In [1]:

```
## Example 11.1
import math
## Given data
h = 30.; ## Height at which the effluent is relaesed
## Calculation of maxima location
sigma_z = h/math.sqrt(2.); ## Vertical dispersion coefficient
## Using the plot given in Figure 11.12 for Type F condition
## The corresponding value with calculated maximum location is noted.
h_max = 1900.;
## Result
print'%s %.2f %s'%(" \n The point of maximum concentration of non-radioactive effluent = ",h_max," m \n");
```

In [2]:

```
## Example 11.2
import math
## Given data
A = 2*10**3; ## Amount of radioactivity release due to Xenon-133 in curie
t = 365.*24.*3600.; ## Time in seconds
Q_dash = A/t; ## Average emission rate of Xenon-133
h = 100.; ## Location of radioactivity release through vent
v_bar = 1.; ## Wind speed in m/sec
## Using the plot given in Figure 11.11 and 11.12 for Type F condition at 100 m
sigma_y = 275.; ## Horizontal dispersion coefficient
sigma_z = 46.; ## Vertical dispersion coefficient
chi = (Q_dash*math.exp(-h**2/(2.*sigma_z**2)))/(math.pi*v_bar*sigma_y*sigma_z); ## Radionuclide concentration in terms of Ci/cm^3
## Using data from Table 11.3
Eg_bar = 0.03; ## Average gamma decay energy per disintegration in MeV
## Calculation
H_dot = 0.262*chi*Eg_bar*t*10**3; ## The units are in mrem/year
## Expressing the dose rate in SI units
H_dot_SI = 2.62*chi*Eg_bar*t*10**3;
## Result
print'%s %.2f %s %.2f %s '%(" \n The external gamma dose rate due to xenon release under type F condition = ",H_dot," mrem/year" or " ",H_dot_SI,"mSv/year \n");
```

In [3]:

```
## Example 11.3
import math
## Using data from Table 11.3
Eg_bar = 0.00211; ## Average gamma decay energy per disintegration in MeV
## Calculation
C_y = 0.262*Eg_bar;
## Result
print'%s %.2e %s'%(" \n The dose rate factor due to krypton exposure = ",C_y," rem*m^3/sec-Ci \n");
```

In [4]:

```
## Example 11.4
import math
## The results given in Example 11.2 are to be used in this problem
chi = 1.5*10**(-10); ## Radionuclide concentration in terms of Ci/cm^3
t = 365.*24.*3600.; ## Time in seconds
## Using data from Table 11.3
Ebeta_bar = 0.146; ## Average gamma decay energy per disintegration in MeV
f = 1.; ## Experimentally detemined factor
## Calculation
H_dot = 0.229*Ebeta_bar*chi*f*t;
## Expressing the result in milli-rem
print'%s %.2f %s'%(" \n The external beta dose rate due to xenon exposure for a year = ",H_dot*10**3," mrem/year \n");
```

In [5]:

```
## Example 11.5
import math
## Given data
A = 1.23; ## Amount of radioactivity release due to I-131 in curie in one year
h = 30.; ## Location of radioactivity release through vent in meter
v_bar = 1.2; ## Wind speed in m/sec
T_12 = 8.04; ## Half life of Iodine 131 in days
T_12b = 138.; ## Biological half life of Iodine 131 in days
zeta = 0.23; ## Fraction of core inventory in MeV
q = 0.23; ## Fraction of Iodine-131 in thyroid
M = 20.; ## Mass of an adult thyroid in grams
## 1.
t = 365.*24.*3600.; ## Time in seconds
Q_dash = A/t; ## Average emission rate of Iodine-131
## Converting days into seconds by using 1 day = 86400 seconds
lambd = 0.693/(T_12*86400); ## Decay constant of Iodine-131
lambda_b = 0.693/(T_12b*86400.); ## Biological decay constant of Iodine-131
## Let the quantity chi*v_bar/Q_bar = x
## Using the plot given in Figure 11.13 for Type E condition at 2000 m
x = 6.*10**(-5);
## Solving for chi
chi = (x*Q_dash)/v_bar;
## As per the standards of International Commission on Radiolgical Protection (ICRP)
B = 2.32*10**(-4); ## Normal breathing rate in m^3/sec
## Calculation
H_dot = (592.*B*zeta*q*chi)/(M*(lambd+lambda_b));
## Result
print'%s %.2e %s'%(" \n The equilibrium dose rate to an adult thyroid = ",H_dot,"sem/sec \n");
## 2.
## Calculation
H = H_dot*t;
## Expressing the result in milli-rem
## Result
print'%s %.2f %s'%(" \n The annual dose to an adult thyroid = ",H*10**3," mrem \n");
```

