# Chapter8-Heat Removal from Nuclear Reactors¶

## Ex1-pg407¶

In :
## Example 8.1
import math

## Given data
P = 3025.;                                 ## Reactor thermal power in MW
w = 136.3*10**6;                           ## Coolant flow rate in lb/hr
## According to Table 1.9
## 1 kW = 3412 Btu/hr
q = P*1000.*3412.;                         ## Converting into Btu/hr
delh = q/w;                              ## Rise in enthalpy
## Using the data from Table IV.1 for temperature 542.6 F
hin = 539.7;                              ## Enthalpy of input water in Btu/lb
## Calculation
hout = hin+delh;                          ## Enthalpy of released water in Btu/lb
## From Table IV.1
## Result
print'%s %.2f %s'%(" \n ",hout," Btu/lb corresponds to 599 F coolant water temperature. \n");


615.42  Btu/lb corresponds to 599 F coolant water temperature.



## Ex2-pg408¶

In :
## Example 8.2
import math

## Given data
P = 6.895;                                ## Pressure of steam in MPa
w = 2.93*10**6;                            ## Steam flow rate in kg/hr
Tin = 190.6+273.;                          ## Inlet temperature in Kelvin

## 1.
## Using the data from Table IV.2
## Result
print(" \n At a pressure of 6.895 MPa the steam temeperature is 284.86 C \n");

## 2.
## Using the data from Table IV.2
hout = 2773.2;                            ## Enthalpy of spent steam in kJ/kg
## Using the data from Table IV.1
hin = 807.8;                              ## Enthalpy of inlet steam at Tin in kJ/kg
## Calculation
q = w*(hout-hin);
## Result
print'%s %.2e %s %.2f %s '%(" \n Reactor power is ",q," J/hr" or "",q/(3600*1000),"MW \n");


At a pressure of 6.895 MPa the steam temeperature is 284.86 C

Reactor power is  5.76e+09  J/hr 1599.62 MW



## Ex3-pg413¶

In :
## Example 8.3
import math

## Given data
n = 193.*204.;                              ## Total number of fuel rods in the reactor
## 1 feet  = 12 inches
R = 67./12.;                                ## Outer radius of the cylinder in feet(ft)
H = 144./12.;                               ## Outer radius of the cylinder in ft
d = 0.42/12.;                              ## Diameter of the fuel rod in ft
a = d/2.;                                  ## Radius of the fuel rod in ft
P = 1893.;                                 ## Reactor thermal power in MW
Ed = 180.;                                 ## Energy deposited locally in the fuel per fission in MW(Assumption)
ER = 200.;                                 ## Recoverable energy per fission in MW(Assumption)

## 1.
## Calculation
## According to Table 1.9
## 1 kW=3412 Btu/hr
q_r = (2.32*P*Ed)/(n*ER);
q_max=(q_r*3412.*1000.)/(2.*H*a**2);
## Result
print'%s %.2e %s '%(" \n Total energy production at the axis = ",q_r*3412*1000," Btu/hr"); print'%s %.2e %s'%(" \n Maximum energy production at the axis =",q_max,"Btu/hr-ft^3 \n");

## 2.
r = 20./12.;                                   ## Distance from the axis in ft
j0 =0.825             ## Bessel function
## Calculation
## According to Table 1.9
## 1 kW=3412 Btu/hr
q_r20 = q_r*j0;
q_max20 = (q_r20*3412.*1000.)/(2*H*a**2);
## Result
print'%s %.2f %s'%(" \n Total energy production at a distance of 20 inches =  ",(q_r20*3412*1000)," Btu/hr");
print'%s %.2f %s'%(" \n Maximum energy production at a distance of 20 inches = ",q_max20," Btu/hr-ft^3 \n");


Total energy production at the axis =  3.43e+05  Btu/hr

Maximum energy production at the axis = 4.66e+07 Btu/hr-ft^3

Total energy production at a distance of 20 inches =   282589.88  Btu/hr

Maximum energy production at a distance of 20 inches =  38447602.54  Btu/hr-ft^3



## Ex4-pg416¶

In :
## Example 8.4
import math

## Given data
P0 = 825.;                                     ## Reactor thermal power in MW
t0 = 1.5*3.16*10**7;                           ## Reactor operation time in seconds
ts = 0.1;                                     ## Reactor shutdown time in seconds

## 1.
## Let P/P0 = q
## From Figure 8.3
q_ts = 0.07;                                  ## Fission product to decay power during shutdown time
q_t0 = 0.0007;                                ## Fission product to decay power after operating time
q = q_ts-q_t0;                                ## Net fission product to decay power
## Calculation
P = q*P0;
## Result
print'%s %.2f %s'%(" \n Decay energy at shutdown = ",P," MW");

