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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+
+clc
+clear
+//Input data
+F=190;//Each fission of U-235 yeilds in MeV
+a=85;//Assuming the Neutrons absorbed by U-235 cause fission in percentage
+b=15;//Non fission capture to produce an isotope U-236 in percentage
+Q=3000;//The amount of thermal power produced in MW
+
+//Calculations
+E=F*1.60*10^-13;//Each fission yields a useful energy in J
+N=1/E;//Number of fissions required
+B=[(10^6)*(N*86400)]/(a/100);//One day operation of a reactor the number of U-235 nuclei burned is in absorptions per day
+M=(B*235)/(6.023*10^23);//Mass of U-235 consumed to produce one MW power in g/day
+M1=M*3;//Mass of U-235 consumed to produce 3000 MW power in g/day
+
+//Output
+printf('The fuel consumed of U-235 per day = %3.1f g/day ',M1)