# Ch-9, Nuclear Power Stations¶

## example 9.1 Page 169¶

In :
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 :
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 :
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 :
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 :
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 :
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 :
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