## One hour after shutdown
ts1 = 3.6*10**3;                                ## Reactor shutdown time in seconds
## Let P/P0=q
## From Figure 8.3
q_ts1 = 0.014;                                 ## Fission product to decay power at shutdown time
q_t0 = 0.0007;                                ## Fission product to decay power after operating time
q1 = q_ts1-q_t0;
## Calculation
P1 = q1*P0;
## Result
print'%s %.2f %s'%(" \n Decay energy one hour after shutdown = ",P1," MW");

## One year after shutdown
ts2 = 3.16*10**7;                               ## Reactor shutdown time in seconds
## Let P/P0=q
## From Figure 8.3
q_ts2 = 0.00079;                               ## Fission product to decay power at shutdown time
## Now the operating time is t0+ts2 which can be denoted by t01
q_t01 = 0.00063;                               ## Fission product to decay power after operating time
q2 = q_ts2-q_t01;
## Calculation
P2 = q2*P0;
## Result
print'%s %.2f %s'%(" \n Decay energy one year after shutdown = ",P2," MW \n");

## 2.
C = 0.88;                                     ## Conversion factor
## Using data from Table II.2
sigma_a25 = 681.;                              ## Microscopic absorption cross section in barns
sigma_f25 = 582.;                              ## Microscopic fission cross section in barns
## At shutdown time
P_29 = (2.28*10**(-3)*C*(sigma_a25/sigma_f25))*P0;
P_39 = (2.17*10**(-3)*C*(sigma_a25/sigma_f25))*P0;
print'%s %.2f %s %.2f %s '%("  \n Decay energy at shutdown with effect of Uranium-239 and Neptunium-239 decay =",P_29," MW" and"",P_39," MW respectively");

## One hour after shutdown
ts1 = 3600.;                                   ## TIme in seconds
P_291 = P_29*math.exp(-4.9*10**(-4)*ts1);
P_391 = P_39*(math.exp(-3.41*10**(-6)*ts1)-(7*10**(-3)*math.exp(-4.9*10**(-4)*ts1)));
print'%s %.2f %s %.2f %s ' %(" \n Decay energy one hour after shutdown with effect of Uranium-239 and Neptunium-239 decay = ",P_291," MW" and "",P_391," MW respectively");

## One year after shutdown
P_292 = 0.;                                    ## Half life of Uranium-239 is 23.5 minutes
P_392 = 0.;                                    ## Half life of Neptunium-239 is 2.35 days
print'%s %.2f %s %.2f %s '%("  \n Decay energy one year after shutdown with effect of Uranium-239 and Neptunium-239 decay = ",P_292," MW "and "",P_392," MW respectively");
## There is a slight deviation in the values as compared with the texbook. This is because of approximation of difference values in the textbook.


Decay energy at shutdown =  57.17  MW

Decay energy one hour after shutdown =  10.97  MW

Decay energy one year after shutdown =  0.13  MW

Decay energy at shutdown with effect of Uranium-239 and Neptunium-239 decay = 1.94  1.84  MW respectively

Decay energy one hour after shutdown with effect of Uranium-239 and Neptunium-239 decay =  0.33  1.82  MW respectively

Decay energy one year after shutdown with effect of Uranium-239 and Neptunium-239 decay =  0.00  0.00  MW respectively


## Ex5-pg426¶

In :
## Example 8.5
import math

## Given data
## 1 feet  = 12 inches
d = 0.42/12.;                              ## Diameter of the fuel rod in feet(ft)
a = d/2.;                                  ## Radius of the fuel rod in ft
b = 0.024/12.;                             ## Thickness of Zircaloy-4 clad in ft
H = 12.;                                   ## Length of fuel rod in ft
T_m = 3970.;                               ## Center temperature of fuel in F

## 1.
## Using the result of Example 8.3
q_max = 4.66*10**7;                        ## Maximum heat flux at the center of the rod in Btu/hr-ft^3
## Calculation
q_bar = (a**2*q_max)/(2*(a+b));
## According to Table 1.9
## 1 kW=3412 Btu/hr
## Result
print'%s %.2f %s %.2f %s ' %("  \n Heat flux of the fuel rod = ",q_bar," Btu/hr-ft^2" or "",(q_bar*1000)/(3412*30.48**2)," W/cm^2 \n");

## 2.
## Using the data from Table IV.6
k_f = 1.1;                                ## Thermal conductivity of fuel rod in Btu/hr-ft-F
k_c = 10.;                                 ## Thermal conductivity of cladding in Btu/hr-ft-F
R_f = 1./(4.*math.pi*H*k_f);                    ## Thermal resistance of fuel in F-hour/Btu
R_c = math.log(1.+(b/a))/(2.*math.pi*H*k_c);         ## Thermal resistance of cladding in F-hour/Btu
## Calculation
T_c = T_m-(q_bar*2*math.pi*(a+b)*H*(R_f+R_c));
## Result
print'%s %.2f %s'%(" \n Outer temperature of cladding = ",math.ceil(T_c)," F \n");