In [6]:

```
## Example 11.6
import math
## Given datah
E = 0.66; ## Energy of gamma ray emitted by caesium in MeV
x = 100.; ## Height of exposure in cm
## Using the data from Table II.5 for air at E = 0.66 MeV
mua_rho = 0.0294; ## The ratio of absorption coefficient to density of air in cm^2/g
## Using the data from Table II.4 for air at E = 0.66 MeV
mu_rho = 0.0775; ## The ratio of attenuation coefficient to density of air in cm^2/g
## Using standard value for density of air
rho = 1.293*10**(-3);
mu = mu_rho*rho;
mux = mu*x;
gamma = 0.57722; ## Euler's constant
E1 = -gamma+math.log(1./mux)+mux; ## Conversion factor
## Using parameter data from Table 11.16
C = 1.41; ## A constant
beta = 0.0857; ## A constant
## Calculation
H_dot_S = 3.39*10**(-2)*E*mua_rho*(E1+(C*math.exp(-(1.-beta)*mux)/(1.-beta)));
## Converting time in hours by 1 hour = 3600 seconds
## Result
print'%s %.2e %s %.2f %s '%(" \n The gamma ray dose rate conversion factor due to caesium-137 = ",H_dot_S," rem*m^2/sec-Ci" or " ",H_dot_S*3600,"rem*m^2/hour-Ci\n");
```

In [7]:

```
## Example 11.7
import math
## Given data
C0 = 6.25*10**6; ## Amount of initial radioactivity release due to I-131 in curie
p = 0.1; ## Leakage rate in percent
t0 = 2*3600.; ## Analysis time in seconds
v_bar = 1.; ## Wind speed in m/sec
## 1.
lambdal = 0.01*p/86400.; ## Decay constant corresponding to leakage rate in seconds (1 day = 86400 seconds)
## Using the data from Example 11.5 for the half life of Iodine-131
T_12 = 8.04; ## Half life of Iodine 131 in days
lambdac = 0.693/(T_12*86400.); ## Decay constant of Iodine-131 (I-131) in seconds
## Using data from Table 11.3
Eg_bar = 0.371; ## Average gamma decay energy per disintegration of I-131 in MeV
## Using the plot given in Figure 11.11 and 11.12 for Type F condition at 100 m
sigma_y = 21.; ## Horizontal dispersion coefficient
sigma_z = 70.; ## Vertical dispersion coefficient
## As lambdac*t << 1,
## Calculation
H = (0.262*Eg_bar*lambdal*C0*t0)/(math.pi*v_bar*sigma_y*sigma_z);
## Result
print'%s %.2f %s'%(" \n The total external dose due to gamma ray exposure = ",H," rem\n")
## 2.
## Using the data given in Example 11.5
B = 2.32*10**(-4); ## Normal breathing rate in m^3/sec
zeta = 0.23; ## Fraction of core inventory in MeV
q = 0.23; ## Fraction of Iodine-131 in thyroid
M = 20.; ## Mass of an adult thyroid in grams
## Calculation
H_dot = (592.*B*zeta*q*lambdal*C0*t0)/(math.pi*v_bar*sigma_y*sigma_z*M);
## Converting the units from rem/sec to milli-rem/hour by multiplying by (1000*3600)
## Result
print'%s %.2e %s %.2f %s '%(" \n The dose rate to an adult thyroid after 2 hours = ",H_dot," rem/sec" or"",math.ceil(H_dot*(1000*3600))," mrem/hour\n");
## 3.
## Using the data given in Example 11.5
T_12 = 8.04; ## Half life of Iodine 131 in days
T_12b = 138.; ## Biological half life of Iodine 131 in days
lambd = 0.693/(T_12*86400.); ## Decay constant of Iodine-131 in sec^(-1)
lambda_b = 0.693/(T_12b*86400.); ## Biological decay constant of Iodine-131 in sec^(-1)
## Calculation
H = H_dot/(lambd+lambda_b);
## Result
print'%s %.2f %s'%(" \n The dose commitment to thyroid by Iodine-131 exposure after 2 hours = ",H," rem \n");
```

