.39, 10.2, 16.2, 28.3, 42.7]);
x = numpy.array([1 ,2 ,4 ,7 ,10, 15, 20]);
leng=len(Bp)
RHS= numpy.zeros(leng);
for i in range(0,7):
RHS[i] = (3.22*10**3*Bp[i]*math.exp(-x[i])/x[i]**2);
pyplot.plot(x,RHS)
pyplot.legend("Conical","2D-CD")
pyplot.xlabel("mu*R")
pyplot.ylabel("RHS")
pyplot.title("Semilog plot of RHS vs mu*R")
pyplot.show()
## From the graph
muR = 6.55; ## This is the value when RHS = 1
## Calculation
R = muR/mu;
## Result
print'%s %.2f %s'%("\n The shield radius required = ",math.ceil(R)," cm \n");
Ex3-pg561¶
In [40]:
## Example 10.3
import math
import warnings
warnings.filterwarnings('ignore')
import numpy
%matplotlib inline
import matplotlib
from matplotlib import pyplot
## Given data
E = 2.; ## Energy of gamma rays in MeV
X_dot = 2.5; ## Exposure rate in mR/hour
phi0 = 10**9; ## Intensity of gamma rays in gamma-rays/cm^2-sec from isotropic point source
## Using the data from Table II.5 for 1 MeV
mua_rho_air = 0.0238; ## The ratio of total attenuation coefficient to density of air in cm^2/g
phi_b = X_dot/(0.0659*E*mua_rho_air); ## Buildup flux in gamma-rays/cm^2-sec
## From standard data tables for concrete
rho = 2.35; ## Density of concrete in g/cm^3
## Using the data from Table 10.3 for concrete at 2 MeV
A1 = 18.089;
A2 = 1-A1;
alpha1 = -0.0425;
alpha2 = 0.00849;
## Using Eq 10.26
#print'%s %.2f %s'%(" \n The equation to calculate thickness is \n %.2E = (%E/2) *(%4.3f*E1(%4.3f*mu*a) %4.3f*E1(%4.3f*mu*a)) \n",phi_b,phi0,A1,(1+alpha1),A2,(1+alpha2));
## Using the data from Table II.4 for E = 1 MeV for concrete
mu_rho = 0.0445; ## The ratio of total attenuation coefficient to density in cm^2/g
mu = mu_rho*rho;
## On solving the right hand side of equation
## RHS = 1.13*10^7*(E1(0.9575*mu*a)-0.94*E1(1.00849*mu*a))
## Let mu*a = x
x = numpy.array([1 ,2 ,4 ,7 ,10, 15, 20]);
leng=len(x)
RHS= numpy.zeros(leng);
for i in range(0,leng):
RHS[i] = 1.13*10**7*(math.exp(-0.9575*x[i])*((1./(0.9575*x[i])+(1/(0.9575*x[i])**3))) - math.exp(-1.00849*x[i])*((1./(1.00849*x[i])+(1./(1.00849*x[i])**3))));
pyplot.plot(x,RHS)
pyplot.legend("Conical","2D-CD")
pyplot.xlabel("mu*R")
pyplot.ylabel("RHS")
pyplot.title("Semilog plot of RHS vs mu*R")
pyplot.show()
## From the graph
mua = 13.6; ## This is the value when RHS = 1
## Calculation
a = mua/mu;
## Result
print'%s %.2f %s'%("\n The concrete thickness = ",a," cm \n");
Ex4-pg575¶
In [4]:
## Example 10.4
import math
## Given data
E = 6.; ## Energy of gamma rays in MeV
phi0 = 10**2; ## Intensity of gamma rays in gamma-rays/cm^2-sec from mono-directional beam
x_w = 100.; ## Thickness of water in cm
x_Pb = 8.; ## Thickness of lead in cm
## Using data from Table II.4 at 6 MeV
mu_w = 0.0275; ## Total attenuation coefficient of water in cm^(-1)
mu_Pb = 0.4944; ## Total attenuation coefficient of lead in cm^(-1)
mua_w = x_w*mu_w; ## Attenuation due to thickness of water
mua_Pb = x_Pb*mu_Pb; ## Attenuation due to thickness of lead
## Case (a) - Water is placed before the lead
print'%s %.2f %s'%(" \n In case (a), Buildup factor is only due to lead measured at ",mua_Pb,"");
## Using the data from Table 10.1 at 6 MeV
B_Pb = 1.86;
phi_b_a = phi0*B_Pb*math.exp(-(mua_w+mua_Pb));
## Using the data from Table II.5 for 6 MeV
mua_rho_air = 0.0172; ## The ratio of total attenuation coefficient to density of air in cm^2/g
## Calculation
X_dot_a = 0.0659*E*mua_rho_air*phi_b_a;
## Result
print'%s %.2f %s'%("\n Exposure rate if water is placed before lead shield = ",X_dot_a*1000," uR/hour \n");
## Case (b) - Lead is placed before water
print'%s %.2f %s %.2f %s'%(" \n In case (b), Buildup factor is due to water and lead measured at ",mua_w,"" and "",mua_Pb," respectively");
## Using the data from Table 10.1 for water at 3.2 MeV,, which is the minimum point of mu_Pb curve
B_w = 2.72;
B_m = B_Pb*B_w;
phi_b_b = phi0*B_m*math.exp(-(mua_w+mua_Pb));
## Calculation
X_dot_b = 0.0659*E*mua_rho_air*phi_b_b;
## Result
print'%s %.2f %s'%("\n Exposure rate if lead is placed before water = ",X_dot_b*1000," uR/hour \n");
## The answer given in the textbook is wrong. This is because the intensity of gamma rays is wrongly taken for calculation.
