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+//Chapter 7 Example 3//
+clc
+clear
+// useful energy=e1,energy in terms of joules=e//
+e1=190;// in MeV//
+e=e1*10^6*1.6*10^-19;
+printf("\n Energy in terms of joules = %.15f J\n",e);
+// number of fisions required to produce one joule=n//
+n=1/e;
+printf("\n No of fissions required = %.3f \n",n);
+// number of nuclei burnt during 1 hr per MW of power=n1,percent of neutrons absorbed=p //
+p=80;// in percent//
+n1=10^6*n*3600/(p/100);
+printf("\n Number of nuclei burnt during 1hr per MW of power  = %.3f absorption/hr \n",n1);
+// Mass of U-235 consumed to produce 1MW of power=m,Avagadro number=A//
+A=6.023*10^23;
+m=235;// Atomic mass of uranium//
+m=n1*m/A;// this is for 1MW//
+m1=m*100// for 100MW//
+printf("\n Fuel Consumption to produce 100MW = %.4f g/hr\n",m1);
+
+
+ 
+\ No newline at end of file