In [8]:

```
## Example 11.8
import math
## Given
E = 2.4; ## Energy of gamma rays in MeV
r = 1000*100; ## Distance from the building where radiation is exposed in cm
t0 = 2*3600; ## Time of exposure in seconds
A = 3*10^7; ## Amount of initial radioactivity release due to Kr-88 in curie
f = 0.4; ## Fraction of disintegrations which release 2.4 MeV gamma rays
C0 = A*f; ## Effective number of curies
T_12 = 2.79; ## Half life of Iodine 131 in hours
lambd = 0.693/(T_12*3600.); ## Decay constant of Iodine-131 in sec^(-1)
## Using the result given in Example 11.7
lambdal = 1.16*10**(-8); ## Decay constant corresponding to fission prouduct release from building
lambdac = lambd+lambdal; ## Total decay constant in sec^(-1)
## Using the data from Table II.4 for air at E = 2.4 MeV
mu_rho = 0.041; ## The attenuation coefficient in cm^2/g
## Using standard value for density of air in g/cm^3
rho = 1.293*10**(-3);
mu = mu_rho*rho;
## Using the data from Table II.5 for air at E = 2.4 MeV
mua_rho = 0.0227; ## The ratio of absorption coefficient to density of air in cm^2/g
print'%s %.2f %s'%(" \n Buildup factor is measured at ",mu*r,"");
## Using Berger's form in Problem 11.9
B_p = 4.7; ## Buildup factor due to a point source
## Calculation
H = (54.*C0*(1.-math.exp(-lambdac*t0))/lambdac)*(E*mua_rho*B_p*math.exp(-mu*r)/r**2);
## Result
print'%s %.2e %s'%(" \n The direct dose due to gamma ray exposure = ",H," rem \n")
## There is a slight deviation in the answer due to approximation of value in the textbook.
```

In [9]:

```
## Example 11.9
import math
## Given data
gammai = 0.0277; ## Fission yield of Iodine-131 in fraction
P = 3200.; ## Thermal power of the plant in MW
## Calculation
alphai = 8.46*10**5*P*gammai;
## Result
print'%s %.2f %s'%(" \n The saturation activity of Iodine-131 during reactor operation = ",alphai," curie \n")
## Using assumption 2 of Nuclear Regulatory Commission (NRC) in calculation of radii of exclusion zone and Low Population Zone (LPZ)
## Due to core meltdown, 25 percent of iodine inventory is released and out of which 91 percent is in elemental form.
Fp = 0.25*0.91; ## Fraction of Iodine-131 released from the fuel into the reactor containment
## As entire iodine escapes through air
Fb = 1.; ## Fraction of 'Fp' which remains airborne and is capable of escaping from the building
## Calculation
C0 = alphai*Fp*Fb;
## Result
print'%s %.2f %s'%(" \n The activity of Iodine-131 in elemental form due to core meltdown = ",C0," curie \n");
```

In [10]:

```
## Example 11.10
import math
## Given data
P = 1000.; ## Electrical power of the plant in Mwe
eta = 0.38; ## Plant efficiency
P_th = P/eta; ## Thermal power of the plant in MW
h = 100.; ## Height of stack in metre
t = 24*365.; ## The number of hours in a year
m0 = 13000.; ## Amount of coal in the plant in Btu/lb
m0_ash = 0.09; ## Fraction of ash in the coal
## 1.
E = P_th*t; ## Energy released in a year in MW-hour
## Converting the units in Btu by using 1 MW-hour = 3.412*10^6 Btu
m = (E*3.412*10**6)/m0;
## Converting the units in g/year by using 1 lb/year = 453.59237 g/year
m = m*453.59237;
## Assuming the fly ash equipment has an efficiency of 97.5% of fly ash removal
eta_flyash = 0.025; ## Only (1-efficiency) is exhausted
m_ash = eta_flyash*m0_ash*m;
## A typical power plant contains 3pCi/g of each nuclide (Radium-226) in one year
A = 3*10**(-12);
## Calculation
A_total = A*m_ash;
## Result
print'%s %.2f %s'%(" \n Total activity of Radium-226 emitted = ",A_total," curie \n")
## 2.
v_bar = 1.; ## Wind speed in m/sec
t = 24.*365.*3600.; ## Analysis time of one year equivalently in seconds
MPC = 3.*10**(-12.); ## Maximum Permissible Concentration in micro-curie/cm^3
Q_bar = A_total/t; ## Emission rate for one year in curie/year
## Let the quantity chi*v_bar/Q_bar = x
## Using the plot for Pasquill F, given in Fig. A.7, Pg no 413 from Slade, D. H., Editor, Meteorology and Atomic Energy-1968. U. S. Atomic Energy Commission Report TID-24190, 1968.
x = 2.5*10**(-6.); ## Maximum value of x at 10^4 m
## Solving for chi
chi = (x*Q_bar)/v_bar;
## Converting the units from Ci/m^3 to micro-curie/cm^3 by multiplying by 10^6/10^6 = 1
print'%s %.2e %s'%(" \n Concentration of Radium-226 present = ",chi," micro-curie/cm^3 \n")
## Let c be the comparison factor
## Calculation
c = MPC/chi;
## Result
print'%s %.2f %s'%(" \n On comparison, the total concentration of Radium-226 is ",c," times smaller than Maximum Permissible Concentration (MPC) \n")
## The comparison factor is slightly different from the value in the textbook. This is because of approximation used in the textbook.
```

