Ch-9, Nuclear Power Stations

example 9.1 Page 169

In [1]:
m=1*10**-3#mass of 1 grm in kgs
c=3*10**8
e=m*c**2 
E=e/(1000*3600)
print "energy equivalent of 1 gram is %dkWh"%E

energy equivalent of 1 gram is 25000000kWh

example 9.2 Page 169

In [2]:
amu=1.66*10**-27#mass equvalent in kgs
c=3*10**8
j=6.242*10**12
e=amu*c**2
E=e*j 
print " energy evalent in joules is %ejoules \n energy equvalent in Mev is %dMeV \n hense shown"%(e,E)

 energy evalent in joules is 1.494000e-10joules 
 energy equvalent in Mev is 932MeV 
 hense shown

example 9.3 Page 169

In [3]:
hm=2.0141
hp=1.007825
hn=1.008665
nm=58.9342
np=28
nn=59
um=235.0439
up=92
un=235
hmd=hp+hn-hm ;nmd=np*hp+(nn-np)*hn-nm ;umd=up*hp+(un-up)*hn-um 
hbe=931*hmd ;nbe=931*nmd; ube=931*umd 
ahbe=hbe/2 ;anbe=nbe/nn ;aube=ube/un 
print "\t(a)\n mass defect is for hydrogen %famu \n total binding energy for hydrogens %fMev \n average binding energy for hydrogen is %fMeV"%(hmd,hbe,ahbe)
print "\n\t(b)\n mass defect is for nickel %famu \n total binding energy for nickel is %fMev \n average binding energy for nickelis %fMeV"%(nmd,nbe,anbe)
print "\n\t(c)\n mass defect of uranium is %famu \n total binding energy uranium is %fMev \n average binding energy uranium is %fMeV"%(umd,ube,aube)

	(a)
 mass defect is for hydrogen 0.002390amu 
 total binding energy for hydrogens 2.225090Mev 
 average binding energy for hydrogen is 1.112545MeV

	(b)
 mass defect is for nickel 0.553515amu 
 total binding energy for nickel is 515.322465Mev 
 average binding energy for nickelis 8.734279MeV

	(c)
 mass defect of uranium is 1.915095amu 
 total binding energy uranium is 1782.953445Mev 
 average binding energy uranium is 7.587036MeV

example 9.4 Page 170

In [4]:
from math import exp
no=1.7*10**24
hl=7.1*10**8
t=10*10**8
lm=0.693/(hl)
lmda=lm/(8760*3600)
ia=lmda*no
n=no*(exp(-lm*t))
print "(lamda) disintegrations per sec is %ebq \n initial activity is lamda*na is %ebq \n final number of atoms is %eatoms"%(lmda,ia,n)

(lamda) disintegrations per sec is 3.095054e-17bq 
 initial activity is lamda*na is 5.261592e+07bq 
 final number of atoms is 6.405500e+23atoms

example 9.5 Page 170

In [5]:
um=5
owp=2.6784*10**15
an=6.023*10**23
na1g=an/235
na5g=an*5/235
p=na5g/owp
print " 1 watt power requvires %efussions per day \n number of atoms in 5 gram is %eatoms \n power is %eMW "%(owp,na5g,p)

 1 watt power requvires 2.678400e+15fussions per day 
 number of atoms in 5 gram is 1.281489e+22atoms 
 power is 4.784533e+06MW

example 9.6 Page 171

In [6]:
pp=235
pe=0.33
lf=1
teo=pp*8760*3600*10**6
ei=teo/pe
nfr=3.1*10**10#fessions required
tnfr=nfr*ei
t1gu=2.563*10**21 #total uranium atoms in 1 grm
fure=tnfr/t1gu
print "total energy input %eWatt sec \n energy input is %eWatt-sec\n total number of fissions required is %efissions \n fuel required is %e grams %dkg"%(teo,ei,tnfr,fure,fure/1000)

total energy input 7.410960e+15Watt sec 
 energy input is 2.245745e+16Watt-sec
 total number of fissions required is 6.961811e+26fissions 
 fuel required is 2.716274e+05 grams 271kg

example 9.7 Page 171

In [7]:
from __future__ import division
from math import log,exp
en=3*10**6
a=12
fen=0.1
Es=2/(12+2/3)
re=exp(Es)
print "(a)\nratio of energies per collision is %f"%(re)
rietf=en/fen
ldie=log(rietf)
nc=ldie/Es
print "(b)\npatio of iniial to final energies is %e \nlogarithemic decrement in energy is %f \nnumber of collisions is %d"%(rietf,ldie,nc)

(a)
ratio of energies per collision is 1.171043
(b)
patio of iniial to final energies is 3.000000e+07 
logarithemic decrement in energy is 17.216708 
number of collisions is 109