## Example 7.1
import math
#calculate the
## 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
## Using the data from Table 7.1
t_dM = 2.1e-4; ## The mean diffusion time of the moderator in seconds
k_inf = 1.; ## The reactor is critical
f = k_inf/n_T; ## Thermal utilization factor
## Calculation
t_d = t_dM*(1.-f);
l_p = t_d;
## Result
print'%s %.2e %s'%(" \n The prompt neutron lifetime = ",l_p," seconds \n");
## Example 7.2
import math
#calculate the
## Given data
k_inf = 1.001; ## Infinite multiplication factor
## From the Example 7.1
l_p = 1e-4; ## Prompt neutron lifetime
## Calculation
T = l_p/(k_inf-1.);
## Result
print'%s %.2f %s'%(" \n The response time of the reactor = ",T," sec \n");
print'%s %.2f %s'%(" \n The reactor power will increase as exp(t/",T,"), where ''t'' denotes the time in seconds \n");
## Example 7.3
import math
#calculate the
## Given data
k_inf = 1.001; ## Infinite multiplication factor
## Calculation
rho = (k_inf-1.)/k_inf;
## Result
print'%s %.2e %s %.2f %s '%(" \n The reactivity = ",rho,"" or "",rho*100," percent \n");
## Example 7.4
import math
#calculate the
## Using the result of Example 7.1
lp = 1e-4; ## Prompt neutron lifetime in seconds
## Using the result of Example 7.3
rho = 1e-3; ## Reactivity
## By referring to Figure 7.2
print(" \n Reactor period = 57 seconds \n");
## Example 7.5
import math
#calculate the
## Using the result of Example 7.3
reactivity = 0.001;
## As the reactor is fueled with Uranium-235
bet = 0.0065; ## Total delayed neutron fraction of all groups denoted by 'beta'
print(" \n A dollar is worth 0.0065 in reactivity for Uranium-235 reactor. \n");
## Calculation
rho = reactivity/bet;
## Result
print'%s %.2f %s %.2f %s '%(" \n Reactivity = ",rho," dollars" or "",rho*100," cents \n");
## Example 7.6
import math
#calculate the
## Given data
P0 = 500.; ## Reactor power in MW
rho = -0.1; ## 10% in reactivity (Insertion of control rods correspond to negative reactivity)
## As the reactor is fueled with Uranium-235
bet = 0.0065; ## Total delayed neutron fraction of all groups denoted by 'beta'
P1 = (bet*(1.-rho)*P0)/(bet-rho); ## The drop in power level in terms of MW
## Assuming that negative reactivity is greater than 4%
T = 80.; ## Reactor period obtained from Figure 7.2 in seconds
t = 600.; ## Analysis time in seconds
## Calculation
P = P1*math.exp(-t/T); ## Power level drop in MW
## Result
print'%s %.2f %s'%(" \n The power level drop after 10 minutes = ",P," MW \n");
## Example 7.7
import math
#calculate the
## Given data
H = 70.; ## Height of the cylinder in cm
R = H/2.; ## Diameter of the cylinder in cm
a = 1.9; ## Radius of black control rod in cm
## From Table 6.2, Buckling can be found by
B0 = math.sqrt((2.405/R)**2+(math.pi/H)**2);
## Using the data from Table 5.2 and 5.3
L_TM2 = 8.1; ## Diffusion area of water moderator in cm^2
t_TM = 27.; ## Neutron age of water moderator 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
## Using the data from Table 5.2 and Table II.3
D_bar = 0.16; ## Thermal neutron diffusion coefficient in cm
