15: Nuclear Energy Sources

Example number 15.1, Page number 290

In [9]:
#import modules
import math
from __future__ import division

#Variable declaration
#m is mass of neutron and M is mass of other neucleus
ma=1;
Ma=2;
mb=1;
Mb=12;
mc=1;
Mc=238;     

#Calculation
eeta1=(4*ma*Ma/((ma+Ma)**2))*100;      #Maximum fraction of KE lost by a neutron for H2(%)
eeta2=(4*mb*Mb/((mb+Mb)**2))*100;      #Maximum fraction of KE lost by a neutron for C12(%)
eeta3=(4*mc*Mc/((mc+Mc)**2))*100;      #Maximum fraction of KE lost by a neutron for U238(%)

#Result
print "Maximum fraction of KE lost by a neutron for H2 is",round(eeta1,1),"%"
print "Maximum fraction of KE lost by a neutron for C12 is",round(eeta2,1),"%"
print "Maximum fraction of KE lost by a neutron for U238 is",round(eeta3,2),"%"
print "answer for eeta2 given in the book is wrong"

Maximum fraction of KE lost by a neutron for H2 is 88.9 %
Maximum fraction of KE lost by a neutron for C12 is 28.4 %
Maximum fraction of KE lost by a neutron for U238 is 1.67 %
answer for eeta2 given in the book is wrong

Example number 15.2, Page number 291

In [13]:
#import modules
import math
from __future__ import division

#Variable declaration
E=200;     #energy released per fission(MeV)
e=1.6*10**-19;    #the charge on electron(C)
Na=6.02*10**26;   #Avgraodo no.(per kg mole)

#Calculation
CE=E*e*10**6;     #conversion in J
RF=1/CE;        #fission rate(fissions/second)
Ekg=CE*Na/235;     #Energy realeased in complete fission of 1 kg(J)

#Result
print "fission rate is",round(RF/10**10,1),"*10**10 fissions/second"
print "Energy realeased in complete fission of 1 kg is",round(Ekg/1e+13,1),"*10**13 J"

fission rate is 3.1 *10**10 fissions/second
Energy realeased in complete fission of 1 kg is 8.2 *10**13 J

Example number 15.3, Page number 291

In [16]:
#import modules
import math
from __future__ import division

#Variable declaration
R=3*10**7;       #rate of energy development(Js)
E=200;           #energy released per fission(MeV)
e=1.6*10**-19;    #the charge on electron(C)
t=1000;           #time(hours)
Ekg=8.2*10**13;   #energy released per kg of U-235

#Calculation
CE=E*e*10**6;     #conversion in J
n=R/CE;         #no of atoms undergoing fission/second
TE=R*t*3600;      #energy produced in 1000 hours(J)
MC=TE/Ekg;        #mass consumed(kg)  

#Result
print "number of atoms undergoing fissions per second is",round(n/1e+17,1),"*10**17"
print "mass consumed is",round(MC,2),"kg"

number of atoms undergoing fissions per second is 9.4 *10**17
mass consumed is 1.32 kg

Example number 15.4, Page number 292

In [18]:
#import modules
import math
from __future__ import division

#Variable declaration
EPF=180;       #Energy consumed per disintegration(MeV)
E=1200;        #average power(kW)
t=10;          #time(hours)
Na=6.02*10**26;      #Avgraodo no.(per kg mole)
e=1.6*10**-19;       #the charge on electron(C)

#Calculation
TE=E*t;    #energy consumed(kWh)
TE=TE*36*10**5;     #conversion(J)
EE=TE/0.2;          #efficient energy
CE=EPF*e*10**6;     #conversion in J
n=EE/CE;
m=235*n/Na*1000;     #mass consumed(gram)

#Result
print "mass consumed is",round(m,2),"gram"

mass consumed is 2.93 gram

Example number 15.5, Page number 292

In [22]:
#import modules
import math
from __future__ import division

#Variable declaration
OE=200;       #output power(MW)
E=200;        #energy released per fission(MeV)
WF=3.1*10**10;     #fission rate(fissions/second)
Na=6.02*10**26;    #Avagadro no.(per kg mole)

#Calculation
IE=OE/0.3*10**6;     #reactor input(W)
TFR=WF*IE;
n=TFR*24*3600;      #no. of U-235 for one day
m=235*n/Na;         #mass required(kg)
   
#Result
print "amount of natural uranium consumed per day is",round(m*100/0.7,3),"kg"

amount of natural uranium consumed per day is 99.577 kg

Example number 15.6, Page number 292

In [25]:
#import modules
import math
from __future__ import division

#Variable declaration
AE=100;       #electrical power(MW)
E=200;        #energy released per fission(MeV)
e=1.6*10**-19;     #the charge on electron(C)
Na=6.02*10**26;    #Avagadro no.(per kg mole)

#Calculation
TE=AE*10**6*24*3600;     #energy consumed in city in one day(J)
EE=TE/0.2;
CE=E*e*10**6;          #conversion in J
n=EE/CE;               #no. of atoms to be fissioned  
m=235*n/Na;           #amount of fuel required(kg)

#Result
print "amount of fuel required is",round(m,2),"kg"

amount of fuel required is 0.53 kg

Example number 15.7, Page number 293

In [32]:
#import modules
import math
from __future__ import division

#Variable declaration
OE=3000;       #output power(MWh)
E=200;         #energy released per fission(MeV)
e=1.6*10**-19;       #the charge on electron(C)
Na=6.02*10**26;      #Avagadro no.(per kg mole)

#Calculation
IE=OE/0.2;        #nuclear energy input(MWh)
TE=IE*36*10**8;     #conversion in J
CE=E*e*10**6;       #conversion in J
n=TE/CE;            #number of nuclides required per day
m=235*n/Na;         #daily fuel requirement(kg)

#Result
print "daily fuel requirement is",round(m,3),"kg or",round(m,3)*1000,"gram"
print "answer given in the book varies due to rounding off errors"

daily fuel requirement is 0.659 kg or 659.0 gram
answer given in the book varies due to rounding off errors

Example number 15.8, Page number 293

In [42]:
#import modules
import math
from __future__ import division

#Variable declaration
OP=32824;    #output power(kW)
E=200;       #energy released per fission(MeV)
Ekg=8.2*10**13;     #energy released per kg of U-235(J)

#Calculation
DOP=OP*1000*24*3600;      #daily output power(J)
IP=DOP/0.2;               #nuclear energy input(J)
DFC=IP/Ekg;            #daily fuel consumption(kg)
DI=DOP/(0.8*4186);     #daily input at 80% efficiency(kcal)
Crqd=DI/(7*10**3);     #Coal required per day(tonnes)

#Result
print "Daily fuel consumption is",round(DFC,3)*1000,"gram"
print "Coal required per day is",int(Crqd),"tonnes"
print "answer for coal required per day Crqd given in the book is wrong"

Daily fuel consumption is 173.0 gram
Coal required per day is 120981 tonnes
answer for coal required per day Crqd given in the book is wrong