Ex5-pg582¶
In [45]:
## Example 10.5
import math
## Given data
fission_density = 4*10**7; ## Fission density in fissions/cm^2-sec
## 1 inches = 2.54 cm
d = 28*2.54; ## Diamaeter of plate in cm
R = d/2.; ## Radius of plate in cm
v = 2.42; ## Number of fission neutrons emitted per fission
x = 75.; ## Distance of point from center of plate in cm
## Using the data from Table 10.4 for removal macroscopic cross section of water
sigma_RW = 0.103; ## Removal macroscopic cross section of water in cm^-1
S = v*fission_density; ## Strength of neutron source in terms of neutrons/cm^2-sec
A = 0.12; ## A constant
## From Figure 10.19
## Let the upper limit of integral be 20
x_limit = 20;
E1_x11 = 0.0000512
E1_x21 = 0.0000205
## Calculation
phi_1 = S*A/2.*(E1_x11-E1_x21);
## Result
print'%s %.2f %s'%(" \n The fast flux without iron shield = ",phi_1," neutrons/cm^2-sec \n");
## 2. Iron slab is inserted in front of the fission plate
## Using the data from Table 10.4 for removal macroscopic cross section of iron
sigma_R = 0.168; ## Removal macroscopic cross section of iron in cm^-1
t = 3*2.54; ## Thickness of iron slab in cm
## Now the analysis is similar to multi layered shielding
x_limit = 20;
E1_x12 = 0.0000124
E1_x22 = 0.0000043
## Calculation
phi_2 = S*A/2*(E1_x12-E1_x22);
## Result
print'%s %.2f %s'%(" \n The fast flux with iron shield = ",phi_2," neutrons/cm^2-sec \n");
Ex6-pg587¶
In [6]:
## Example 10.6
import math
## Given data
## Assuming average energy produced per fission reaction is 200 MeV
P = 250.*10**3; ## Power of research reactor in Watts
P_fission = 200*10**6*1.6*10**(-19); ## Energy produced in a fission reaction in terms of joule
f = 0.75; ## Metal volume fraction
## In this problem, both reflector and shield act as a composite shield
a = 150.+15.; ## Net shield distance in cm
## 1 litre = 1000 grams
V = 32.*1000.; ## Core volume in gram
fission_density = (P/P_fission)*(1./V);
v = 2.42; ## Number of fission neutrons emitted per fission
S = fission_density*v; ## Neutron source strength in neutrons/cm^3-sec
## Assuming spherical shape
## Volume of sphere = (4/3)*pi*(radius)^3
R = ((3.*V)/(4.*math.pi))**(1./3.);
## Using the data from Table 10.4 for removal macroscopic cross section
sigma_R = 0.174; ## Removal macroscopic cross section of uranium in cm^-1
sigma_RW = 0.103; ## Removal macroscopic cross section of water in cm^-1
A = 0.12; ## A constant
alpha = ((1.-f)*sigma_RW)+(f*sigma_R); ## A parameter
## Calculation
theta = (S*A/(4*alpha))*(math.ceil(R)/(math.ceil(R)+a))**2*math.exp(-sigma_RW*a)*(1-math.exp(-2*alpha*math.ceil(R)));
## Converting into equivalent dose rate by referring Figure 9.12
## Fast neutron flux of 6.8 neutrons/cm^2-sec is equivalent to 1 mrem/hr
H_dot = theta/6.8;
## Result
print'%s %.2f %s'%(" \n Fast neutron dose rate near the surface of the shield = ",H_dot," mrem/hour \n ");
Ex7-pg593¶
In [7]:
## Example 10.7
import math
## Given data
E = 14.; ## Energy of neutrons in MeV
phi0 = 10**9; ## Intensity of neutrons in neutrons/cm^2-sec from isotropic point source
## 1 feet = 30.48 cm
d = 10*30.48; ## Distance of concrete wall from a point source in cm
## As Intensity obeys inverse square law
I = phi0/(4.*math.pi*d**2); ## Intensity of neutron beam in terms of neutrons/cm^2-sec
