# Chapter7-The Time Dependent Reactor¶

## Ex1-pg332¶

In :
## 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");


The prompt neutron lifetime =  1.08e-04  seconds



## Ex2-pg333¶

In :
## 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");


The response time of the reactor =  0.10  sec

The reactor power will increase as exp(t/ 0.10 ), where ''t'' denotes the time in seconds



## Ex3-pg337¶

In :
## 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");


The reactivity =  9.99e-04  0.10  percent



## Ex4-pg340¶

In :
## 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");


Reactor period = 57 seconds



## Ex5-pg341¶

In :
## 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");


A dollar is worth 0.0065 in reactivity for Uranium-235 reactor.

Reactivity =  0.15  dollars 15.38  cents



## Ex6-pg345¶

In :
## 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");


The power level drop after 10 minutes =  0.02  MW



## Ex7-pg351¶

In :
## 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");


The worth of a black control rod =  0.07    6.53  percent



## Ex8-pg354¶

In :
## 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.


The total worth of the control rods =  0.29  292.66  percent



## Ex9-pg358¶

In :
## 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.


The total worth of the control rods =  0.04  4.17  percent



## Ex10-pg360¶

In :
## 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.


The worth of the control rods using one group theory =  0.07  6.91  percent



## Ex11-pg362¶

In :
## 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');


 The length of control rod to be inserted =  21.30  cm



## Ex12-pg364¶

In :
## 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");

 The minimum concentration of boric acid =  893.00 ppm

The shim system must contain  5.11  g/litre of boric acid to hold down the reactor.



## Ex13-pg370¶

In :
## 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');

 The prompt temperature coefficient =  -1.38e-05  per K



## Ex14-pg388¶

In :
## 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");


The reactivity due to poisons =  -0.01  -0.84  percent