initial commit / add all books

7 files changed, 124 insertions, 0 deletions

diff --git a/1985/CH10/EX10.1/Chapter10_example1.sce b/1985/CH10/EX10.1/Chapter10_example1.sce
new file mode 100755
index 000000000..dd2bad9cf
--- /dev/null
+++ b/1985/CH10/EX10.1/Chapter10_example1.sce
@@ -0,0 +1,16 @@
+clc

+clear

+//Input data

+mU235=235.044//Mass of U235 in a.m.u

+mXe135=134.907//Mass of Xe135 in a.m.u

+mMo98=97.906//Mass of Mo98 in a.m.u

+mn=1.008665//Mass of neutron in a.m.u

+

+//Calculations

+LHS=mU235+mn//The total mass of the reactants in a.m.u

+RHS=mMo98+mXe135+3*mn//The total mass of the products in a.m.u

+md=LHS-RHS//Mass defect in a.m.u

+E=(md*934.18)//Energy released in MeV

+

+//Output

+printf('The energy released in the nuclear fission reaction is %3i MeV',E)

diff --git a/1985/CH10/EX10.2/Chapter10_example2.sce b/1985/CH10/EX10.2/Chapter10_example2.sce
new file mode 100755
index 000000000..5e0e5fcd6
--- /dev/null
+++ b/1985/CH10/EX10.2/Chapter10_example2.sce
@@ -0,0 +1,14 @@
+clc

+clear

+//Input data

+E=200//Energy released in the fission of U235 in MeV

+e=1.6*10^-19//Charge of electron in Coulumb

+A=6.023*10^23//Avagadros number 

+a=235//U235

+

+//Calculations

+x=(A/a)//Number of atoms in 1 gram of U235

+E=((x*E*e*10^6)/(3.6*10^6))/10^4//Energy released by 1 gm of U235 in kWh

+

+//Output

+printf('Energy released by 1 gm of U235 is %3.2f*10^4 kWh',E)

diff --git a/1985/CH10/EX10.3/Chapter10_example3.sce b/1985/CH10/EX10.3/Chapter10_example3.sce
new file mode 100755
index 000000000..2202fcd30
--- /dev/null
+++ b/1985/CH10/EX10.3/Chapter10_example3.sce
@@ -0,0 +1,12 @@
+clc

+clear

+//Input data

+Ef=200//Energy released per fission in MeV

+Er=32*10^6//Energy produced by the reactor in W

+e=1.6*10^-19//Charge of electron in Coulumb

+

+//Calculations

+n=(Er/(Ef*10^6*e))/10^18//Number of U235 nuclei needed to produce an energy of 32*10^6 J/s *10^18

+

+//Output

+printf('%3.0f*10^18 U235 nuclei are needed to produce an energy of 32*10^6 J/s',n)

diff --git a/1985/CH10/EX10.4/Chapter10_example4.sce b/1985/CH10/EX10.4/Chapter10_example4.sce
new file mode 100755
index 000000000..c42178540
--- /dev/null
+++ b/1985/CH10/EX10.4/Chapter10_example4.sce
@@ -0,0 +1,15 @@
+clc

+clear

+//Input data

+E=100*10^3//Energy produced by the reactor in W

+e=1.6*10^-19//Charge of electron in Coulumb

+A=6.023*10^23//Avagadros number 

+a=235//U235

+

+//Calculations

+Er=200//Let the energy released per fission be 200 MeV,

+n=(E/(Er*10^6*e))//The number of U235 nuclei needed to produce 100kW of energy

+m=((a*n)/(A*1000))/10^-9//Mass of 'n' atoms of U235

+

+//Output

+printf('The reactor consumes %3.5f*10^-9 kg of U235 in one second',m)

diff --git a/1985/CH10/EX10.5/Chapter10_example5.sce b/1985/CH10/EX10.5/Chapter10_example5.sce
new file mode 100755
index 000000000..15d505a6e
--- /dev/null
+++ b/1985/CH10/EX10.5/Chapter10_example5.sce
@@ -0,0 +1,20 @@
+clc

+clear

+//Input data

+n=30//Efficiency of the reactor in percent

+Ef=200//Energy released per fission in MeV

+E=200//Energy needed to the city in MW

+e=1.6*10^-19//Charge of electron in Coulumb

+A=6.023*10^23//Avagadros number 

+a=235//U235

+

+//Calculations

+E1=E*10^6//Energy required to the city in J/s

+E2=E1*24*60*60//Energy required to the city for one day in J

+I=(E2/n)*100//Useful input in J

+Ef2=(Ef*10^6*e)//Energy released per fission in J

+n=(I/Ef2)//Number of nucei required to produce 'I' J of energy 

+m=((a*n)/(A*1000))//Mass of 'n' atoms of U235 in kg

+

+//Output

+printf('The amount of fuel required for one day operation of he reactor is %3.4f kg',m)

diff --git a/1985/CH10/EX10.6/Chapter10_example6.sce b/1985/CH10/EX10.6/Chapter10_example6.sce
new file mode 100755
index 000000000..3e56d7197
--- /dev/null
+++ b/1985/CH10/EX10.6/Chapter10_example6.sce
@@ -0,0 +1,22 @@
+clc

+clear

+//Input data

+mH=2.01478//Mass of Hydrogen (1H2) in a.m.u

+mHe=4.00388//Mass of Helium (He4) in a.m.u

+n=20//Efficiency in percent

+O=10000//Output of the reactor in kW

+e=1.6*10^-19//Charge of electron in Coulumb

+A=6.023*10^23//Avagadros number 

+

+//Calculations

+md=(2*mH-mHe)//Mass defect in a.m.u

+E=(md*931.48)//Energy released in MeV

+O1=(O*1000)//Output of the reactor in J/s

+E1=(O1*24*60*60)//Energy released by the reactor in one day in J

+I=(E1/n)*100//Useful input in J

+N=(I*2/(E*10^6*e))//Number of deuterons required to release an energy of 'I' J

+m=((2*N)/A)//Mass of 'N' atoms of 1H2 in gm

+

+//Output

+printf('The reactor consumes %3.3f*10^-3 kg of deuteron in one day',m)

+

diff --git a/1985/CH10/EX10.7/Chapter10_example7.sce b/1985/CH10/EX10.7/Chapter10_example7.sce
new file mode 100755
index 000000000..e849b0f8a
--- /dev/null
+++ b/1985/CH10/EX10.7/Chapter10_example7.sce
@@ -0,0 +1,25 @@
+clc

+clear

+//Input data

+mH1=1.007825//Mass of 1H1 in a.m.u

+mH2=2.014102//Mass of 1H2 in a.m.u

+mHe3=3.01603//Mass of 2He3 in a.m.u

+mHe4=4.002603//Mass of 2He4 in a.m.u

+

+//Calculations

+//For Eq.(i)

+md1=(2*mH1)-mH2//Mass defect in a.m.u. Mass defect in the textbook is wrong since 2*1.007825 is taken as 2.014650 instead of 2.015650

+E1=md1*931.48//Energy released in MeV

+

+//For Eq.(ii)

+md2=(mH1+mH2)-mHe3//Mass defect in a.m.u

+E2=md2*931.48//Energy released in MeV

+

+//For Eq.(iii)

+md3=(2*mHe3-mHe4-2*mH1)//Mass defect in a.m.u. Mass defect in the textbook is wrong since 2*1.007825 is taken as 2.014650 instead of 2.015650

+E3=md3*931.48//Energy released in MeV

+

+E=(E1+E2+E3)//Total energy released in the above reactions in MeV

+

+//Output

+printf('Total energy released in the above reactions is %3.4f MeV',E)