H_dot = 1.; ## The required dose equivalent rate in mrem/hour
## From Figure 10.23(b)
H0_dot = H_dot/I; ## The dose equivalent rate
## Result
print'%s %.2f %s'%(" \n The reduced dose equivalent rate due to concrete wall is = ",H0_dot," mrem/hr \n");
## By looking into Figure 10.23(b) on thickness axis
print(" \n The concrete slab thickness is = 70 cm \n");
Ex8-pg598¶
In [8]:
## Example 10.8
import math
## Given data
R = 7.*30.48; ## Distance of core from the center of shield in cm
## Assuming average energy produced per fission reaction is 200 MeV
P = 10.; ## Power of teaching reactor in Watts
P_fission = 200*10**6*1.6*10**(-19); ## Energy produced in a fission reaction in terms of joule
fission_rate = P/P_fission; ## Number of fission reactions
## By assuming that the gammma rays are of equal energy of 1 MeV (Group 1) and looking into Table 10.5
E1 = 1.; ## Energy of gamma rays in MeV (Assumed)
chi_pn1 = 5.2; ## Number of prompt gamma rays emitted per fission with energy 2 MeV
S1 = chi_pn1*fission_rate; ## Source strength in gamma rays/sec
## Using the data from Table II.4 for E = 1 MeV for water
mu1 = 0.0996; ## Mass attenuation coefficient at 1 MeV in cm^-1
print'%s %.2f %s'%(" \n Buildup factor is due to water measured at ",mu1*R,"");
## Using the data from Table 10.2 at 1 MeV
B_p1 = 373.;
phi_b1 = (S1/(4.*math.pi*R**2))*B_p1*math.exp(-mu1*R); ## Buildup flux
## Using the data from Table II.5 for 1 MeV
mua_rho_air1 = 0.028; ## The ratio of total attenuation coefficient to density of air in cm^2/g
## Calculation
X_dot1 = 0.0659*E1*mua_rho_air1*phi_b1;
print'%s %.2f %s'%("\n Exposure rate due to group 1 = ",X_dot1," mR/hour \n");
## By assuming that the gammma rays are of equal energy of 2 MeV (Group 2) and looking into Table 10.5
E2 = 2.; ## Energy of gamma rays in MeV (Assumed)
chi_pn2 = 1.8; ## Number of prompt gamma rays emitted per fission with energy 2 MeV
S2 = chi_pn2*fission_rate; ## Source strength in gamma rays/sec
## Using the data from Table II.4 for E = 2 MeV for water
mu2 = 0.0493; ## Mass attenuation coefficient at 2 MeV in cm^-1
print'%s %.2f %s'%(" \n Buildup factor is due to water measured at ",mu2*R,"");
## Using the data from Table 10.2 at 2 MeV
B_p2 = 13.1;
phi_b2 = (S2/(4.*math.pi*R**2))*B_p2*math.exp(-mu2*R); ## Buildup flux
## Using the data from Table II.5 for 2 MeV
mua_rho_air2 = 0.0238; ## The ratio of total attenuation coefficient to density of air in cm^2/g
## Calculation
X_dot2 = 0.0659*E2*mua_rho_air2*phi_b2;
print'%s %.2f %s'%("\n Exposure rate due to group 2 = ",X_dot2," mR/hour \n");
## By assuming that the gammma rays are of equal energy of 4 MeV (Group 3) and looking into Table 10.5
E3 = 4.; ## Energy of gamma rays in MeV (Assumed)
chi_pn3 = 0.22; ## Number of prompt gamma rays emitted per fission with energy 4 MeV
S3 = chi_pn3*fission_rate; ## Source strength in gamma rays/sec
## Using the data from Table II.4 for E = 4 MeV for water
mu3 = 0.0339; ## Mass attenuation coefficient at 4 MeV in cm^-1
print'%s %.2f %s'%(" \n Buildup factor is due to water measured at",mu3*R,"");
## Using the data from Table 10.2 at 4 MeV
B_p3 = 5.27;
phi_b3 = (S3/(4.*math.pi*R**2))*B_p3*math.exp(-mu3*R); ## Buildup flux