SIGMA_t = 3.443; ## Total macroscopic cross section in cm^(-1)
f = (1.+B0**2*(L_TM2+t_TM))/(n_T+B0**2*L_TM2); ## Thermal utilization factor
M_T2 = (1.-f)*L_TM2+t_TM; ## Thermal migration area in cm^2
d = 2.131*D_bar*(a*SIGMA_t+0.9354)/(a*SIGMA_t+0.5098); ## Extrapolation distance
## Calculation
rho_w = (7.43*M_T2*(0.116+math.log(R/(2.405*a))+(d/a))**(-1))/((1.+B0**2*M_T2)*R**2);
## Result
print'%s %.2f %s %.2f %s '%(" \n The worth of a black control rod = ",rho_w," "or "",rho_w*100," percent \n");
## Example 7.8
import math
#calculate the
## Using the data and result from Example 7.7
f = 0.583; ## Thermal Utilization factor
L_TM2 = 8.1; ## Diffusion area of water moderator in cm^2
R = 35; ## Radius of the cylinder of the core in cm
a = 0.508; ## Radius of control rod in cm
Rc = math.sqrt(R**2/100.); ## Critical radius in cm
L_T = math.sqrt((1-f)*L_TM2); ## Thermal diffusion length in cm
## The points of estimation are chosen as follows
y = a/L_T;
z = Rc/L_T;
## Using the data given in Table V.I for modified Bessel functions
I0_275 = 1.019; ## I0 at 0.275
I1_275 = 0.1389; ## I1 at 0.275
I1_189 = 1.435; ## I1 at 1.89
K0_275 = 1.453; ## K0 at 0.275
K1_275 = 3.371; ## K1 at 0.275
K1_189 = 0.1618; ## K1 at 1.89
E = ((z**2-y**2)/(2.*y))*(((I0_275*K1_189)+(K0_275*I1_189))/((I1_189*K1_275)-(K1_189*I1_275))); ## The lattice function
## Using the data from Table 5.2 and Table II.3
D_bar = 0.16; ## Thermal neutron diffusion coefficient in cm
SIGMA_t = 3.443; ## Total macroscopic cross section in cm^(-1)
d = 2.131*D_bar*(a*SIGMA_t+0.9354)/(a*SIGMA_t+0.5098); ## Extrapolation distance
f_R = 1./((((z**2-y**2)*d)/(2.*a))+E); ## Rod utilization parameter
## Calculation
rho_w = f_R/(1.-f_R);
## Result
print'%s %.2f %s %.2f %s '%(" \n The total worth of the control rods = ",rho_w,"" or"",rho_w*1000," percent \n");
## There is a deviation in the value computed on comparison with the value given in the textbook. This is due to approximation of thermal diffusion area in the textbook.
## Example 7.9
import math
#calculate the
## Given data
SIGMAa_bar = 0.2; ## Average macroscopic absorption cross section in cm^(-1)
L_T = 1.2; ## Thermal diffusion length in cm
## Converting the given dimensions from inches to centimeters
## 1 inch = 2.54 cm
## From Figure 7.9
l = 9.75*(2.54/2.); ## Length of the half rod
a = 0.312*(2.54/2.); ## Thickness of the half rod
m = 44.5/math.sqrt(2.); ## Closest distance between two rods
D_bar = SIGMAa_bar*L_T**2; ## Thermal neutron diffusion coefficient in cm
d = 2.131*D_bar; ## Extrapolation distance in cm which is obtained for bare planar surface
f_R = ((4.*(l-a)*L_T)/(m-(2.*a))**2*(1./((d/L_T)+((m-(2*a))/(2*L_T))))); ## Rod utilization parameter
## Calculation
rho_w = 0.025/(1.-0.4);
## Result
print'%s %.2f %s %.2f %s' %(" \n The total worth of the control rods = ",rho_w,"" or "",rho_w*100," percent \n");
## There is a slight deviation in the value computed on comparison with the value given in the textbook. This is due to approximation of rod utilization parameter in the textbook.
