# Chapter10-Principles of Nuclear Reactors¶

## Ex1-pg-pg280¶

In [1]:
##Chapter 10, Example 10.1, Page 280
import math
## Thermal utilization factor
Summation = ((0.0055*103.4)+(0.720*687)+(99.2745*2.73))/100
sigma = 0.0034
f = 7.662/(7.662+(sigma*450.))
print'%s %.2f %s'%("Total thermal macroscopic = ",Summation," N^U cm^1\n")
print'%s %.2f %s'%(" f = \n",f,"")
## Answer may vary due to round off error

Total thermal macroscopic =  7.66  N^U cm^1

f =
0.83


## Ex2-pg280¶

In [2]:
##Chapter 10, Example 10.2, Page 280
import math
## Thermal fission factor
neeta = (2.42*587.)/(687. +(2.73*0.98/0.02))
print'%s %.2f %s'%(" Thermal fission factor = \n",neeta,"")
## Answer may vary due to round off error

 Thermal fission factor =
1.73


## Ex3-pg282¶

In [3]:
##Chapter 10, Example 10.3, Page 282
import math
## Find the probability
P = math.exp(-6.85*10**-4*368.)
Pnl = 1./(1.+(578.*6.85*10**-4))
print'%s %.2f %s'%("Fast-neutron nonleakage probability =  \n",P,"")
print'%s %.2f %s'%(" Thermal-neutron nonleakage probability =  \n",Pnl,"")
## Answer may vary due to round off error

Fast-neutron nonleakage probability =
0.78
Thermal-neutron nonleakage probability =
0.72


## Ex4-pg283¶

In [4]:
##Chapter 10, Example 10.4, Page 283
import math
## k of a homogeneous
f = 687./(687. +(0.0034*40000.))
k = 2.07*f
print'%s %.2f %s'%("f =\n",f,"")
print'%s %.2f %s'%(" k =  \n",k,"")
##Answer may vary due to round off error

f =
0.83
k =
1.73


## Ex5-pg284¶

In [5]:
##Chapter 10, Example 10.5, Page 284
import math
L = 578.
T = 368.
Bc = 6.358*10**-4
R = math.sqrt(math.pi**2/Bc)
print'%s %.2f %s'%(" R = ",R," cm \n")
##Answer may vary due to round off error

 R =  124.59  cm



## Ex6-pg285¶

In [6]:
##Chapter 10, Example 10.6, Page 285
import math
## mass of U235
m = (((4./3.)*math.pi*125**3*1.60)*235.)/(40000.*12.)
print'%s %.2f %s'%(" m = ",m*10**-3," kg \n")
##Answer may vary due to round off error

 m =  6.41  kg



## Ex7-pg285¶

In [7]:
##Chapter 10, Example 10.7, Page 285
import math
## Keff
Keff = 1./(1.-0.0065*0.1)
print'%s %.2f %s'%(" Keff =  \n",Keff,"")
##Answer may vary due to round off error

 Keff =
1.00


## Ex8-pg293¶

In [8]:
##Chapter 10, Example 10.8, Page 293
import math
##Resulting reactor period
bt = 0.0065
dt = 0.00065
T = (bt*12.8)/dt
Pt = 10000.
P0 = 10.
t = T*math.log(Pt/P0)
print'%s %.2f %s'%(" Resulting reactor period = ",T," sec \n")
print'%s %.2f %s'%(" t = ",t," sec\n")
##Answer may vary due to round off error

 Resulting reactor period =  128.00  sec

t =  884.19  sec