Heat flux of the fuel rod =  365929.49  Btu/hr-ft^2 115.44  W/cm^2

Outer temperature of cladding =  650.00  F



## Ex6-pg431¶

In :
## Example 8.6
import math

## Given data
h = 7500.;                                     ## Heat transfer coefficient in Btu/hr-ft^2-F
## Using the result of Example 8.5
q_bar = 3.66*10**5;                            ## Heat flux of the fuel rod in Btu/hr-ft^2
T_c = 650.;                                    ## Outer temperature of cladding in F
## Calculation
T_b = T_c-(q_bar/h);
## Result
print'%s %.2f %s'%(" \n Temperature of water with respect to the midpoint of the hottest fuel rod = ",T_b," F \n");


Temperature of water with respect to the midpoint of the hottest fuel rod =  601.20  F



## Ex7-pg435¶

In :
## Example 8.7
import math

## Given data
T_b0 = 543.;                                ## Temperature of inlet coolant in F
w = 3148.;                                 ## Coolant rate per channel in lb/hr

## 1.
V_f = 1.15*10**(-2);                       ## Volume of fueled portion in ft^2
## Using the result of Example 8.3
q_max = 4.66*10**7;                        ## Maximum heat flux at the center of the rod in Btu/hr-ft^3
## From Table IV.3
c_p = 1.3;                                ## Specific heat at constant pressure in Btu/lb-F
## Calculation
T_bmax = T_b0+((2*q_max*V_f)/(math.pi*w*c_p));
## Result
print'%s %.2f %s'%(" \n Exit temperature of the coolant = ",T_bmax," F \n");

## 2.
## Using the data of Example 8.6
h = 7500.;                                 ## Heat transfer coefficient in Btu/hr-ft^2-F
## Using the data of Example 8.5
d = 0.42/12.;                              ## Diameter of the fuel rod in feet(ft)
a = d/2.;                                  ## Radius of the fuel rod in ft
b = 0.024/12.;                             ## Thickness of Zircaloy-4 clad in ft
H = 12.;                                   ## Length of fuel rod in ft
A = 2*math.pi*(a+b)*H;                        ## Area of the assumed cylinder in ft^2
R_h = 1/(h*A);                            ## Convective resistance in F-hour/Btu
alpha = math.pi*w*c_p*R_h;                    ## A parameter

## Using the result from Example 8.5
R_f = 6.03*10**(-3);                       ## Thermal resistance of fuel
R_c = 1.43*10**(-4);                       ## Thermal resistance of cladding
R = R_f+R_c+R_h;                          ## Total resistance
bet = math.pi*w*c_p*R;                        ## A parameter denoted by 'beta'
## Calculation
T_cmax = T_b0+((q_max*V_f*R_h)*((1+math.sqrt(1+alpha**2))/alpha));
T_mmax = T_b0+((q_max*V_f*R)*((1+math.sqrt(1+bet**2))/bet));
## Result
print'%s %.2f %s %.2f %s '%("  \n Maximum temperature of cladding and fuel = ",math.ceil(T_cmax)," F" and "",T_mmax," F respectively. \n");


Exit temperature of the coolant =  626.37  F

Maximum temperature of cladding and fuel =  649.00  3941.65  F respectively.



## Ex8-pg438¶

In :
## Example 8.8
import math

## Given data
d = 0.42;                                 ## Diameter of the fuel rod in inches
b = 0.024;                                ## Thickness of Zircaloy-4 clad in inches
v = 15.6*3600.;                            ## Speed of fluid in feet/hour
a = (d/2.)+b;                              ## Radius of fuel rods in inches
P = 2000.;                                 ## Pressure of water in psi
T = 600.;                                  ## Water temperature in F
## Using the data from example 8.5
s = 0.6;                                  ## Pitch of square array in inches
D_e = 2.*((s**2-(math.pi*a**2))/(math.pi*a));        ## Equivalent diameter in inches
## Converting the units in terms of feet
D_e = D_e/12.;
## Using tha Table IV.3 at given T and P value
rho = 42.9;                               ## Density of fluid in ft/hr
mu = 0.212;                               ## Viscosity of fluid in lb/hr-ft
## Calculation
Re = (D_e*v*rho)/mu;
## Result
print'%s %.2f %s'%(" \n Reylonds number = ",Re," \n");
if Re >= 10000:
print(" \n As the reylonds number is greater than 10000, the flow is turbulent. \n");

## The value is different as compared to the textbook value. This is due to approximation of Reynolds number in the textbook.


Reylonds number =  484329.34

As the reylonds number is greater than 10000, the flow is turbulent.