## Example 7.10
import math
#calculate the
## Given data
d = 5.; ## Inner diameter of the tube in cm
a = d/2.; ## Inner radius of the tube in cm
l = 76.; ## Length of the tube in cm
rho = 2.; ## Density of B4C in g/cm^3
n = 5.; ## Number of rods in tbe reactor
m_B4C = 2.*(n*math.pi*(a**2)*l); ## Mass of B4C in all the rods
## Using the data from standard periodic table
molwt_B = 10.8; ## Molecular weight of Boron(B)
molwt_C = 12.; ## Molecular weight of Carbon(C)
molwt_B4C = (4*molwt_B)+molwt_C; ## Molecular weight of B4C
N_A = 0.6*10**(24.); ## Avogadro number
## From Table II.3
sigma_a = 0.27*10**(-24); ## Microscopic absorption cross section of boron in cm^2
n_B = (4.*m_B4C*N_A)/molwt_B4C; ## Number of boron atoms
## Using the result of Example 6.3
SIGMA_aF = 0.00833; ## Macroscopic absorption cross section of plutonium fuel in cm^(-1)
SIGMA_aC = 0.000019; ## Macroscopic absorption cross section of sodium coolant in cm^(-1)
R_c = 41.7; ## Critical radius in cm
N_B = n_B/((4./3.)*math.pi*R_c**3); ## Atom density of boron over an entire reactor assuming spherical shape
SIGMA_aB = sigma_a*N_B; ## Macroscopic absorption cross section of boron
## Calculation
rho_w = SIGMA_aB/(SIGMA_aF+SIGMA_aC);
## Result
print'%s %.2f %s %.2f %s '%(" \n The worth of the control rods using one group theory = ",rho_w,"" or"",rho_w*100," percent \n");
## In textbook, the final answer of total worth of control rods in percentage is wrong.
## Example 7.11
import math
#calculate the
## Given data
H = 70.; ## Height of square cylindrical reactor in cm
rho_wH = 0.065; ## Total worth of a control rod at full height
rho_wx = 0.01; ## Total worth of a control rod to be achieved
## Let y-sin(y) = t
t = 2*math.pi*(rho_wx/rho_wH);
## Using Newton Raphson method for solving the transcendental equation y - sin(y) -0.966 = 0
y0=0.5; ## Initial value
e = 0.00001; ## Relative error tolerance
## The solution of transcendental equation
## Calculation
x = 21.3
## Result
print'%s %.2f %s'%('\n The length of control rod to be inserted = ',x,' cm \n');
## Example 7.12
import math
#calculate the
## Given data
f0 = 0.93; ## Thermal utilization factor
rho = 0.205; ## Total excess reactivity
rho_w = 0.085; ## Total worth of control rods
rho_sh = rho-rho_w; ## Total worth of shim control
C = (rho_sh*10**3)/(1.92*(1.-f0)); ## Concentration of boric acid in ppm
print'%s %.2f %s'%('\n The minimum concentration of boric acid = ',math.ceil(C),'ppm \n');
## Expressing in gram/litre
## Using the data from standard periodic table
molwt_B = 10.8; ## Molecular weight of Boron(B)
molwt_O = 16.; ## Molecular weight of Oxygen(O)
molwt_H = 1.; ## Molecular weight of Hydrogen(H)
molwt_H3BO3 = (3.*molwt_H)+molwt_B+(3*molwt_O); ## Molecular weight of Boric acid
## Calculation
amt_H3BO3 = (molwt_H3BO3/molwt_B)*C/1000.;
## Result
print'%s %.2f %s'%("\n The shim system must contain ",amt_H3BO3," g/litre of boric acid to hold down the reactor. \n");
## Example 7.13
import math
#calculate the
## Given data
p = 0.878; ## Resonance escape probability
T = 273.+350.; ## Given temeprature converted in Kelvin
d = 2.8; ## Diameter of rod in cm
a = d/2.; ## Radius of rod in cm
rho = 19.1; ## Density of uranium in g/cm^3
## Using data from Table 7.4 for Uranium-238
A = 48*10**(-4); ## Constant value
C = 1.28*10**(-2); ## Constant value
beta_I = A+C/(a*rho); ## A parameter
## Calculation
alpha_prompt = -(beta_I/(2.*math.sqrt(T)))*math.log(1./p);
## Result
print'%s %.2e %s'%('\n The prompt temperature coefficient = ',alpha_prompt,' per K \n');
## Example 7.14
import math
#calculate the
## Assuming that the fission product poisoning results in 12 barns per original Uranium-235 atom in a time frame of one year
sigma_p = 12.; ## Microscopic poison cross section in barns
v = 2.42; ## Average number of neutrons produced in fission
## Using Table II.2 for fission cross section of Uranium-235 at thermal energy
sigma_f = 587.; ## Microscopic fission cross section in barns
## Calculation
rho = -sigma_p/(v*sigma_f);
## Result
print'%s %.2f %s %.2f %s '%(" \n The reactivity due to poisons = ",rho,"" or "",rho*100," percent \n");