In [11]:

```
## Example 11.11
import math
## Given data
Qy_bar = 1.04*10**(-2); ## Emission rate for one year in curie/year
## Let (chi/Q_bar) = d which is called 'Dilution factor'
d = 4*10**(-8); ## Dilution factor in year/cm^3
vd = 0.01; ## Experimentally determined constant
## 1.
T_12 = 8.04; ## Half life of Iodine 131 in days
T_12f = 14; ## First order half life of Iodine 131 in days
## Converting days into years by using 1 year = 365 days
lambd = 0.693/(T_12/365.); ## Decay constant of Iodine-131
lambdaf = 0.693/(T_12f/365.); ## First order decay constant of Iodine-131
## Calculation
Cf = (Qy_bar*d*vd)/(lambd+lambdaf);
## Expressing the result in micro-curie
Cf = Cf*10**6;
## Result
print'%s %.2e %s'%(" \n The activity of I-131 on the vegetation = ",Cf," micro-curie/m^2 \n");
## 2.
## The proportionality factor has a value 9*10^(-5) Ci/cm^3 of milk per Ci/m^2 of grass
## Calculation
Cm = 9*10**(-5)*Cf;
## Result
print'%s %.2e %s'%(" \n The concentration of I-131 in the milk = ",Cm," micro-curie/m^2 \n");
## 3.
MPC = 3*10**(-7); ## Maximum Permissible Concentration in micro-curie/cm^3
## Calculation
H_dot = (2270.*Cm)/MPC;
## Result
print'%s %.2e %s'%(" \n The annual dose rate to an infant thyroid by consuming radiated milk = ",H_dot," mrem/year \n");
```

In [12]:

```
## Example 11.12
import math
## Given data
Qy_bar = 0.197; ## Emission rate for one year in micro-curie/year
## Let (chi/Q_bar) = d which is called 'Dilution factor'
d = 6.29*10**(-16); ## Dilution factor in year/cm^3
MPC_w = 6*10**(-5); ## Maximum Permissible Concentration (MPC) of iron in micro-curie/cm^3
Cw = Qy_bar*d; ## The concentration of Fe-59
## For fish
Rs_fish = 110.; ## Consumption rate in g/day
## Using the data from Table 11.15 for saltwater concentration of fish for iron
CF_fish = 1800.; ## Concentration Factor of fish
Cs_fish = CF_fish*Cw; ## Activity of fish
H_dot_fish = (Cs_fish*Rs_fish*500.)/(MPC_w*2200.); ## Dose rate for fish
## For mollusks
Rs_mollusk = 10.; ## Consumption rate in g/day
## Using the data from Table 11.15 for saltwater concentration of mollusk for iron
CF_mollusk = 7600.; ## Concentration Factor of mollusk
Cs_mollusk = CF_mollusk*Cw; ## Activity of mollusk
H_dot_mollusk = (Cs_mollusk*Rs_mollusk*500.)/(MPC_w*2200.); ## Dose rate for mollusk
## For crustaceans
Rs_crustacean = 10.; ## Consumption rate in g/day
## Using the data from Table 11.15 for saltwater concentration of crustacean for iron
CF_crustacean = 2000.; ## Concentration Factor of crustacean
Cs_crustacean = CF_crustacean*Cw; ## Activity of crustacean
H_dot_crustacean = (Cs_crustacean*Rs_crustacean*500.)/(MPC_w*2200.); ## Dose rate for crustacean
## Calculation
H_dot = H_dot_fish+H_dot_mollusk+H_dot_crustacean;
## Result
print'%s %.2e %s'%(" \n The annual dose rate to GI tract by consuming fish = ",H_dot_fish," mrem/year");
print'%s %.2e %s'%(" \n The annual dose rate to GI tract by consuming mollusk = ",H_dot_mollusk," mrem/year");
print'%s %.2e %s'%(" \n The annual dose rate to GI tract by consuming crustaceans = ",H_dot_crustacean," mrem/year");
print'%s %.2e %s'%(" \n The annual dose rate to GI tract by consuming seafood = ",H_dot," mrem/year \n");
## The answer for annual dose rate to GI tract by consuming fish is wrong in the textbook. This is because the value of fish consumption rate is wrongly considered.
```