## Using the data from Table II.5 for 4 MeV
mua_rho_air3=0.0194; ## The ratio of total attenuation coefficient to density of air in cm^2/g
## Calculation
X_dot3 = 0.0659*E3*mua_rho_air3*phi_b3;
print'%s %.2f %s'%("\n Exposure rate due to group 3 = ",X_dot3," mR/hour \n");
## By assuming that the gammma rays are of equal energy of 6 MeV (Group 4) and looking into Table 10.5
E4 = 6; ## Energy of gamma rays in MeV (Assumed)
chi_pn4 = 0.025; ## Number of prompt gamma rays emitted per fission with energy 4 MeV
S4 = chi_pn4*fission_rate; ## Source strength in gamma rays/sec
## Using the data from Table II.4 for E = 6 MeV for water
mu4 = 0.0275; ## Mass attenuation coefficient at 6 MeV in cm^-1
print'%s %.2f %s'%(" \n Buildup factor is due to water measured at ",mu4*R,"");
## Using the data from Table 10.2 at 6 MeV
B_p4 = 3.53;
phi_b4 = (S4/(4.*math.pi*R**2))*B_p4*math.exp(-mu4*R); ## Buildup flux
## Using the data from Table II.5 for 4 MeV
mua_rho_air4=0.0172; ## The ratio of total attenuation coefficient to density of air in cm^2/g
## Calculation
X_dot4 = 0.0659*E4*mua_rho_air4*phi_b4;
print'%s %.2f %s'%("\n Exposure rate due to group 3 = ",X_dot4," mR/hour \n");
##Calculation
X_dot = X_dot1+X_dot2+X_dot3+X_dot4;
## Result
print'%s %.2f %s'%("\n The total exposure rate due to all groups = ",X_dot," mR/hour \n");
Ex9-pg603¶
In [9]:
## Example 10.9
import math
## Given data
## Assuming average energy produced per fission reaction is 200 MeV
P = 55.; ## Power density of reactor in watts/cm^3
rho_eff_U = 0.33; ## Effective density of uranium in g/cm^3
rho_eff_W = 1-rho_eff_U; ## Effective density of water in g/cm^3
t_i = 3.; ## Average time spent by water in the reactor in seconds
t_0 = 2.; ## Average time spent by water in the external coolant circuit in seconds
## 1 eV = 1.6*10^(-19) J
P_fission = 200.*10**6*1.6*10**(-19); ## Energy produced in a fission reaction in terms of joule
fission_density = P/P_fission; ## Number of fission reactions
v = 2.42; ## Number of fission neutrons emitted per fission
S = v*fission_density; ## Strength of neutron source in terms of neutrons/cm^2-sec
## Atom density of oxygen at normal density of 1 g/cm^3 is
rho = 1.; ## Density of water in g/cm^3
N_A = 6.02*10**(23); ## Avogadro's constant
M = 18.; ## Molecular weight of water
atom_density = (rho*N_A)/M;
## Using the data from Table 10.7
sigma_r = 1.9*10**(-5)*10**(-24); ## Reaction cross section in cm^2
T_12 = 7.1; ## Half life of the given reaction in seconds
lambd = 0.693/T_12; ## Decay constant of the given reaction in seconds^(-1)
sigma_act = rho_eff_W*atom_density*sigma_r; ## Average macroscopic activation cross section
## Using the data from Table 10.4
sigma_RW = 0.103; ## Removal cross section of water in cm^-1
sigma_RU = 0.174; ## Removal cross section of Uranium in cm^-1
sigma_R = (sigma_RU*rho_eff_U)+(sigma_RW*rho_eff_W); ## Removal cross section of mixture
## Let activation rate given by (sigma_act*phi_av) be denoted as AR
AR = (sigma_act*S)/sigma_R;
## Calculation
alpha = AR*(1.-math.exp(-t_i*lambd))/(1.-math.exp(-(t_i+t_0)*lambd));
## 1 curie = 3.7*10^(10) disintegrations/sec
## Result
print'%s %.2e %s %.2f %s '%("\n Equilibrium activity of water due to neutron capture of oxygen = ",alpha," disintegrations/cm^3-sec" and " ",math.ceil(alpha*10**6/(3.7*10**10))," uCi/cm^3 \n");