## Ex9-pg440¶

In :
## Example 8.9
import math

## Using the data from Example 8.8
s = 0.6;                                  ## Pitch of square lattice in inches
d = 0.42;                                 ## Diameter of the fuel rod in inches
b = 0.024;                                ## Thickness of Zircaloy-4 clad in inches
a = (d/2)+b;                              ## Radius of fuel rods in inches
D_e = 0.0427;                             ## Equivalent diameter in feet
Re = 484329;                              ## Reynolds number
PD = s/(2*a);                             ## The ratio of pitch to diameter of fuel rod
## For a square lattice
C = 0.042*PD-0.024;                       ## A constant

## According to Table IV.3
c_p = 1.45;                               ## Specific heat at constant pressure in Btu/lb-F
mu = 0.212;                               ## Viscosity of fluid in lb/hr-ft
k = 0.296;                                ## Conductivity of fluid in Btu/hr-ft F
Pr=(c_p*mu)/k;                          ## Prandtl number
## The constants are assumed as
m = 0.8;
n = 1/3;
## Calculation
h = C*(k/D_e)*(Re)**m*Pr**n;
## Result
print'%s %.2f %s'%(" \n Heat transfer coefficient = ",h," Btu/hr-ft^2-F\n");
## The value is different as compared to the textbook value. This is due to approximation of Reynolds number in the textbook and in this problem actual value is considered.


Heat transfer coefficient =  7309.11  Btu/hr-ft^2-F



## Ex10-pg448¶

In :
## Example 8.10
import math

## Using the data from Example 8.3 to 8.8
P = 2000.;                                 ## Pressure in psi
v = 15.6;                                 ## Coolant velocity in ft/sec
D_e = 0.0427;                             ## Equivalent diameter in ft
d = 0.42;                                 ## Diameter of the fuel rod in inches
b = 0.024;                                ## Thickness of Zircaloy-4 clad in inches
a = (d/2.)+b;                              ## Radius of fuel rods in inches
T_b = 600.;                                ## Bulk temeperature in F

## 1.
## Using Bernath correlation
## Calculation
T_wc = 102.6*math.log(P)-((97.2*P)/(P+15.))-(0.45*v)+32.;
## Result
print'%s %.2f %s'%(" \n Cladding temeperature = ",T_wc," F\n");

## 2.
D_i = (2.*math.pi*a)/(math.pi*12.);                  ## Heated perimeter is (2*%pi*a)/12 in feet
## Calculation
h_c = 10890.*((D_e)/(D_e+D_i))+((48.*v)/D_e**0.6);
## Result
print'%s %.2f %s'%(" \n Heat transfer coefficient = ",h_c," Btu/hr-ft^2-F\n");

## 3.
## Calculation
q_c = h_c*(T_wc-T_b);
## Result
print'%s %.2e %s'%(" \n Critical heat flux = ",q_c," Btu/hr-ft^2\n");
## In the textbook, the unit of critical heat flux is wrong.



Heat transfer coefficient =  10658.76  Btu/hr-ft^2-F

Critical heat flux =  1.15e+06  Btu/hr-ft^2



## Ex11-pg454¶

In :
## Example 8.11
import math

## Given data
m_lbar = 0.457;                        ## Average linear density of UO2 in lb/ft
sigma = 0.0122;                        ## Standard deviation of set of measured linear densities of UO2
## Calculation
F_Eml = 1.+(3.*sigma)/m_lbar;
print'%s %.2f %s'%(" \n Engineering subfactor for the amount of fissile material = ",F_Eml," \n");


Engineering subfactor for the amount of fissile material =  1.08



## Ex12-pg457¶

In :
## Example 8.12
import math

q_max = 539000.;                               ## Maximum heat flux  Btu/hr-ft^2
F = 2.8;                                      ## Hot channel factor
P = 3000.;                                     ## Reactor thermal power in MW
## Expressing in Btu/hr
## According to Table 1.9, 1 kW = 3412 Btu/hr
P = P*3.412*10**6;                             ## Reactor thermal power in Btu/hr
l = 12.;                                       ## Length of fuel rod in ft
d = 0.5/12.;                                   ## Diameter of fuel rod in ft
r = d/2.;                                      ## Radius of fuel rod in ft

## 1.
q_av = q_max/F;
## Calculation
A = P/q_av;
## Result
print'%s %.2f %s'%(" \n Total heat transfer area = ",A," ft^2\n");

## 2.
A_one = 2.*math.pi*r;                              ## The total surface area of one fuel rod
## Calculation
n = A/A_one;
## Result
print'%s %.2f %s'%(" \n Number of fuel rods = ",n," \n");
## The value is different as compared to the textbook value. This is due to approximation of total heat transfer area in the textbook and in this problem actual value is considered. As total heat transfer area is further used to calculate number of fuel rods, therefore the difference in value exists.


Total heat transfer area =  53174.03  ft^2

Number of fuel rods =  406219.64