Answer:
[tex]C_{lifetime} = 2,618,017,174\,USD[/tex]
Explanation:
The energy contained per second in the fuel is:
[tex]\dot E_{fuel} = \frac{800\times 10^{6}\,W}{0.29}[/tex]
[tex]\dot E_{fuel} = 2.758\times 10^{9}\,W[/tex]
It is know that fission of a gram of uranium liberates [tex]5.8\times 10^{8}\,J[/tex]. Mass flow rate is computed herein:
[tex]\dot m_{U}= \frac{2.758\times 10^{9}\,W}{5.8\times 10^{8}\,\frac{J}{g}}[/tex]
[tex]\dot m_{U} = 4.755\,\frac{g}{s}[/tex]
The needed quantity of fuel per second is:
[tex]\dot m_{fuel} = \frac{4.755\,\frac{g}{s} }{0.04}[/tex]
[tex]\dot m_{fuel} = 118.875\,\frac{g}{s}[/tex]
The annual fuel consumption is now obtained:
[tex]\Delta m_{fuel,year} = (0.119\,\frac{kg}{s})\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{24\,h}{1\,day} )\cdot(\frac{365\,days}{1\,year} )[/tex]
[tex]\Delta m_{fuel,year} = 3,752,784\,\frac{kg}{year}[/tex]
The cost of fuel for the lifetime of the plant is:
[tex]C_{lifetime} = (3,752,784\,\frac{kg}{year} )\cdot (\frac{0.453\,lb}{1\,kg} )\cdot [(20\,years)\cdot(\frac{12\,USD}{1\,lb} )+(20\,years)\cdot (\frac{65\,USD}{1\,lb} )[/tex]
[tex]C_{lifetime} = 2,618,017,174\,USD[/tex]
